Mechanics III - Dynamics

Transcription

1 Institute of Mechanical Systems Chair in Nonlinear Dynamics Prof. George Haller Mechanics III - Dynamics Institute of Mechanical Systems Swiss Federal Institute of Technology (ETH) Zurich ased on the lecture notes by Prof. George Haller Edited by: Lucas Liebenwein David Öttinger Dr. Paolo Tiso Version from September 16, 2015

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3 Contents Nomenclature iii 0 Introduction and Newton s three axioms 1 I Dynamics of a single particle 3 I.1 Linear momentum principle I.2 Angular momentum principle I.3 Work-energy principle I.3.1 Work-energy principle for conservative systems I.4 Example problems I.5 Single particle motion in curvilinear coordinates I.6 Particle motion in non-inertial frames II Dynamics of system of particles 37 II.1 Forces II.2 Constraints II.2.1 Holonomic constraints II.2.2 Forces and constraints II.3 Degrees of freedom II.4 Center of mass II.5 Linear momentum principle II.6 Angular momentum principle II.7 Work-energy principle II.7.1 Work-energy principle for rigid body systems II.7.2 Work-energy principle for conservative systems II.8 Example problems III Kinematics of planar rigid bodies 63 III.1 Rigid body III.2 Angular velocity III.3 Velocity transport formula III.4 Example problems III.5 Instantaneous center of rotation IV Kinetics of planar rigid bodies 73 IV.1 Linear momentum principle IV.2 Angular momentum principle IV.3 Moment of inertia and angular momentum transfer formula IV.4 Work-energy principle i

4 IV.5 Degrees of freedom IV.6 Example problems V Kinematics of three dimensional rigid bodies 111 V.1 asic concepts V.1.1 Degrees of freedom V.1.2 Relative velocities V.2 Three dimensional rotation V.3 Angular velocity of time-dependent rotations VI Kinetics of three-dimensional rigid bodies 127 VI.1 Linear momentum principle VI.2 Angular momentum principle VI.2.1 Moment of inertia tensor VI.3 Work-energy principle VI.4 Rotating frames VI.4.1 Rotation transformation VI.4.2 Differentiation of quantities in rotating frames VI.5 Euler equations for spinning tops VII Vibrations 153 VII.1 One-DOF vibrations VII.1.1 Free vibrations VII.1.2 Forced vibrations VII.2 Multi-DOF vibrations VII.2.1 Free, undamped vibrations VII.2.2 Modal decomposition of free, undamped vibrations VII.2.3 Multi-DOF forced, damped vibrations

5 Nomenclature Acronyms and abbreviations AM AMP CM dim DOF LM LMP ODE PDE w.r.t. angular momentum angular momentum principle center of mass dimensions degrees of freedom linear momentum linear momentum principle ordinary differential equation partial differential equation with respect to Latin symbols a acceleration [ m /s 2 ] A amplitude of particular solution [m] rigid body [-] c damping coefficient [ kg /s] C damping matrix [ kg /s] dm infinitesimal mass element [kg] da infinitesimal area element [m 2 ] dv infinitesimal volume element [m 3 ] D Lehr s damping [-] [ i, j, k] inertial frame [-] [ e 1, e 2, e 3 ] principal frame [-] E total mechanical energy [J] f 0 normalized force vector [ m /s 2 ] F resultant force [N] F ext resultant external force [N] F int resultant internal force [N] F np resultant non-potential force [N] F p resultant potential force [N] iii

6 F 0 force vector [N] G gyroscopic matrix [ kg /s] H angular momentum w.r.t. [ kg m2 /s] I mass moment of inertia w.r.t. [kg m 2 ] I mass moment of inertia tensor w.r.t. [kg m 2 ] K stiffness coefficient [ N /m] K stiffness matrix [ N /m] K ij force exerted by particle j on particle i [N] m mass of a particle [kg] M mass of a rigid body [kg] M mass matrix [kg] M resultant torque w.r.t. [N m] M ext resultant external torque w.r.t. [N m] M int resultant internal torque w.r.t. [N m] N follower force matrix [ N /m] O(3) orthogonal group in 3D [-] P linear momentum [ kg m /s] r position vector [m] r A vector pointing from point A to point [m] ṙ A relative velocity of point w.r.t. point A [ m /s] SO(3) special orthogonal group in 3D [-] T kinetic energy of a system [J] u eigenvector [-] U modal matrix [-] v velocity [ m /s] v /A relative velocity of point w.r.t. point A [ m /s] V potential energy of a system [J] V magnification factor [-] W 12 work done on a system between state 1 and 2 [J] W12 ext external work done on a system between state 1 and 2 [J] W12 int internal work done on a system between state 1 and 2 [J] x deviation from stable equilibrium [m] x deviation from stable equilibrium [m] y modal coordinate vector [m] Greek symbols δ characteristic damping [ 1 /s] η frequency ratio [-] λ eigenvalues [ 1 /s 2 ]

7 ϕ phase shift [-] ρ density of a rigid body [ kg /m 3 ] or [ kg /m 2 ] ϱ vector from reference point to generic position r [m] ω angular velocity [ rad /s] ω forcing frequency [ rad /s] ω 0 natural frequency [ rad /s] Ω angular velocity matrix [ rad /s] Indices C h p reference point center of mass homogeneous solution to a linear ODE particular solution to a linear ODE

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9 Chapter 0 Introduction and Newton s three axioms In previous mechanics courses you have learned how to describe the velocity of a rigid body and about the concept of forces and moments (Mechanics I - Statics) as well as how a body deforms under a certain stress (Mechanics II - Strength of Materials). In contrast, this course treats the dynamics of a system, i.e. how a system reacts/moves when certain forces/moments are applied to it. Think, for example, about a car engine where the piston in each cylinder is displaced due to the forces exerted on it that arise after the combustion. Another common example is the rotation of a shaft where certain moments are applied to it or the motion of a car. In summary, Dynamics investigates how bodies move under the forces applied to it. In the 17 th and 18 th century, Sir Isaac Newton ( ) proposed a relation between the motion of a particle and the forces exerted on that particle when he observed apples falling down a tree due to gravity. He summarized his observations in three axioms, commonly called Newton s three axioms or Newton s laws. ased on these three axioms, Leonhard Euler, Joseph-Louis de Lagrange, William Rowan Hamilton and others extended Newton s theory to today s understanding of dynamics. These extended axioms cover systems of rigid bodies, not just particles. Newton s original three axioms are stated below. Newton s three axioms Axiom 1. A force-free particle is at rest or moves uniformly along a straight line. Axiom 2. The resultant force F (t) acting on a particle is proportional to its acceleration a(t). The factor of proportionality is the mass m: F (t) = ma(t). Axiom 3. Action and reaction forces are equal and act in opposite directions. As Newton s laws are based on a vectorial description of the system, a reference frame is needed. Newton s laws are not valid in an arbitrary reference frame, but in inertial frame 1

10 2 as defined below. Take, e.g. a candle attached to a rotating platform and is described w.r.t. to a frame rotating with the platform. In this frame of the platform, the flame of the candle will exhibit motions that cannot be explained by substituting active forces into Newton s 2 nd axiom, evaluated in the frame of the platform. The frame of the rotating platform is therefore not an inertial reference frame. Definition. An inertial frame is a reference frame in which Axioms 1-3 hold true. In this course, we will start out by extending Newton s laws to describe the motion of particles, namely we will derive the linear momentum principle, the angular momentum principle and the work-energy principle. A crucial step in a dynamics problem is to find the equation of motion that governs the system s motion, which will be in general a nonlinear multi-dimensional 2 nd order ordinary differential equation (ODE) in time. First, single particles are treated in Chapter I and then the derived equations are generalized to an arbitrary number of particles in Chapter II. In the second part of the course we will investigate the dynamics of a rigid body. A rigid body is characterized by the fact that any two points of the rigid body remain separated by a constant distance. This means there is no deformation and the underlying equations can be greatly simplified. The kinematics of a planar rigid body are treated in Chapter III and the kinetics of a planar rigid body in Chapter IV. Similarly for three dimensional rigid bodies, we also examine the kinematics, see Chapter V, and the kinetics, Chapter V. In the last part of the course, see Chapter VII, we introduce the concept of vibrations, i.e. small oscillations around a stable equilibrium. Since only small motions are considered, the equation of motion may be linearized, allowing us to use tools from linear algebra to closely examine the characteristics of oscillations near the equilibrium.

11 Chapter I Dynamics of a single particle In this chapter, we begin by describing the motion of a single particle. For that we apply Newton s three axioms and then find a relation between the forces applied to a particle and its motion. As a direct consequence of Newton s 2 nd law, we state the principle of linear momentum in Section I.1. Afterwards we also state Newton s 2 nd law in terms of the angular momentum (see Section I.2) and in terms of energy (see Section I.3). In Section I.4 these principles are applied to various examples. Finally, we see how Newton s principles are written in curvilinear coordinates (see Section I.5) and in non-inertial frames (see Section I.6). Figure I.1: A particle of mass m is shown moving along a trajectory r(t), subjected to the force F (t). We describe the system w.r.t. the inertial frame of (x, y, z) coordinate along the unit vectors [ i, j, k ]. The vector r(t 0 ) refers to the initial position of the particle. Consider a single particle of mass m as shown in Figure I.1, moving in time t along a trajectory r(t) R 3, with velocity v(t) = d dt r(t) = ṙ, 3

12 4 I.1. Linear momentum principle and acceleration a(t) = d2 r(t) = r. dt2 All quantities are expressed w.r.t. an inertial frame [ i, j, k ], i.e. r(t) = x(t) i + y(t) j + z(t) k, v(t) = ẋ(t) i + ẏ(t) j + ż(t) k, a(t) = ẍ(t) i + ÿ(t) j + z(t) k. Further, the particle is subjected to n forces F i R 3 of all forces as the resultant force F, i = 1,..., n. We refer to the sum F = n F i. i=1 I.1 Linear momentum principle (Newton s 2 nd Law) To express Newton s 2 nd law in a mathematical form, we first define the linear momentum. Definition. The linear momentum of a particle is defined as P := mv. (I.1) When differentiating the linear momentum w.r.t. time we obtain d dt P = d (mv) = ma = F, dt which we call the linear momentum principle (LMP). The principle of linear momentum (LMP) for a single particle Ṗ = ma = F (I.2) P... linear momentum of the particle m... mass of the particle a... acceleration of the particle F... resultant force acting on the particle Note: Conservation of linear momentum: F = 0, P = const.. momentum is conserved under vanishing resultant force. Therefore, linear

13 Chapter I. Dynamics of a single particle 5 Example I.1: lock sliding without friction Figure I.2: A block of mass m sliding without friction on a horizontal plane. Consider a block of mass m that slides without friction as shown in Figure I.2. The mass is subjected to gravity, and the ground exerts a normal reaction force N on the block. The normal reaction force must equal the force of gravity, because there is no vertical displacement. Thus, we find that the resultant force is equal to 0, i.e. F = mg + N = 0. Hence, linear momentum is conserved and the velocity of the block remains constant. I.2 Angular momentum principle We still consider a particle of mass m and with trajectory r(t) as shown in Figure I.3. We now introduce an arbitrary, possibly moving point in space with the trajectory r (t). The vector ϱ poitnting from the point to the particle is ϱ (t) = r(t) r (t). (I.3) Furthermore, we define the angular momentum H of the particle w.r.t. and the resultant torque M of the resultant force F w.r.t. as follows:

14 6 I.2. Angular momentum principle Figure I.3: A particle of mass m is moving along r(t). The reference point at r (t) is arbitrary and may move. Definition. The angular momentum H (t) of a particle w.r.t a point is defined as the cross product of the vector ϱ between the point and the particle with the linear momentum P, H (t) := ϱ (t) P (t). (I.4) Here, the point can be an arbitrary point in space, either fixed or moving. Definition. A force F i acting on a particle at position r(t) produces a torque or moment M i w.r.t. a point of the form M i = ϱ F i, (I.5) where denotes the cross product. The resultant torque M w.r.t. is then the sum of all torques w.r.t. : M := n M i = i=1 n n ϱ F i = ϱ F i, i=1 i=1 which is equal to the torque of the resultant force F w.r.t. point : M := ϱ F. (I.6) Similarly to the LMP, we can now relate the time derivative of the angular momentum H to the resultant torque M. We first differentiate the angular momentum H in time

15 Chapter I. Dynamics of a single particle 7 using the product rule, d dt H = d ( ) ϱ dt P, Ḣ = ϱ P + ϱ Ṗ. Then we apply the linear momentum principle (I.2), Ḣ = ϱ P + ϱ F. Next we replace ϱ F by (I.6) and ϱ by (I.3) to obtain Ḣ = ϱ P + ϱ F, = ϱ P + M, = (ṙ ṙ ) P + M, = (v v ) P + M. Furthermore, we use the fact the vector product between two parallel vectors vanishes, v P = v (mv) = 0, to simplify the above equation to Ḣ = v P v P + M, = v P + M. This equation represents the angular momentum principle (AMP). The principle of angular momentum (AMP) for a single particle Ḣ + v P = M (I.7) H v... angular momentum of the particle w.r.t. the point.... velocity of the point P..... linear momentum of the particle M... resultant torque w.r.t. the point Specifically, Ḣ = M if and only if either of the following two conditions is met: v = 0 ( is fixed) v v Note: Conservation of angular momentum: If M = 0 and (v = 0 or v v), then Ḣ = 0 and hence H = const.. Therefore the angular momentum is conserved.

16 8 I.2. Angular momentum principle Example I.2: String pulled slowly from below Figure I.4: The mass m shown here is constrained to move on the plane and is attached to a string pulled downwards by a force F. Note that since the string is massless, we have F = F. A mass m moves on a plane while attached to a taut string, as shown in Figure I.4. The string runs through a hole in the plane at 0 and the other end of the string is subjected to a force F (t). The initial length l(0) of the planar section of the string and the initial velocity v(0) perpendicular to the string of the mass m are given. Indicate with l(t) the length of the string at an arbitrary time t. What is the velocity v(t) of the mass m for any given time? Solution: First, we pick the reference point 0 with velocity v 0, then compute the moment of the resultant force w.r.t. as M (t) = r(t) F (t) = 0, since r F. Here r(t) denotes the vector between the hole and the mass m, where r(t) = l(t). Note that the moment coming from the gravitational force and the moment produced by the reaction force arising due to gravity cancel each other out and hence do not contribute to the resultant torque M w.r.t.. y the AMP, we therefore obtain that H = const. and hence conservation of angular momentum holds for all times. In the following, we denote the angle between the vectors r and

17 Chapter I. Dynamics of a single particle 9 v by α. We can write: H (0) = H (t) r(0) mv 0 = r(t) mv(t) l(0) v(0) sin α(0) k = l(t) v(t) sin α(t) k }{{}}{{} =1 since r(0) v 0 =1 since r(t) v t t v(t) = l(0) v(0) l(t) Thus, we can determine the velocity at time t using conservation of angular momentum. I.3 Work-energy principle Figure I.5: A particle of mass m subjected to the force F (t) moving from position r 1 to r 2. Suppose that a particle of mass m experiences a force F (t) as it moves along its path r(t) as shown in Figure I.5. etween the positions r 1 = r(t 1 ) and r 2 = r(t 2 ), the work W 12 done by F (t) on the particle is W 12 = r2 r 1 F dr, (I.8) where a b indicates the scalar product between the vectors a and b. Now, we can use the fact that dr = v dt and substitute to (I.8), W 12 = t2 t 1 F v dt. Using the linear momentum principle (I.2), we express the force F replaced in terms of the velocity v to obtain W 12 = t2 t 1 d (mv) v dt, dt

18 10 I.3. Work-energy principle then evaluate the integral using the chain rule as, W 12 = t2 t 1 d dt (1 m v v) dt, 2 = 1 2 m v m v 1 2. (I.9) The term 1 2 m v 2 is just the kinetic energy T of a particle, a measure of the amount of work one has to put into a particle to accelerate it to the velocity v. Definition. The kinetic energy T of a particle of mass m and velocity v is defined as T := 1 2 m v 2. (I.10) From (I.9) it follows that the change in kinetic energy (T 2 T 1 ) of a particle is equal to the work W 12 done by all forces on the particle (work-energy principle). Work-energy principle for a single particle W 12 = T 2 T 1 (I.11) W work done on the particle between r 1 and r 2 T i.... kinetic energy of particle associated with velocity v i I.3.1 Work-energy principle for conservative systems To further simplify the work-energy principle (I.11) we now introduce the concept of a potential force. Definition. A potential force (or conservative force) F i is a force that can be expressed as the gradient of a scalar function V i (r), called potential or (potential function), F i = V i r = V i(r). V i (r) is the gradient of the function V i (r), defined as V i (r) = the Cartesian coordinate system [x, y, z]. (I.12) ( ) T Vi x, V i y, V i z in

19 Chapter I. Dynamics of a single particle 11 Note: For a given potential force F i the potential V i (r) is determined only up to a constant C since ( ) F i = V i (r) = V i (r) + C. The constant C can be arbitrarily chosen, implying we can select the potential to be 0 at any desired position r 0. Example I.3: Which forces are conservative? We review here typical forces encountered in dynamical systems and state whether they are conservative, or not. Remember that there are an infinite number of potential functions since the constant C can be arbitrarily set. (a ) Gravity Figure I.6: A particle of mass m subjected to gravity. When a particle is subjected to gravity, see Figure I.6, a force F = mg = mg j is exerted onto the particle. This force is conservative and the corresponding potential is V (r) = mgy. Indeed, we have V (r) = 0 i mg j = F. (b ) Linear spring force Figure I.7: A block attached to a spring with stiffness k.

20 12 I.3. Work-energy principle A block of mass m attached to a (linear) spring (see Figure I.7), is subjected to the spring force F = kx i caused by the spring elongation. This force is also conservative and the corresponding potential function is V (r) = 1 2 kx2. Indeed, we can write V (r) = kx i 0 j = F. (c ) Viscous damping force Figure I.8: A block attached to a damper with damping coefficient c. A (linear) damper exerts the force F = cẋ i on a particle (see Figure I.8). This force is not potential since we cannot find a potential V (r), F (ṙ) V (r). Indeed, the gradient of an r-dependent function v(r) cannot depend on ṙ. We now introduce the definition of conservative systems. Definition. A conservative system is a system where all forces are either potential or do no work. To capture the effect of all potential forces acting on a system we introduce the potential energy of a system.

21 Chapter I. Dynamics of a single particle 13 Definition. The potential energy V (r) of a mechanical system is defined as the sum of the potentials V i (r) of all potential forces F i acting on the system, V (r) := n V i (r), i = 1,..., n. (I.13) i=1 The resultant force F p := n i=1 F i of all potential forces can then be expressed as F p = d V (r) = V (r). dr (I.14) We now recall (I.9), which describes the work done by the resultant force F along a certain path: W 12 = r2 r 1 F dr. We would like to evaluate this expression for a conservative system. The resultant force F of any system can be split into a potential part F p and a non-potential part F np, F = F p + F np, which can be inserted into the above expression for W 12 to obtain: W 12 = = r2 r 1 (F p + F np ) dr, r2 r 1 F p dr + r2 r 1 F np dr. y definition, in a conservative mechanical system, only potential forces do work. Therefore and hence F np dr = 0 W 12 = r2 r 1 F p dr. Using the fact that the potential forces F p can be expressed with the potential energy V (r) of the system as F p = d dr V (r)

22 14 I.4. Example problems we can evaluate the work integral as W 12 = = r2 r 1 F dr = r2 r 1 r2 r 1 d ( V ) dr dr d dr V dr = V (r 1 ) V (r 2 ) = V 1 V 2. (I.15) Combining the result (I.15) with (I.11) yields the work-energy principle for conservative systems, V 1 V 2 = T 2 T 1. Work-energy principle for a conservative system The total mechanical energy E := T + V of a conservative systems is conserved along all motions, T 1 + V 1 = T 2 + V 2 (I.16) or equivalently, d dt E(t) = 0. E... total mechanical energy T i... kinetic energy of the system in state i V i... potential energy of the system in state i Note: (I.15) implies that, for a conservative system, the work done to move a particle from position r 1 to r 2 does not depend on the path taken. This is not true for nonconservative systems, where the work done does depend on the path taken. Consider, e.g., a non-conservative system subjected to friction where any displacement results in a loss of energy. I.4 Example problems Example I.4: Car stunt in a circular ramp with foot off the pedal Consider a car of mass m that approaches a looping and attempts to run through it without falling off, as shown in Figure I.9. Suppose that the car rolls without friction and can be modeled as a point mass. The ramp is massless and has a radius R. Further, the car is subjected to gravity and the driver does not accelerate the car once it enters the loop (foot is off the pedal). For all computations, the inertial frame [ i, j, k ] will be used.

23 Chapter I. Dynamics of a single particle 15 (a ) What is the minimum entry velocity so that the car does not fall off? (b ) What is the maximum g-force acting on the driver? Figure I.9: A car of mass m is approaching a vertical, looping track. Point 1 denotes the entry point and point 2 is the turn-over point. (a ) Minimum entry velocity i Free-body diagram (FD) The only two forces acting on the car are the constraint force N and the gravity mg (potential force) as seen in the free-body diagram in Figure I.10. The constraint force N does no work since it always remains normal to the track, and therefore N dr = 0. Figure I.10: The free-body diagram of the system is shown. The blue line is tangent to the track (black=dashed line). From this it follows that the system (particle motion) is conservative and the work-energy principle for conservative systems (I.16) can be applied. The potential energy V (r) of the

24 16 I.4. Example problems system is the potential of the gravitational force, V (r) = V (y) = mgy. The critical point where the car is most likely to fall off is at the turning point of the ramp, point 2. This is why we apply the conservation of energy between the critical point 2 and the entry point 1: T 1 + V 1 = T 2 + V 2, T 1 = T 2 + V 2, 1 2 m v 1 2 = 1 2 m v mg 2R, (I.17) where V 1 = 0 since y = 0 at position 1. From (I.17) a relation between v 1 and v 2 can be determined, v 2 = v1 2 4gR. (I.18) This relation is only valid if the car is still on the track. At any time the track pushes the car with a force N. As long as N > 0 the car remains on the track. The limiting case is when N = 0. ii Linear momentum principle Figure I.11: The car at an arbitrary position on the looping, indicated by the angle ϕ. To relate the velocities v 1 and v 2 to the normal force N and therefore get a condition for the minimal entry velocity we apply the linear momentum principle Ṗ = F = N + mg. (I.19) We use the normal unit vector e N = cos ϕ i + sin ϕ j to express the trajectory r(t) of the car, r(t) = R e N. The angle ϕ(t) is defined in Figure I.11. The speed of the car is v = v = R ϕ.

25 Chapter I. Dynamics of a single particle 17 To solve (I.19) we must first evaluate P and Ṗ : P = mv = m d dt r, (I.20) = m d (R cos ϕ i + R sin ϕ j), dt = m R ϕ( sin ϕ i + cos ϕ j), Ṗ = m R ( ϕ cos ϕ + ϕ 2 cos ϕ) i + m R ( ϕ cos ϕ ϕ 2 sin ϕ) j. (I.21) Now we project (I.19) onto e N by using the dot product between the vectors: Ṗ = N + mg, Ṗ e N = (N + mg) e N. We now insert in (I.21) to get m R ϕ 2 = N mg sin ϕ, m v2 R = N mg sin ϕ. When evaluated at ϕ = π 2 and solved for N 2 = N(ϕ = π 2 ) we obtain N 2 = m ( v2 2 R g). (I.22) For the car to be still on the track (barely), it has to be N 2 0. This yields the minimal speed v 2 using (I.22) v 2 g R. Using (I.18) v 2 can be replaced with the necessary initial speed v 1 : v g R g R, v 1 5 g R (I.23) Using realistic values for a 2R = 40 ft = m high ramp and g = 9.81 m /s 2 yields v m /s m = 299 m2 /s 2, m /s = mph. This movie shown during the lecture suggests the car must be faster than 36 mph (v 1 36 mph), which corresponds to a height of 35 ft.

26 18 I.4. Example problems (b ) Maximal g-force For this part we look for the maximal g-force a driver of mass m d experiences during the ride. The normal force N(ϕ) that is exerted on the passenger can be expressed using (I.22), however, for any arbitrary angle ϕ, N(ϕ) = m d v 2 (ϕ) R m d g sin ϕ. (I.24) In order to obtain v(ϕ) the conservation of energy is again applied but this time for any arbitrary angle ϕ, T ϕ + V ϕ = T 1 + V 1, T ϕ = T 1 + V 1 }{{} V (y=0)=0 V ϕ, 1 2 m dv 2 (ϕ) = 1 2 m dv 2 1 m d gr(1 + sin ϕ), v 2 (ϕ) = v 2 1 2gR(1 + sin ϕ). (I.25) Plugging (I.25) into (I.24) gives us N(ϕ) = m ( ) d v1 2 2gR(1 + sin ϕ) m d g sin ϕ, R = m d R v2 1 2m d g 3m d g sin ϕ for N(ϕ). Now the maximum over the desired interval is evaluated yielding max N(ϕ) = m d ( v2 1 ϕ [ π 2, 3π 2 ] R + g) for sin ϕ = 1, respectively ϕ = π 2 or ϕ = 3π 2, i.e. at the entry of the loop. When plugging in the minimal necessary entry speed from (I.23) we obtain max N(ϕ) = m d (5g + g) = 6m d g ϕ [ π 2, 3π 2 ] for the maximal force the driver feels. The passenger experiences "6g"; a typical person can handle about 5g without passing out. How a roller coaster loop is designed in practice! The curvature of the loop increases linearly to ease the transition into the loop (Euler spiral). The loop then has a similar curvature as the loop shown in Figure I.12. Also, e.g. highway exits are designed in the same way.

27 Chapter I. Dynamics of a single particle 19 Figure I.12: A roller-coaster looping is shown that has a linear increasing curvature to ease the transition. Example I.5: Motion of a rocking chair (a) (b) Figure I.13: (a) An actual rocking chair. (b) The corresponding mechanical model. The chair is modeled as point mass supported by a weightless ground structure of a circular profile of radius R. The rocking chair shown in Figure I.13 is modeled as a single point mass m with a weightless ground structure that models the bottom part of the chair. The mass of the chair as well as of the person sitting on it is concentrated in C and the chair "rocks without slipping", i.e. pure rolling. The curvature of the bottom part has radius R and the mass m is at

28 20 I.4. Example problems a distance a away from the origin, i.e. the center of the circle, of the coordinate system [ i, j, k ]. The system is subjected to gravity. (a ) What is the equation of motion for the point mass m in the given configuration? (b ) Calculate the contact forces between the chair and the ground. (c ) Determine a comfortable design for the rocking chair, i.e. find an R for the chair to rock in an adequate pace. (a ) Equation of motion i Free-body diagram Consider the free-body diagram shown in Figure I.14. Point hereby denotes the moving (!) point of contact between the rocking chair and the ground, N is the normal reaction force. Note: N mg common mistake from statics. S is the tangential reaction force (friction), which is necessary to prevent the chair from sliding on the ground (S 0). Figure I.14: The free-body diagram for the mechanical model of the chair at an arbitrary angle ϕ is shown. Note that the orange arrow indicates the position of the contact point when ϕ = 0. ii Angular momentum principle Since we are interested in finding the equation of motion but not the forces it is preferable to choose a reference point where the resultant torque does not include unknown forces. This is why as reference point is chosen. The only force producing a torque about is known (gravitational force), whereas S and N produce no torque w.r.t.. The AMP about point is Ḣ + v P = M. (I.26) To evaluate (I.26) all corresponding quantities must be found first. The velocity of the contact point is v = R ϕ i.

29 Chapter I. Dynamics of a single particle 21 The velocity v is the time derivative of the arclength Rϕ covered by. More information on rolling without slipping is provided in Example III.2. All other quantities are found as well: P = mv C = m d dt r C = m d ( (Rϕ a sin ϕ) i + (R a cos ϕ) j ( dt ) = m ϕ (R a cos ϕ) i + a sin ϕ j, H = r C P = r C (mv C ), ( ) ( ) = m a sin ϕ i + (R a cos ϕ) j ϕ (R a cos ϕ) i + a sin ϕ j, ( ) = m ϕ a 2 R 2 + 2aR cos ϕ k, ( Ḣ = m(2ar cos ϕ a 2 R 2 ) ϕ 2maR sin ϕ ϕ 2) k, v P = R ϕ i (mv C ) = arm ϕ 2 sin ϕ k, M = mga sin ϕ k. Inserting all the terms into (I.26) and simplifying it yields the equation of motion, m(2ar cos ϕ a 2 R 2 ) ϕ mar ϕ 2 sin ϕ mga sin ϕ = 0, (R 2 + a 2 2aR cos ϕ) ϕ + ar sin ϕ ϕ 2 + ag sin ϕ = 0 (I.27) The equation of motion is a 2 nd order nonlinear ordinary differential equation (ODE). Since the ODE is nonlinear it can be hard or even impossible to find an analytical solution. To solve the ODE it can be linearized and then solved analytically or solved numerically, e.g. using MATLA s ODE45 solver. For that the equation needs to be first transformed into a system of 1 st -order ODEs (state space representation), see below. The states x 1 = ϕ and x 2 = ϕ are introduced to be able to solve the system numerically, ( ) x1 x =. x 2 ), The state space representation then is ẋ = f(x), ) x 2 (ẋ1 ẋ = = ẋ 2 a sin x 1(Rx g). R 2 + a 2 2aR cos x 1 The initial conditions for the system are ( ) ϕ0 x(0) =. ϕ 0 (b ) Contact forces To find the contact forces the linear momentum principle Ṗ = F (I.28)

30 22 I.4. Example problems is applied. Note: We could also apply the AMP again w.r.t. a point where the forces N and S produce a torque but this would unnecessarily increase the complexity of the derivation. i LMP in x-direction Evaluating (I.28) in x-direction gives us the tangential reaction force S: mẍ C = S, x C = Rϕ a sin ϕ, S = m d ( ) (R a cos ϕ) ϕ dt (I.27) is used to eliminate ϕ from (I.29) yielding S = m ( ( ) = m a sin ϕ ϕ 2 + (R a cos ϕ) ϕ. (I.29) a sin ϕ ϕ 2 (R a cos ϕ)(ar sin ϕ ϕ2 + ag sin ϕ) R 2 + a 2 2aR cos ϕ ii LMP in y-direction Similarly, evaluating (I.28) in y-direction yields an equation for N: my C = N mg, y C = R a cos ϕ, ( ) N = m(ÿ C + g) = m a cos ϕ ϕ 2 + a sin ϕ ϕ 2 + g ) (c ) Comfortable chair A rocking chair is considered comfortable when it rocks with about 30 rocks (periods) per minute. To obtain a relation between the geometric parameters R and a the equation of motion (I.27) is first linearized around an equilibrium. Afterwards the eigenfrequency of the linearized equation is set to the desired rocking frequency. One equilibrium is ϕ = 0 (plugging ϕ 0 into (I.27) yields ϕ = 0 and ϕ = 0). As long as only small oscillations around the equilibrium are considered it is a valid assumption to work with linearized equations. For that, (I.27) is Taylor-expanded and only the linear terms are kept. The equation of motion is (R 2 + a 2 2aR cos ϕ) ϕ + ar sin ϕ ϕ 2 + ag sin ϕ = 0 and the Taylor-expansion of the equation is ( ) ( ) R 2 + a 2 2aR + O(ϕ 2 ) ϕ + ar ϕ + O(ϕ 3 ) ( ) ϕ 2 + ag ϕ + O(ϕ 3 ) = 0. When dropping all terms of 2 nd or higher order and only keeping linear terms we obtain (R a) 2 ϕ + agϕ = 0 (I.30) for the linearized equation of motion around the equilibrium position ϕ = 0. Now we will bring it in the normalized form to read the eigenfrequency from the equation by dividing

31 Chapter I. Dynamics of a single particle 23 by (R a) 2 : (R a) 2 ϕ + ag ϕ = 0, ag ϕ + (R a) 2 ϕ = 0, ϕ + ω 2 0 ϕ = 0. The eigenfrequency of the linearized equation is ω 0 = ag R a (I.31) which can be set to the desired rocking frequency. 30 rocks per minute correspond to ω 0 = 2π 60/30 s 1 = πs 1. (I.32) We use the results obtained from (I.31) and (I.32) to relate the radius R and the distance a of a comfortable rocking chair. The relation can be used as a design principle. ag R = a + π The units of the quantities are [R] = m, [a] = m, [g] = m /s 2 and [π] = s 1. Example I.6: Design of a curb-tipped rollover test for cars (a) (b) Figure I.15: (a) A car hitting a curb and starting to roll-over. (b) The car is modeled as point mass with a weightless frame. The surface is supposed to be smooth and the height between the street and the curb (marked red) is negligible. The curb point is denoted by. A car of mass m is skidding sideways with skidding velocity v 0 into a curb and then rollsover when hitting the curb, see Figure I.15. To assess the damage on the car, a crash test must be properly designed. The question in this case is what the minimal skidding

32 24 I.4. Example problems velocity v 0 is causing the car to roll over. For this purpose the car is modeled as a point mass m at the center of mass C located in the geometric center of the weightless frame with dimensions a and b. Further, any motion tangent to the curb is neglected. The contact point where the car hits the curb is smooth and the height between the curb and the street is negligible as well. The collision is assumed to be perfectly inelastic (no rebound) and gravity acts upon the car. We will divide the situation into two phases, a phase shortly before and after the collision and a phase starting after the collision describing the roll-over. (a ) Phase 1: Collision Collisions are intuitively not easy to understand and there exist various misconceptions about collisions. In general, energy is not conserved during a collision, also energy is not transformed into angular momentum. Similarly, linear energy is not transformed into rotational energy and linear momentum is not transformed into angular momentum. i Free-body diagram Let us consider the forces acting on the particle in the moment t = 0 of the collision as shown in Figure I.16. The car is subjected to gravity mg as well as the normal force N. From the moment the car hits the curb the force F (t) is exerted as well. Neither the direction nor the magnitude of F (t) is known. (a) (b) Figure I.16: The free-body diagram and the velocity after the impact is shown. (a) The free-body diagram depicts the situation when the car hits the curb. (b) After the collision, the car rotates around the contact point and the velocity is denoted as v(0 + ). As the car rotates around, the velocity vector v(0 + ) is perpendicular to the vector r C. ii Linear momentum principle The LMP for the sketched situation is Ṗ + F = mg + N + F. However, F (t) is unknown and complicated to express. Therefore, we will not apply the LMP. iii Angular momentum principle

33 Chapter I. Dynamics of a single particle 25 The force F does not produce a torque about point : M = r (N + mg). Therefore we can apply the AMP without knowledge of F (t). Since v = 0 we obtain Ḣ + v P = M, Ḣ = M (I.33) for the AMP about point. Next we will integrate (I.33) for t [0, 0 + ], where t = 0 denotes the instance just before the collision and t = 0 + just after it: Ḣ dt = H (0 + ) H (0 ) = M dt, r C (N + mg) dt. (I.34) Now we will evaluate the right hand side of (I.34) and let the integration interval go to zero. Since N and mg are bounded in time, 0 + lim r C (N + mg) dt = 0. Therefore, the angular momentum w.r.t. is conserved during the collision, H (0 + ) = H (0 ). (I.35) The angular momenta before and after the collision can be expanded to and H (0 ) = r C P (0 ) = r C (mv 0 ) = a 2 mv 0 k H (0 + ) = r C P (0 + ) = r C (mv + ) = mv + a 2 + b 2 with the velocities v 0 = v(0 ) just before and v + = v(0 + ) after the collision, see Figure I.16. The initial velocity v 0 is given. As stated before in Figure I.16 r C v +. and the vector product r C (mv + ) simply eyields 2 k. r C (mv + ) = m r C v + k = mv+ a 2 + b 2 Inserting all the terms into (I.35) gives 2 k. a 2 mv 0 k = mv + a 2 + b 2 2 which can be solved for the velocity v + after the collision, v + = a a 2 + b 2 v 0 k, (I.36)

34 26 I.4. Example problems (b ) Phase 2: Roll-over After the collision the point mass rotates around in a circular motion starting from position 1 with initial velocity v +. For the car to roll over the point mass needs to overcome the curb point, see position 2 in Figure I.17. This can only happen for a sufficiently large velocity v +, otherwise the car rolls back. The goal is to find the necessary velocity v + by applying the appropriate principle. i Free-body diagram After the collision the only forces that are acting on the system are gravity and the force K at the contact point, see Figure I.17. During this phase angular momentum about is not conserved anymore since mg will produce a torque about and the time interval regarded is not infinitesimally small (which was the reason why conservation of angular momentum about could be applied before). Therefore we cannot relate position 1 and 2 to each other without solving the equation of motion, which is rather cumbersome. Figure I.17: Free-body diagram for the situation after the collision. The zero potential level is on the ground. Recall that the height difference between the ground and the curb is neglected. ii Conservation of energy Instead we can exploit the fact that the system is conservative since gravity is a potential force and the constraint force K does no work because K dr = 0 during the roll-over. The total mechanical energy of the system is conserved between position 1 and 2 (work-energy principle for conservative systems). The conservation of energy states T 1 + V 1 = T 2 + V 2 (I.37) where V (y) = mgy is the potential energy of the system (zero-level assumed to be on the ground). We are interested in the limiting case, where v + is just large enough to cause the car to roll-over, then T 2 = 0

35 Chapter I. Dynamics of a single particle 27 since v 2 = 0. Evaluating (I.37) yields 1 2 mv2 + + mg a a 2 = 0 + mg 2 + b 2 2, which can be solved for v +, v + = g( a 2 + b 2 a). (I.38) Combining (I.38) with the relation (I.36) for the velocities before and after the collision gives the minimum necessary skidding speed that results in roll-over is obtained: v crit 0 = ( ) ( ) g a 2 + b 2 a 1 + b2 a 2 (I.39) We see from this result that flat cars (b a) are more stable, since v crit 0 is larger for them. This also corresponds to our intuition and experience from driving that flat cars with a lower center of mass are more stable. I.5 Single particle motion in curvilinear coordinates In some situations it is preferable to write the momentum principles w.r.t. a different coordinate system such as curvilinear coordinates. This does not introduce any new principle but simply provides an alternative way to approach a problem. For many problems dealing with rotational motions this approach can reduce the complexity of the derivation. In the following we will give an general approach of how the momentum principles can be transformed to curvilinear coordinates. Consider the position vector r in Cartesian coordinates, x r = y. z We could also express r as a function of some curvilinear coordinates c R 3 as r = f(c). The linear momentum principle states that m d2 r = F (r, ṙ, t) dt2 (I.40) or expressed in curvilinear coordinates m (f, d2 dt 2 f(c) = F ddt ) f, t.

36 28 I.5. Single particle motion in curvilinear coordinates Note: Since f = f(c(t)) is not explicitly dependent on time but only implicitly, f can be differentiated in time by applying the generalized chain rule (absolute derivative), ( d f dt f(c) = cf ċ = c 1 f c 2 f c 3 ) ċ = f 1 f 1 c 1 c f 3 f 3 c 1 c 3 c f hereby denotes the Jacobian of f for the curvilinear coordinates c and ċ = d dt c the time derivative of the coordinates c. Referring back to (I.40) we can now differentiate w.r.t. time once, m d2 dt 2 f(c) = F, m d ( c f ċ ) = F, dt and twice to obtains the linear momentum principle in curvilinear coordinates ( ( 2 m cf ċ ) ) ċ + c f c = F. (I.41) Another way of writing this down is by index notation where we implicitly sum over every free index, i.e. an index that only appears on one side of the equation, e.g. ċ 1 ċ 2 ċ 3. α i β i = i α i β i. Further, we denote the partial derivative of f i w.r.t. c j as f i,j = f i c j. Thus, (I.41) becomes ) m (f i,jk ċ j ċ k + f i,j c j = F i i = 1, 2, 3 (I.42) where i denotes the component index. Summarizing: The principle of linear momentum in curvilinear coordinates ( ( 2 m cf ċ ) ) ċ + c f c = F (f, ddt ) f, t or in index notation: ) m (f i,jk ċ j ċ k + f i,j c j = F i (f, ddt ) f, t i = 1, 2, 3 (I.43) (I.44)

37 Chapter I. Dynamics of a single particle 29 c curvilinear coordinates R 3 f(c)... position of particle expressed in curvilinear coordinates c c f... Jacobian of f in c m..... mass of the particle F..... resultant force acting on the particle Example I.7: Cylindrical coordinates A common example for curvilinear coordinates are cylindrical coordinates, see Figure I.18, ϱ c = θ. z We can now express the position vector r in terms of the cylindrical coordinates c, x ϱ cos θ r = y = ϱ sin θ, z z and use (I.43) or (I.44) to apply the LMP in cylindrical coordinates. We can solve problems with high rotational symmetry like Example I.5 with much less calculation effort then. Figure I.18: The cylindrical coordinates are indicated w.r.t. the inertial frame. I.6 Particle motion in non-inertial frames Sometimes it is convenient to describe the motion of a system w.r.t. a moving frame of reference. Consider two reference frames, an inertial frame [x 1, x 2, x 3 ] with origin O and a frame [y 1, y 2, y 3 ] with origin O t (t) moving relatively to the inertial frame as depicted in Figure I.19. Since the frame [y 1, y 2, y 3 ] is not an inertial frame we cannot apply the linear momentum principle directly. However, we can express the LMP in the inertial frame

38 30 I.6. Particle motion in non-inertial frames Figure I.19: The motion of a particle can be described by the vector x w.r.t. an inertial reference frame [x 1, x 2, x 3 ] or by the vector y w.r.t. a moving, non-inertial frame [y 1, y 2, y 3 ]. [x 1, x 2, x 3 ] and then relate the two frames to each other to obtain the LMP expressed in the non-inertial frame. The idea is somewhat similar to what was shown in Section I.5. The general form of transformation from the y-frame to the y-frame is x = Q (t)y + b(t). (I.45) where b(t) R 3 is the translational vector between the two origins of the coordinate system. y is the position of the particle in the y-frame and x the particle s position in the inertial frame x. Q (t) R 3 3 takes into account that the two frames can be rotated w.r.t. each other. Q (t) is actually not any matrix in R 3 3 but enjoys some properties, namely: Q T = Q 1, ( ) det Q = 1. The group of matrices fulfilling above specifications is called the special orthogonal group SO(3). They describe rotations in 3D-space, an operation that is length, angle and orientation preserving, also see Section VI.4. Note: Suppose that the x and y-frame share the same origin. Then: x = Q (t)y rotates a vector represented in the y-frame as y into its representation x in the x-frame,where the x and y-frame are rotated w.r.t. each other. y = Q 1 (t)x = Q T (t)x does the opposite, i.e. it rotates the vector from the x-frame to the y-frame. In case the two reference frames are not only rotated to each other but also shifted we need the additional translation vector b to describe the transformation, see (I.45). The vector b is represented in the inertial frame x.

39 Chapter I. Dynamics of a single particle 31 Next, we will insert the transformation (I.45) into the LMP, mẍ = F (x, ẋ, t), which gives us m d2 dt 2 ( Q y + b ) = F ( Q y + b, d dt (Q y + b), t). Differentiating the above equation twice we obtain m d2 dt 2 ( Q y + b ) m d dt ( Q y + Q ẏ + ḃ)m( Q y + 2 Q ẏ + Q ÿ + b) = F. Multiplying from the left with Q T and rearranging some terms we find the LMP expressed in the y-frame, mÿ = Q T F mq T Q y 2mQ T Q ẏ mq T b. Each term in the above equation has a clear physical meaning. Q T F are simply the active forces on the particle represented in the moving frame. mq T Q y is called the centrifugal effect or centrifugal (pseudo-) force. A particle in a rotating frame can experience this effect. 2mQ T Q ẏ represents the Coriolis effect that arises when a particle moves w.r.t. a rotating frame. mq T b are the inertial "forces" due to the linear acceleration of the moving frame. The principle of linear momentum in a non-inertial frame mÿ = Q T F mq T Q y 2mQ T Q ẏ mq T b (I.46) The transformation of the vector y into its inertial representation x is x = Q (t)y + b(t). (I.47) m... mass of the particle y... position of the particle expressed in the non-inertial frame y x... position of the particle expressed in the inertial frame x Q... rotation matrix from y to x b.... linear translation of the frames Q T F active forces expressed in the moving frame mq T Q y.... centrifugal effect 2mQ T Q ẏ... Coriolis effect mq T b Inertial forces

40 32 I.6. Particle motion in non-inertial frames Although only the term Q T F describes forces that are actively exerted onto the particle, all other terms are present as well due to the motion of the reference frame. It is anyhow important to understand that the extra terms are not actual forces in a strict sense. They solely arise due to the fact that the reference frame used is non-inertial. Example I.8: Planar motion in a steadily rotating frame (a) (b) Figure I.20: (a) The motion of a particle in the xy-frame can be described by a steadily rotating frame. (b) The Coriolis force always acts perpendicular to the axis of rotation and the velocity. We look at a particle of mass m that moves in the x,y - plane. The particle is subjected to forces represented by the resultant F (t) as depicted in Figure I.20. The inertial frame is denoted by [x 1, x 2 ]. An additional frame [y 1, y 2 ] steadily rotates with angular velocity Ω = Ω x 3 w.r.t. to the inertial frame. The particle s position in the inertial frame x is represented by ( ) x1 x = x 2 and in the moving frame y by ( ) y1 y =. y 2 We can relate x and y by the transformation x = Q y with ( ) cos Ωt sin Ωt Q =. sin Ωt cos Ωt Note: b = 0 here since both coordinate frames share the same origin. In the following, we are interested in computing the "forces" due to motion of the reference frame. With (I.46) we can compute each term individually. The active forces are Q T F.

41 Chapter I. Dynamics of a single particle 33 The centrifugal forces F cf can be computed explicitly: F cf = mq T Q y = mq T ( Ω 2 Q )y, = mω 2 Q T Q y. Exploiting the fact that Q 1 = Q T and therefore Q T Q = I we obtain F cf = mω 2 y (I.48) As can be seen from (I.48) the centrifugal force acts parallel to y, drawing the particle away from the center 0. This effect is used by many machines (centrifuges). Similarly, we can compute the Coriolis force F cor : F cor = 2mQ T Q ẏ, ( ) ( ) cos Ωt sin Ωt sin Ωt cos Ωt = 2mΩ ẏ, sin Ωt cos Ωt cos Ωt sin Ωt ( ) 0 1 = 2mΩ ẏ. 1 0 (I.49) The matrix ( ) represents a counter-clockwise rotation by π 2. Instead of using this matrix we can also use the vector product x 3 ẏ = ( ) ẏ to express the same rotation. Going back to (I.49) gives ( ) 0 1 F cor = 2mΩ ẏ, 1 0 = 2mΩx 3 ẏ, F cor = 2mΩ ẏ (I.50) for the Coriolis force. We can see that the Coriolis force F cor is orthogonal to the axis of rotation as well as the velocity, see Figure I.20. Example I.9: How non-inertial is the earth as a frame? In most practical cases we assume the earth to be an inertial frame. However, this is not true since the earth rotates around its own axis and encircles the sun, see Figure I.21. We now want to estimate the error we make when considering the earth

42 34 I.6. Particle motion in non-inertial frames as an inertial frame. For that we can use the results obtained from the previous Example I.8 to estimate the Coriolis force F cor and the centrifugal force F cf due to the rotation of the earth. Figure I.21: The earth itself is a rotating frame with angular velocity Ω. The angular velocity of the earth is Ω = 2π rad = h s const. and its radius R = [m]. Inserting these values into (I.50) we obtain an estimation for the Coriolis force, F cor = 2m Ω v, = m v, where v denotes the velocity relative to the earth. Using (I.48) we also estimate the centrifugal force acting on a mass located on earth, F cf = 2mΩ 2 y, = 2m Ω 2 R, = (7.3) m = m. We here assumed the mass to be on the earth s surface that is why we set y = R. The centrifugal force, which draws the mass away from the center of earth, is much smaller then the gravitational force drawing it towards the center of earth making it negligible, F cf = m 9.81m = g m.

43 Chapter I. Dynamics of a single particle 35 Likewise, for ordinary velocities, the Coriolis force can be neglected. However, for large masses,like in the study of ocean or wind flows, these effects cannot be neglected. Also, in our calculation we neglected the fact that not only the earth moves but also our galaxy and hence our solar system is not inertial either.

44 36 I.6. Particle motion in non-inertial frames

45 Chapter II Dynamics of system of particles In this chapter we generalize the results from Chapter I for system made of an arbitrary number of n particles. For that, we first introduce in Section II.1 the concept of internal and external forces. Then we discuss how particles can be constrained to each other (Section II.2) and what this implies on the degrees of freedom of the system (Section II.3). The concept of the center of mass as a distinct geometric point for each particle system is explained in Section II.4. Finally, we generalize the momentum and work-energy principles for a system of particles (Section II.5 - Section II.7) and show various examples in Section II.8. Figure II.1: The system shown consists of n particles, each subjected to forces F i (t). The reference point at r (t) can be arbitrarily chosen. A system of particles can be described as follows. Consider n particles described w.r.t. an inertial frame [ i, j, k ] as shown in Figure II.1. The particle i, i = 1,..., n, of the system has mass m i and is located at position r i. The resultant force acting on m i is denoted by F i. Furthermore, we can again choose an arbitrary reference point at position r and denote the distance to particle i as ϱ i (t) = r i r. 37

46 38 II.1. Forces II.1 Forces In this section we examine the resultant force F i on a particle i of a particle system more closely. This will later help us to simplify the momentum principles for a system of particles. One can split the resultant force into two parts, the external forces F ext i and the internal forces F int i, F i = F ext i + F int i. (II.1) Whether, a force is classified as external or internal depends on where the reaction force acts, see definitions below. Definition. An external force is a force whose reaction force acts outside of the considered system, i.e. a force acting on particle i whose reaction force does not act on any particle j of the system, i, j = 1,..., n. Definition. An internal force is a force whose reaction force acts inside of the considered system, i.e. a force acting on particle i whose reaction force acts on any particle j of the system, i, j = 1,..., n. We can split the internal force F int i other particles, F int i = n j=1,j i R ij, acting on m i into its contributions from each of the (II.2) where R ij is the force exerted by m j on m i. y Newton s 3 rd law the action and reaction forces act along the same but opposite direction, therefore R ij = R ji. (II.3) When we sum over the internal forces of all particles in the system we obtain n i=1 F int i = n n i=1 j=1,j i Using (II.3) this evaluates to n n i=1 j=1,j i R ij = 0. R ij. Therefore, the sum over all internal forces of the system does not create a net force acting on the system.

47 Chapter II. Dynamics of system of particles 39 Resultant force on a system of particles The resultant force acting on a system of n particles F = n i=1 F ext i + n i=1 F int i is equivalent to the resultant force F ext of all external forces, F = F ext = n i=1 F ext i, (II.4) since F ext i F int i n i=1 F int i = External forces acting on particle i... Internal forces acting on particle i A similar result can be found for the torque w.r.t. an arbitrary point resulting from all internal forces K ij, i, j = 1,..., n. We will show the case for two masses m i and m j, n = 2, but the result can be generalized to any number of particles. Consider the torque M int i w.r.t. resulting from the force K ij that particle j exerts on particle i, M int i = ϱ i R ij, where ϱ i is defined as ϱ i = r i r. Similarly, R ji creates a moment about, M int j = ϱ j R ji. The result torque M int M int = n i=1 M int i = of all internal forces is n or for the case of two masses n i=1 j=1,j i M int = ϱ i R ij + ϱ j R ji. ϱ i R ij y Newton s 3 rd law we can show that the resultant internal torque M int is 0: M int = ϱ i R ij + ϱ j ( R ij ) = (ϱ i ϱ j ) R ij = 0. (II.5)

48 40 II.2. Constraints We used the fact that (ϱ i ϱ j ) R ij yielding 0 for the vector product between those two vectors, see Figure II.2. Figure II.2: Internal forces for a system of two particles. We can see that it is indeed true that (ϱ i ϱ j ) R ij. Resultant torque about for a system of particles The resultant torque about an arbitrary point for a system of particles M = M int + M ext = n i=1 M int i + n i=1 M ext i is equivalent to the resultant torque of all external forces M = M ext = n i=1 M ext i = n i=1 ϱ i F ext i (II.6) since M int = n i=1 M int i = n n i=1 j=1,j i ϱ i R ij = 0. M int i M ext i... The torque about from the internal forces F int i... The torque about from the external forces F ext i on particle i on particle i II.2 Constraints When we look at the motion of a particle, the motion is often limited in some way. For example, the particle is only allowed to move in a certain direction or must stay on the ground (2D-motion). This kind of limitation on the motion is called a constraint. We have briefly touched upon various types of constraints for a single particle. This section treats the concept a bit more formally and gives a few common examples of constraints. The material presented in this section is actually not only valid for particles but any type of dynamical system. To begin with, we define a constraint as follows.

49 Chapter II. Dynamics of system of particles 41 Definition. A constraint is a scalar relation that limits the possible displacements of particles, i.e. prohibiting the particle to move arbitrarily in space. Specifically, with a system of particles one can distinguish between constraints acting on the whole system, constraints acting on single particles and constraints that act between particles. Formally, however, they can all be treated in the same way. Example II.1: Different types of constraints In the following common examples of constraints are listed. (a) (b) (c) (d) Figure II.3: Typical constraints. (a) A particle is constrained to be at a fixed distance r away from the link. (b) A particle is constrained to move within a rotating tube whose orientation e(t) is prescribed. (c) The particle is constrained to move on the plane z = 0. (d) The rolling constraint in 2D states that the vertical position of the center of mass is fixed. (a ) Rigid link A rigid link as shown in Figure II.3 (a) prescribes a fixed distance r between two particles i and j, i.e. r 2 = (x i x j ) 2 + (y i y j ) 2 + (z i z j ) 2. (II.7) The above equation is a scalar equation and therefore a rigid link corresponds to one constraint. (b ) Rotating tube

50 42 II.2. Constraints II.2.1 We consider a mass m i in a tube that rotates about the origin O in 3D, see Figure II.3 (b). The current orientation, described by the unit vector e(t) in the direction of the tube, is prescribed. The distance of the mass m i from the origin is denoted by r i (t). We can now express the position (x i, y i, z i ) T of the mass in terms of the constrained orientation of the tube as x i y i = r i (t)e(t). z i This equation corresponds to two independent (scalar) constraints. (c ) Planar motion A particle of mass m i is constrained to only move in the plane z = 0, see Figure II.3 (c). This corresponds to one constraint,i.e. z i = 0, where z i denotes the position of the particle in z. (d ) Planar rolling A cylinder, whose mass m i is concentrated in C, roles on the plane z = 0 in x-direction as shown in Figure II.3 (d). The y-position y i of C is constrained to be y i = R, since the cylinder is rigid and does not penetrate into the ground. This corresponds to one constraint. Holonomic constraints We distinguish between two kinds of constraints, i.e. holonomic and non-holonomic. A holonomic constraint only depends on the current position of the particle/particles that is/are subjected to the constraint (and possibly on the current time). All constraints presented in Example II.1 are holonomic. A non-holonomic constraint may also dependent on other quantities such as the velocity of the particle. Definition. A holonomic constraint is a constraint that only depends on the current position of the involved particles and the current time. A non-holonomic constraint may dependent on time derivatives of the motion as well. II.2.2 Forces and constraints So far we have only treated constraints from a purely kinematic approach, i.e. we considered the consequences on the motion of the particle. However, in order to keep the motion of

51 Chapter II. Dynamics of system of particles 43 the particle constrained a force must be exerted on the particle. Specifically, we can relate a force to each constraint. This type of force that arises due to a constraint is called a constraint force. Forces that are exerted on a particle but not due to a constraint are called active forces. Definition. A constraint force is a force that arises to keep the motion constrained according to the imposed constraints. An active forces, on the other hand, is actively exerted on the body independently of the constraints. Example II.2: Particle moving in a plane A particle of mass m is constrained to move in a plane and is also subjected to gravity as shown in Figure II.4. In this case, the normal force N that is exerted from the ground on the particle is a constraint force. It arises because the particle is restricted to move on the plane. The gravitational force mg, however, is an active force since it acts on the particle independent of the constraints. Figure II.4: A particle constrained to move in the plane. The active gravity force G and the constraint force N are acting on it. Note that for constrained systems, the reaction forces must be included in the principle of linear and angular momentum. As they are in general dependent on the motion, they must be eliminated if one is interested in the equation of motion. This will be shown through several examples. Note: This approach of canceling out the constraint forces plays a significant role in analytical dynamics, like Lagrangian mechanics. This is not treated in this course. II.3 Degrees of freedom To describe the position of a particle or a system of particles we need a set of coordinates. Typically, a particle is described by (x, y, z) T in 3D or by (x, y) T in 2D. This corresponds to three/two variables for each particle that are used to fully determine the position. Similarly, one might choose to use cylindrical coordinates instead (see Section I.5) to describe the position of the particle. In any case, we still need three/two variables. We call this set of variables to fully describe the motion the generalized coordinates.

52 44 II.3. Degrees of freedom Definition. A parameter/variable out of a set of parameters that is used to describe the configuration of a system is called a generalized coordinate. This can, e.g., be an angle or a distance. Now consider the case where the particle is subjected to a holonomic constraint. In this case, the generalized coordinates are not independent anymore since they are coupled by the constraint equation. So actually, one of the generalized coordinates becomes redundant since it is a function of the others by the constraint. Similarly, any other additional holonomic constraint further reduces the number of independent generalized coordinates. The independent generalized coordinates needed to describe the system are called degrees of freedom. Definition. The degrees of freedom (DOF) of a system are defined as the number of independent generalized coordinates necessary to describe the motion of the system completely. The DOF of a particle system are equivalent to DOF = dim n k, (II.8) where dim denotes the dimension of space (2 or 3), n is the number of particles in the system and k is the number of independent holonomic constraints applied to the system. Note: For an unconstrained system the DOF correspond to DOF= dim n, stating that for each particle we need dim generalized coordinates to describe its motion. Note: It is important to understand that only holonomic constraints reduce the degrees of freedom. The reason is that a holonomic constraint gives us an explicit relation between two generalized coordinates. On the other hand, a non-holonomic constraint only tells us how one generalized coordinate can evolve in time under consideration of the others. This, however, does not limit the motion of the particle. Example II.3: Double planar pendulum Consider a double pendulum in 2D with two particles of mass m 1 and m 2 as shown in Figure II.5. oth masses are linked to each other and the first mass m 1 is attached to a hinge. For the given configuration determine the DOF of the system. Then, pick a sufficient number of independent generalized coordinates to fully describe the motion. (a ) Degrees of freedom We first need to determine the number of independent holonomic constraints k.

53 Chapter II. Dynamics of system of particles 45 Figure II.5: The considered double pendulum is shown that consists of two masses m 1 and m 2 linked to each other. Mass m 1 is attached to the hinge via a rigid link. Similarly, mass m 2 is attached to mass m 1 via a rigid link. Therefore k = 2 and we can use (II.8) to determine the DOF, DOF = dim n k = = 2. We see from above equation that the system has two degrees of freedom. (b ) Generalized coordinates Since we have two degrees of freedom, we need two independent generalized coordinates to describe the motion of the double pendulum. For example, we can pick ϕ 1 and ϕ 2 that describe the orientation of mass m 1 and m 2 from the horizontal orientation as generalized coordinates. Or one could also pick the vertical positions y 1, y 2 of each mass instead. II.4 Center of mass When we refer to a mechanical system we often describe it using the center of mass r C. This point is uniquely defined as the geometric point where the total mass moment of the system is zero, see definition below. In our case it will help us in the definition of the momentum principles for a system of particles. Definition. The center of mass is defined as the geometric point w.r.t. which the total mass moment of the system is zero. Note: The center of mass is per definition the point for which the resultant torque of the gravitational forces vanishes.

54 46 II.4. Center of mass Figure II.6: The center of mass, denoted as r CM = r C in the figure, is shown for an arbitrary particle system. Only one generic particle m i is shown. According to above definition the center of mass r C for a system of particles is n m i (r i r C ) = 0, i=1 (II.9) where r i denotes the position of particle i with mass m i, see Figure II.6. Further, we can define the total mass M of a system of n particles as the sum of all individual masses, i.e. n M = m i. (II.10) i=1 y rearranging (II.9) we obtain an equation for the center of mass for a system of particles: n m i (r i r C ) = 0, i=1 n m i r i i=1 n m i r C = 0, i=1 n m i r i Mr C = 0, i=1 r C = 1 M n m i r i. i=1 Center of mass for a system of particles r C = 1 M n m i r i i=1 (II.11) r C... center of mass M... total mass of the particle system m i... mass of particle i r i... position of particle i

55 Chapter II. Dynamics of system of particles 47 II.5 Linear momentum principle Next, we will rewrite the linear momentum principle for a system of n particles. The linear momentum P of a particle system is defined as the sum of the individual linear momenta P i, i.e. P := n P i = n m i v i. (II.12) i=1 i=1 From (II.10) we know that n m i r i = Mr C, i=1 which can be differentiated in time, d dt n m i r i = d dt Mr C = i=1 n m i v i = Mv C, (II.13) i=1 where v C denotes the velocity of the center of mass. With (II.13) we can rewrite (II.12) as P = n m i v i = Mv C. i=1 Definition. The linear momentum P of a particle system with n particles, total mass M and center of mass r C is defined as the sum of the individual linear momenta P i of each particle, P := n P i = n m i v i, (II.14) i=1 i=1 which is equal to P := Mv C. (II.15) Similarly, the linear momentum principle for a particle system is the linear momentum principle (I.2) for a single particle Ṗ i = F i = F ext i + F int i (II.16) summed over all particles, Ṗ = n Ṗ i = n F i. (II.17) i=1 i=1 We can now use (II.4) to simplify (II.17) to Ṗ = n F i = i=1 n i=1 (F ext i + F int i ) = n i=1 F ext i = F ext.

56 48 II.6. Angular momentum principle The principle of linear momentum for a system of particles Ṗ = ma C = n i=1 F ext i = F ext (II.18) P..... linear momentum of the particle system M.... total mass of the system a C F ext i.... acceleration of the center of mass... external forces acting on particle i F ext... resultant external force acting on the system Note: If F ext = 0 then linear momentum is conserved, i.e. P = const. and v C = const. In this case, however, in general a particle i does not have constant velocity, v i const., when linear momentum is conserved. II.6 Angular momentum principle Let us now refer to Figure II.7 and define the angular momentum for a system of particles: Definition. The angular momentum for a system of n particles w.r.t. an arbitrary, moving point is defined as the sum of all individual angular momenta H i, H := n H i = i=1 n ϱ i P i, i=1 (II.19) where ϱ i (t) = r i (t) r (t) describes the distance between the point and particle i. Differentiating (II.19) in time yields d dt H = n ϱ i P i + i=1 n ϱ i Ṗ i. i=1 Next, we replace ϱ i in the first term of the above equation with ϱ i = ṙ i ṙ = v i v

57 Chapter II. Dynamics of system of particles 49 Figure II.7: The necessary quantities to define the angular momentum w.r.t. an arbitrary point at r (t). giving Ḣ = = n n (v i v ) P i + ϱ i Ṗ i, i=1 n v i P i v i=1 = v P + i=1 n P i + n ϱ i Ṗ i. i=1 i=1 i=1 n ϱ i Ṗ i, (II.20) We here used the definition of linear momentum (II.12) and the fact that for each particle i v i P i = mv i, which means v i P i = 0. In the second term of (II.20) we can replace Ṗ i by (II.16), i.e. Ḣ = v P + Inserting in (II.6) M = M ext = n i=1 n ϱ i Ṗ i = v P + i=1 ϱ i F ext i = n i=1 ϱ i (F ext i n i=1 ϱ i (F ext i + F int i ), + F int i ). which is the resultant torque for a system of particles. The principle of angular momentum is therefore: Ḣ = v P + M ext. The principle of angular momentum for a system of particles Ḣ + v P = M ext (II.21)

58 50 II.7. Work-energy principle H.... angular momentum of a particle system w.r.t. v..... velocity of point P M ext linear momentum of a particle system... resultant external torque w.r.t. point Note: If M ext v = 0 or = C or v v C, = 0 and one of the three following conditions is fulfilled, i.e. then angular momentum is conserved, H = const. II.7 Work-energy principle Consider an arbitrary particle i that is part of a system of n particles. It is subjected to its resultant force F i (t). The work W i,12 done between positions r i,1 = r i (t 1 ) and r i,2 = r i (t 2 ) can be expressed as W i,12 = ri,2 r i,1 F i dr i. When we now sum over all particles we get the total work W 12 done on all particles between states 1 and 2, W 12 = n i=1 ri,2 r i,1 F i dr i. (II.22) It is useful at this point to define the kinetic energy T : Definition. The kinetic energy T of a system of n particles is defined as the sum of the individual kinetic energies of all particles, i.e. T := n i=1 1 2 m i v i 2. (II.23) Therefore, (II.22) becomes W 12 = T 2 T 1, (II.24) where T i = T (t i ). The derivation of this result was obtained equivalently to what was shown in Section I.3. This is the general work-energy principle for any system of particles. However, it is not very practical to use because it is very cumbersome to compute the total work done since it includes both internal and external forces.

59 Chapter II. Dynamics of system of particles 51 Work-energy principle for a system of particles W 12 = T 2 T 1 (II.25) W work done on the particles between states 1 and 2 T i.... kinetic energy of the particle system associated with state i II.7.1 Work-energy principle for rigid body systems As has been shown before, the force F i acting on particle i may be split into an internal and external contribution. Similarly, we can divide the work W 12, n ri,2 n ri,2 n ri,2 W 12 = F i dr i = F int i dr i + F ext i dr i, r i,1 i=1 = W int 12 + W ext 12. i=1 r i,1 This is done to see whether we can actually further simplify the work-energy principle, which would facilitate computations in practice. For the next steps, we will only consider the internal work W12 int and first replace dr i = v i dt, n t2 W int 12 = i=1 t 1 From (II.2) we know that n F int i = R ij, j=1,j i F int i v i dt. where K ij denotes the force particle j exerts on particle i. We can replace F int i the resulting double sum to obtain W int 12 = = = t2 n t 1 i=1 t2 t 1 t2 t 1 n F int i v i dt, n i=1 j=1,j i n i=1 j<i R ij v i dt, From Newton s 3 rd law we know that R ij = R ji, i=1 r i,1 and rewrite n (R ij v i + R ji v j ) dt. (II.26) so we can rewrite (II.26) to t2 n n W12 int = R ij (v i v j ) dt. t 1 i=1 j<i (II.27)

60 52 II.7. Work-energy principle Unfortunately, no further simplifications can be made for the general case! However, we can now assume the system to be a rigid body system of particles, see definition below, to simplify the equations. Definition. A rigid body system of particles is a system where any particle is at a constant distance with all the other particles, i.e. d dt r i r j 2 = 0 i, j. (II.28) This is equivalent to say that an arbitrary particle is connected to all the others via rigid links. Note: For any particle system that is not a rigid body system one must go back to (II.24) to compute the work done. From here on, we discuss the case of rigid body systems. When (II.28) is evaluated we obtain, n d dt ( ) (r i r j ) (r i r j ) = 2 (r i r j ) (v i v j ) = 0. (II.29) This result implies that (r i r j ) (v i v j ). Also recall that (r i r j ) R ij for a rigid body system and therefore, together with (II.29), we obtain R ij (v i v j ) = 0. (II.30) Specifically, by using (II.30) the right hand side of (II.27) vanishes, and W int 12 = 0, W 12 = W ext 12. The internal work for a rigid body system of particles is zero. This is the main result we conclude from this section. Work-energy principle for a rigid body system of particles W 12 = W ext 12 = T 2 T 1 (II.31) W work done on the particles between states 1 and 2 W ext work done by the external forces between states 1 and 2 T i..... kinetic energy of the particle system associated with state i

61 Chapter II. Dynamics of system of particles 53 II.7.2 Work-energy principle for conservative systems Similar to Section I.3.1 we can define a potential energy of the system to capture the effect of all potential forces, internal or external. Definition. The potential energy V (r 1,..., r n ) of a system of n particles is defined as the sum of the potentials V k of all potential forces F k acting on the system, internally or externally(!), k = 1,..., K K V (r 1,..., r n ) := V k. k=1 (II.32) Note: For a rigid body system of particles we only need to consider the potentials of external forces, since the potential of the internal force R ij that particle j exerts on particle i and the potential of R ji cancel each other out. In case R ij and R ji are not potential forces they do not contribute to the potential energy anyway. Therefore we get V = K k=1 V ext k. (II.33) As in Section I.3.1 we can write that the total mechanical energy E of a conservative system, which is a system where all forces are either potential or do no work, is conserved. Work-energy principle for a conservative system of particles The total mechanical energy E := T + V of a conservative system of particles is conserved along all motions, T 1 + V 1 = T 2 + V 2 (II.34) or equivalently, d dt E(t) = 0. E... total mechanical energy T i... kinetic energy of the system associated with state i V i... potential energy of the system at state i Note: For a rigid body system to be conservative it is sufficient that all external forces F ext i are either potential or do no work. We have already shown before that the internal work vanishes for such a system and therefore does not contribute to the overall work. For more information on potential forces, potentials, etc..., please refer to Section I.3.1.

62 54 II.8. Example problems II.8 Example problems Example II.4: Elastic collision of a system of particles with a wall Figure II.8: A rigid body system of three particles approaching a rigid wall with a velocity V perpendicular to it. Consider a rigid body system of three particles uniformly approaching a wall with velocity v bef as shown in Figure II.8. The particles are connected to each other via rigid bars and they all hit the wall at the same time. The collision with the wall is assumed to be perfectly elastic, which means the collision forces derive from elastic potentials, and therefore the total mechanical energy is conserved over the collision. Further, the particles are not subjected to any external forces. (a ) Apply the conservation of energy for the system and see what conclusions can be drawn from it. (b ) Qualitatively discuss the case when the particles are not connected via rigid bars but via springs and indicate the consequences on the conservation of energy. (a ) Conservation of energy The system is not subjected to any forces before and after the collision. Also the collision forces do no work. Therefore, the potential energy of the system before V bef and after V af the collision is 0, V bef = V af = 0. The conservation of energy states T bef + V bef = T af + V af, and therefore T bef = T af since the potential energy is zero for both states. Inserting the definition of kinetic energy gives m i v i 2 bef = 1 2 i=1 3 m i v i 2 af i=1

63 Chapter II. Dynamics of system of particles 55 and since the velocities are constant before and after the collision (uniform approach) we can conclude that 1 3 m i v 2 bef = m i v 2 af 2, i=1 i=1 1 2 M vbef 2 1 = 2 M vaf 2, 1 2 M v C 2 bef = 1 2 M v C 2 af. (II.35) In the last step we used the fact that v C = v i, i because all particles move with the same velocity and thus the center of mass must also move with the same velocity. From (II.35) it follows that v C bef = v C af. (II.36) Therefore the absolute value of the velocity of the center of mass remains the same over the collision. The system "bounces" back. (b ) Non-rigid system If we replace the rigid links with springs as depicted in Figure II.9 we can still apply conservation of energy since spring forces are potential and we can simply include them in the potential energy of the system. However, this means v C needs not to be constant anymore since we have T + V = const and not T = const. as before. Figure II.9: If we replace the rigid links with springs, the potential of the springs must be taken into account and therefore v C is not constant anymore. Example II.5: Two-dimensional collision Figure II.10: Two particles of mass m 1 and m 2 are approaching each other and will eventually collide.

64 56 II.8. Example problems We look at two particles of mass m 1 and m 2 that are initially moving with velocities v 1, respectively v 2 as illustrated in Figure II.10. At time t the two particles collide. At the impact, the collision forces are impulsive and they only act in direction normal to the collision but not in tangential direction since we assume a smooth surface. Furthermore, the external forces remain bounded for all times (gravity,etc... ). In the following we are interested in the velocities u 1 and u 2 at time t + after the impact. (a ) Linear momentum principle We first evaluate the linear momentum principle for the whole system, i.e. particles 1 and 2 combined, Ṗ = 2 i=1 F ext i, which we can now integrate between t and t + giving us t + t Ṗ dt = P (t + ) P (t ) = t + 2 t i=1 t + t 2 i=1 F ext i dt, F ext i dt. Since all external forces are bounded and the time interval [t, t + ] is infinitesimally small the right hand side of above equation vanishes, t + lim t + t t 2 i=1 F ext i dt = 0. (II.37) Note: The impulsive, unbounded collision forces are internal and therefore do not need to be included in the above equation! We can deduce from (II.37) that the linear momentum of the whole system is conserved during collision, P (t + ) = P (t ) (II.38) Next, we will divide our equations into a component normal to the collision surface (superscript n) and a component tangential to it (superscript t) and evaluate the linear momentum principle accordingly. In Figure II.11 the respective velocities in normal and tangential direction are shown. i Tangential direction For the tangential direction we assumed that there are no collision forces acting. Therefore, there are no impulsive internal forces acting in tangential direction and the linear momentum is not only conserved for the overall system but also for each particle individually, i.e. P t 1 = const., P t 2 = const.

65 Chapter II. Dynamics of system of particles 57 (a) (b) Figure II.11: The situation just before and just after the collision is sketched. (a) The two masses are shown just before the collision. Note that v1 n > vn 2 otherwise there is no collision. (b) After the collision it must be true that u n 2 un 1 otherwise mass m 1 will penetrate into mass m 2. From above equations it follows that the velocity in tangential direction does not change during the collision. u t 1 = v t 1 u t 2 = v t 2 (II.39) (II.40) The mathematical derivation works equivalently to (II.37). ii Normal direction For the normal direction only the linear momentum of the overall system is conserved, since there is a collision force acting internally in normal direction. We evaluate (II.38) in normal direction and get P n (t ) = P n (t + ), m 1 v n 1 + m 2 v n 2 = m 1 u n 1 + m 2 u n 2. (II.41) (b ) Coefficient of restitution With (II.41) we only get one equation for the two unknowns u n 1 and un 2. We now introduce the parameter e, the coefficient of restitution, to describe the imperfection of the collision. The parameter e is defined as the ratio between the relative normal velocity before and after the collision, i.e. relative normal velocity after e = relative normal velocity before = un 2 un 1 v2 n vn 1, e = un 2 un 1 v n 1 vn 1 (II.42) e [0, 1] depends on various physical parameter, such as surface quality, material, shape,... For e = 0, called perfectly inelastic, both particles move with the same normal velocity after

66 58 II.8. Example problems the impact, while e = 1 corresponds to the perfectly elastic case. This is the only case for which energy is conserved, otherwise energy is not conserved! This is the reason why we did not use the conservation of energy. We can now use (II.41) and (II.42) to obtain an equation for u n 1 and un 2 as a function of e, u n 1 = m 1 em 2 m 1 + m 2 v n 1 + (1 + e)m 2 m 1 + m 2 v n 2 u n 2 = (1 + e)m 1 m 1 + m 2 v n 1 + m 2 em 1 m 1 + m 2 v n 2 (II.43) (II.44) Note: You can reuse this result for any 2D collision problem that meets the assumptions stated at the beginning, (two particles, non-deformable, impulsive collision force, nonimpulsive external forces, smooth surface,... ). Example II.6: Dynamics of a double pendulum Consider a double pendulum as shown in Figure II.12 swinging in the [ i, j ] - plane that is subjected to gravity. oth masses m 1 and m 2 are linked to each other and one is also linked to a smooth joint. The links consist of rigid, massless bars. (a ) Draw a free-body diagram and decide how many generalized coordinates are needed to describe the problem. Extra: Show that a massless bar can only support tangential forces. (b ) Find the equation of motion by applying the appropriate principles. (a) (b) Figure II.12: The considered double pendulum is shown. (a) The double pendulum and the according generalized coordinates can be seen. (b) The free-body diagram of the system when we cut free each particle individually is shown.

67 Chapter II. Dynamics of system of particles 59 (a ) Free-body diagram and generalized coordinates i Free-body diagram When separating the bodies we find the forces shown in Figure II.12. Mass m 1 is subjected to gravity m 1 g, to the constraint force R exerted by the joint and to the internal force K 12 between the two particles. Similarly, mass m 2 experiences m 2 g and K 21. ii Generalized coordinates As discussed before in Example II.3 a double pendulum in 2D possesses two degrees of freedom. We will choose ϕ 1 and ϕ 2 as generalized coordinates to describe the orientation from the horizontal position. iii Massless bars A massless bar can only support forces in tangential direction. Furthermore, those forces will be collinear, as for the above link with the forces K 12 and K 21. To proof this we will apply the linear and angular momentum principle for a massless beam with m = 0 and the hypothetical center of mass C is assumed to be in the geometric middle. Now let us also assume that the bar is subjected to the normal forces N 12 and N 21 at the respective ends r 1 and r 2. When we now apply the linear momentum principle and take into account that the bar is massless, we get: Ṗ = m r C = N 12 + N 21, 0 = N 12 + N 21, N 12 = N 21. (II.45) The angular momentum principle about C yields Ḣ C + v C P = r 1 N 12 + r 2 N 21, 0 = r 1 N 12 + r 2 N 21. (II.46) The left hand side is 0 because the bar is massless. Combining (II.45) and (II.46) gives us (r 1 r 2 ) N 12 = 0, which can only be true when N 12 = N 21 = 0. (II.47) (b ) Equation of motion To find the equation of motion we apply the angular momentum principle about O and the work-energy principle. We could also apply the linear momentum principle but with this choice the constraint force R appears in the equations. In other words, applying the LMP would give us an additional equation but also an additional unknown, namely R.

68 60 II.8. Example problems i Angular momentum principle w.r.t. O As stated before, we apply the angular momentum principle about O since the constraint force R does not produce a torque about O. Also, v = O making the evaluation of the AMP easier. The AMP states Ḣ + v P = M ext, First, we find H to be Ḣ = M ext. (II.48) H = r 1 P 1 + r 2 P 2 (II.49) with r 1 = l 1 (sin ϕ 1 i cos ϕ 1 j), r 2 = (l 1 sin ϕ 1 l 2 sin ϕ 2 ) i (l 1 cos ϕ 1 + l 2 cos ϕ 2 ) j, which we found from geometry. Differentiating (II.49) w.r.t. time yields Ḣ = ṙ 1 P 1 + ṙ 2 P 2 + r 1 Ṗ 1 + r 2 Ṗ 2, which can be simplified to since Ḣ = r 1 Ṗ 1 + r 2 Ṗ 2, = r 1 (m 1 r 1 ) + r 2 (m 2 r 2 ) (II.50) ṙ i P i = mṙ i. The only forces that contribute to the resultant torque M ext are the gravitational forces, M ext = r 1 ( m 1 g j) + r 2 ( m 2 g j). (II.51) Plugging (II.50) and (II.51) into (II.48) gives us m 1 r 1 r 1 + m 2 r 2 r 2 = m 1 g(r 1 j) m 2 g(r 2 j) (II.52) a 2 nd order nonlinear, ordinary differential equation (ODE) in ϕ 1 and ϕ 2. Note: (II.52) is a scalar equation since it only possesses a compenent in k - direction (the other two directions yield the trivial result 0 = 0). We are left to compute the expressions r 1 and r 2, see below, ṙ 1 = l 1 ϕ 1 (cos ϕ 1 i + sin ϕ 1 j), r 2 = l 1 ϕ 2 (cos ϕ 1 i + sin ϕ 1 j) l 1 ϕ 2 1(sin ϕ 1 i cos ϕ 1 j), ṙ 2 = (l 1 ϕ 1 cos ϕ 1 l 2 ϕ 2 cos ϕ 2 ) i + (l 1 ϕ 1 sin ϕ 1 + l 2 ϕ 2 sin ϕ 2 ) j, r 2 = l 1 ( ϕ 1 cos ϕ 1 ϕ 2 1 sin ϕ 1 ) i l 2 ( ϕ 2 cos ϕ 2 ϕ 2 2 sin ϕ 2 ) i + l 1 ( ϕ 1 sin ϕ 1 + ϕ 2 1 cos ϕ 1 ) j + l 2 ( ϕ 2 sin ϕ 2 + ϕ 2 2 cos ϕ 2 ) j.

69 Chapter II. Dynamics of system of particles 61 ii Work-energy principle Since there is either a rigid link between particles or there is no internal work between particles we can use the work-energy principle for a rigid body system. Note, however, that the considered double pendulum is not a rigid body system of particles. The work-energy principle for rigid body states that W ext 12 = T 2 T 1. Since all external forces are potential (gravity) or do no work (constraint force R is normal to velocity v 1 ) the system is also conservative. Therefore, the total mechanical energy E = T + V must be conserved, T + V = const. or d dt (T + V ) = 0. The kinetic energy of the system is T = 1 2 m 1 ṙ m 2 ṙ 2 2 and the potential energy is V = m 1 gl 1 cos ϕ 1 m 2 g(l 1 cos ϕ 1 + l 2 cos ϕ 2 ) (II.53) with the zero-level of the potential energy at O. All necessary quantities have already been computed and need to be inserted into (II.53) to obtain the second 2 nd nonlinear ODE, describing the motion. iii Equations of motion We now have the full set of ODEs to describe the problem, i.e. a two-dimensional 2 nd order, nonlinear ODE system in ϕ 1 and ϕ 2 : ( ) (m 1 + m 2 )l 1 m 2 l 1 cos 2 (ϕ 2 ϕ 1 ) ϕ 1 = 1 2 l 1m 2 ϕ 2 1 sin ( 2(ϕ 2 ϕ 1 ) ) + m 2 g sin ϕ 2 cos(ϕ 2 ϕ 1 ) + m 2 l 2 ϕ 2 2 sin(ϕ 2 ϕ 1 ) (m 1 + m 2 )g sin ϕ 1 ( ) (m 1 + m 2 )l 2 m 2 l 2 cos 2 (ϕ 2 ϕ 1 ) ϕ 2 = 1 2 l 2m 2 ϕ 2 2 sin ( 2(ϕ 2 ϕ 1 ) ) ( + (m 1 + m 2 ) g sin ϕ 1 cos(ϕ 2 ϕ 1 ) ) l 1 ϕ 2 1 sin(ϕ 2 ϕ 1 ) g sin ϕ 2 One can now solve the system numerically; for this, the system has to be first transformed into a four-dimensional system of 1 st ODEs with the state vector (ϕ 1, ϕ 1, ϕ 2, ϕ 2 ) T. Note: The above system is already highly complex to solve although it just consists of two particles. Hence, we recommend to solve problems with even more particles by taking the system apart and applying the principles for each particle individually instead of applying them to the overall system.

70 62 II.8. Example problems

71 Chapter III Kinematics of planar rigid bodies In the previous chapters we have only treated systems consisting of particles. A particle is an abstraction characterized by the fact that it has no physical dimensions. This is, of course, not true for any kind of real physical system. To model realistic settings, we are up to now forced to reduce the dynamics of a system to the motion of its center of mass. What can be done when we actually want to describe the entire motion of the system? In practice, many systems can be described as two-dimensional rigid bodies. This is still not the general three-dimensional case but often sufficient. We will introduce the concept of a rigid body for two dimensions in Section III.1. Section III.2 and III.3 tackle the kinematics of the planar rigid body. Multiple examples related to the kinematics 1 are discussed in Section III.4. We then shortly introduce the concept of the instantaneous center of rotation in Section III.5. Figure III.1: A planar rigid body is shown. Two arbitrary points A, and their corresponding position vectors r A and r are shown as well. III.1 Rigid body To begin with, we need to define a rigid body. Consider a body and two arbitrary points A, on the body at positions r A, r described w.r.t. the inertial frame [ i, j ], as 1 In kinematics one aims to express the motion of a system, whereas in kinetics we describe the relation between forces acting on the system and the motion of the system. 63

72 64 III.2. Angular velocity illustrated in Figure III.1. Further, we denote the vector from A to as r A = r r A. For a rigid body the distance between the two arbitrary points A, must be constant: r A = const. Definition. A rigid body is a body for which the distance r A between two arbitrary points A, remains constant for all times, i.e. r A = const. (III.1) or d dt r A = 0. (III.2) III.2 Angular velocity We are now interested in relating the velocities v A and v of two arbitrary points A and by exploiting the definition of a rigid body. First, we can rewrite condition (III.1) as r A 2 = r A r A = const. and then differentiation it in time yields 2 ṙ A r A = 0. Therefore ṙ A remains orthogonal to r A for all times, ṙ A r A. (III.3) ecause of this, we can write ṙ A as the vector product between some vector ω A = ω A k and r A, ṙ A = ω A r A (III.4) where we defined ω A to simply point in k-direction such that we fulfill the orthogonality condition (III.3), see Figure III.2. We can get the magnitude of ω A by taking the modulus from (III.4), ṙ A = ω A r A, ṙ A = ω A r A sin π 2, ω A = ṙ A r A.

73 Chapter III. Kinematics of planar rigid bodies 65 Figure III.2: The relative velocity vector ṙ A is orthogonal to ω A and r A. One can now show that independently of the choice of A and, we obtain the same vector ω A. In other words, ω A is a unique quantity for a rigid body. We call ω A ω = ω k (III.5) the angular velocity of. Hence, (III.4) becomes ṙ A = ω r A A, (III.6) How to identify the unique angular velocity ω of? To find the angular velocity of we can apply the following steps. First, we put a fixed mark ("straight line") on the body and one in the inertial frame, see Figure III.3 (a). θ denotes the angle between the marks. When the body rotates the angle θ changes and the time derivative of θ gives us the angular velocity, ω = θ. In vectorial form this is ω = θ k. (III.7) The above result is independent of the choice we make for the marks. In fact, consider two different marks, where the new mark on has an angle α w.r.t. the old one and the new mark in the inertial frame is rotated by β compared to the old one, see Figure III.3 (b). The angle θ between the new marks and the old angle θ are related to each other by θ = θ + α β. When we now take the time derivative we get θ = θ and therefore ω = θ = θ is indeed unique.

74 66 III.3. Velocity transport formula (a) (b) Figure III.3: A simple method to show the uniqueness of the angular velocity ω. (a) Choose two marks, one body-fixed and one inertial. (b) Place a new mark on, and a new mark on the inertial frame. The angles θ and θ are just shifted by constant values and hence θ = θ. III.3 Velocity transport formula We now come back to our initial question of how the velocities v A, v of two points A, are related to each other. Let us first write the position r of point as r = r A + r A and then differentiate it in time ṙ = ṙ A + ṙ A. (III.8) From (III.6) we know that ṙ A = ω r A and we can therefore write (III.8) as v = v A + ω r A. The above formula is called the velocity transport formula. It relates the velocities of any two points of the rigid body with the help of the angular velocity ω. It contains two terms: v A and ω r A. We define the second part as v /A := ω r A (III.9) to highlight the velocity of w.r.t A, which can also be interpreted as a rotation of around A. In other words, if v A = 0, point rotates around point A. Note: (Important!) You cannot refer to ω as ω w.r.t. a point or an axis since ω is the same for any axis or point for the same rigid body. Statements as "the angular velocity w.r.t. point P " do not make sense.

75 Chapter III. Kinematics of planar rigid bodies 67 The velocity transport formula for a rigid body v = v A + ω r A (III.10) v.... velocity of an arbitrary point v A.... velocity of an arbitrary point A ω..... angular velocity of r A... vector pointing from A to Note: The velocity transport formula is of course only valid if A and are of the same rigid body. III.4 Example problems Example III.1: Slider-crank mechanism Consider a system of three rigid bodies, consisting of two cranks and one slider as shown in Figure III.4. Such mechanism is used to transform a linear, periodic motion of the slider into a rotary motion. The same mechanism is also used in the cylinders of car engines, where the piston is connected to the crank shaft via a rod to transform the linear motion of the piston into a rotary motion of the crank shaft. We now want to know the velocity v for a given instantaneous velocity v 0 of the slider in the shown configuration. Figure III.4: A slider-crank mechanism consisting of three rigid bodies, two cranks and one slider. Point as well as point A are part of the crank A of length l and angular velocity ω 1.

76 68 III.4. Example problems We can apply the velocity transport formula between A and : v = v A + ω 1 r A. (III.11) Point A does not move, therefore v A = 0. We denote the angle between the horizontal position and the current position of crank A as α. We can then write ω 1 and r A as ω 1 = α k, r A = l cos α i + l sin α j. When inserting the above into (III.11) we obtain v = ( α k) ( l cos α i + l sin α j). (III.12) Since is also part of the longer crank C of length 2l we can also apply the velocity transport formula between and C, v = v C + ω 2 r C. (III.13) Note that ω 1 ω 2 as they belong to two different bodies! Using the angle β between the horizontal and the current position of crank C we can express ω 2 and r C, ω 2 = β k, r C = 2l cos β i + 2l sin β j. v C is given as v C = v 0 = v 0 i. Finally, we get v = v 0 i + ( β k) (2l cos β i + 2l sin β j). (III.14) We now equate (III.12) and (III.14), which yields: ( α k) ( l cos α i + l sin α j) = v 0 i + ( β k) (2l cos β i + 2l sin β j), which we can rearrange such that all terms in i are on side and all terms in j are on the other side, i(l α sin α + v 0 + 2l β sin β) = j( αl cos α + 2l β cos β). Since i j both sides of above equation must be zero giving us two equations in α, α, β and β: l α sin α + v 0 + 2l β sin β = 0, αl cos α + 2l β cos β = 0. (III.15) (III.16)

77 Chapter III. Kinematics of planar rigid bodies 69 From (III.16) we get β = α cos α 2 cos β and from (III.15) together with above result for β α = v 0 l(sin α + cos α tan β). Using the geometry of the system we can relate α and β to each other by 2l sin β a = l sin α. We can the express sin β and cos β as sin β = a + l sin α, 2l (a + l sin α)2 cos β = 1 4l 2. We use above relations to get α as a function of α, i.e. α = f(α) = v 0 l 1. a + l sin α sin α sin α + cos α 4l 2 (a + l sin α) 2 (III.17) Finally, we use (III.12) and plug in (III.17) to get v : v = v 0 4l 2 + (a + l sin α) 2 (sin α i + cos α j) sin α 4l 2 + (a + l sin α) 2 + (a + l sin α) cos α (III.18) Result (III.18) is linear in v 0 but highly nonlinear in the geometry. Example III.2: Two-dimensional rolling Consider a disk that rolls without slipping along i as shown in Figure III.5. We want to know the velocity of the center C and of the point A, where A denotes a fixed point on the disk that is currently in contact with the ground in the shown position. On the other hand, A denotes the point on the ground at the location of instantaneous contact between the disk and the ground. The rolling without slipping conditions states that the covered arc length Rϕ must be equal to the distance s A that point A covered, i.e. s A = Rϕ. From this we can get the velocity of A by differentiating w.r.t. time v A = R ϕ i.

78 70 III.4. Example problems Figure III.5: The disk depicted in the figure rolls without slipping on the ground. A denotes the current point of contact on the disk, whereas A is the current point of contact on the ground. Note: When applying the velocity transport formula to the disk we cannot include A since it is not part of the disk. From the rolling without slipping constraint it also follows that v C = v A = R ϕ i. (III.19) To obtain the velocity v A of the point A on the disk, we can apply the velocity transport formula between C and A, v A = v C + ω r CA, = R ϕ i + ( ϕ k) ( R j), = R ϕ i R ϕ i, = 0. We replaced v C by (III.19) and found r CA from geometry. The angular velocity ω is simply ω = ϕ k. From the velocity transport formula we obtained v A = 0 This result also implies that the force associated with the rolling friction does no work. Further, we call A the instantaneous center of rotation. With A we can simply compute the velocity of any point as v = v A + ω r A = ω r A. We can also use the instantaneous center of rotation to construct the velocity v graphically, see Figure III.6. We simply exploit the fact that v r A and compute the length of v as v = ω r A = v 0 R r A.

79 Chapter III. Kinematics of planar rigid bodies 71 Figure III.6: Since the velocity v of must be perpendicular to the vector r A, where A denotes the instantaneous center of rotation, we can construct the velocity graphically. Example III.3: Ladder sliding down a wall Consider a ladder sliding down a wall, see Figure III.7 (a). We are interested in whether there exists an instantaneous center of rotation for the given configuration and where it is located. y definition, for O to be an instantaneous center of rotation it must hold: v O = 0. We can now apply the velocity transport formula between and O as well as A and O: v = v O + ω r O = ω r O, v A = v O + ω r OA = ω r OA. From above equation we obtain that v A r 0A and v r 0. (III.20) (III.21) We can construct the instantaneous center of rotation O graphically with above conditions. Note that it might not fall on the body, see Figure III.7 (a). However, if O is not part of the body we are not allowed to use the velocity transport formula in the first place. In order to justify the use of the velocity transport formula, though, we extend the body such that O is included. Then we find the instantaneous center of rotation to be part of the extended body. In other words, we make sure that the instantaneous center of rotation moves with the body. As for the previous example, we can now also construct the velocity v P of any point P graphically, see Figure III.7 (b), and find the magnitude of v P to be v P = ω r OP = v A r OA r OP, where ω was computed using some known velocity v A and the velocity transport formula between O and A.

80 72 III.5. Instantaneous center of rotation (a) (b) Figure III.7: A ladder sliding down a wall is shown in the above figure. (a) When knowing the velocity of two points the instantaneous center of rotation can be constructed graphically. (b) With this knowledge also the velocity of any other point can be determined. III.5 Instantaneous center of rotation For a planar rigid body, we can actually always find an instantaneous center of rotation for any type of motion, although it sometimes does not lie within the body itself. The reason is that in 2D we can always find a point O via the velocity transport formula for which v O = O, also see (III.20) and (III.21). Definition. The instantaneous center of rotation O of a planar rigid body is defined as the point O for which v O = 0. The extended body is hereby chosen such that it includes the original body as well as the instantaneous center of rotation O.

81 Chapter IV Kinetics of planar rigid bodies ased on the knowledge from 2D-kinematics we will now derive the momentum and energy principles for a planar rigid body. First, we derive the linear momentum principle for planar rigid bodies in Section IV.1 and the angular momentum principle in Section IV.2. We will also take a more closer look at how to compute the angular momentum in Section IV.3, since the general formulation is computationally quite involved. The work-energy principle for a general and a conservative system respectively, is discussed in Section IV.4. Subsequently, we will look at the degrees of freedom of a system (Section IV.5) and finally, we apply the new principles to multiple examples in Section IV.6. IV.1 Linear momentum principle Figure IV.1: To derive the linear momentum principle the body is divided into infinitesimal areas of mass m i. To derive the linear momentum principle for a planar rigid body of mass M in the [ i, j ] - plane, we will first divide it into N areas of mass m i at position r i, see Figure IV.1. For sufficiently small m i 0 we can approximate the mass elements as particles and therefore we can write the linear momentum P as the linear momentum of a system of N particles and take the limit m i 0, N : P = lim N m i 0 N v i m i = i=1 v dm, (IV.1) where v = v(x, y) is the velocity for each point. Further, dm denotes the infinitesimal 73

82 74 IV.1. Linear momentum principle mass element that can also be expressed in terms of the area density ρ = ρ(x, y) and the infinitesimal surface element da, i.e. dm = ρ da and specifically for Cartesian coordinates [x, y] dm = ρ da = ρ dx dy. Note: Since is a planar rigid body, i.e. two dimensional, ρ is an area density with units [ρ] = kg /m 2 and we only integrate over the surface. As we did before for a system of particles, we can also define a center of mass C. We will again start with the formula for the center of mass for a particle system and take the limit m i 0, N to obtain a formula for the center of mass for a planar rigid body, r C = 1 lim N M m i 0 N r i m i = 1 M i=1 r dm. Center of mass of a planar rigid body r C = 1 M r dm = 1 M r ρ da (IV.2) r..... position vector dm... infinitesimal mass element ρ..... area density of, [ρ] = kg /m 2 da... infinitesimal surface element With (IV.2) we can write the linear momentum (IV.1) as follows, P = M 1 M v dm = M 1 M d dt r dm = M d dt 1 M where we denote the velocity of the center of mass as v C. r dm = M d dt r C = Mv C, Note: We assumed that the mass distribution does not change in time because otherwise we are not allowed to change the order of integration and differentiation in above derivation. Definition. The linear momentum of a planar, two dimensional rigid body of mass M whose center of mass moves with velocity v C is defined as P := Mv C. (IV.3)

83 Chapter IV. Kinetics of planar rigid bodies 75 In the same manner we can also derive the linear momentum principle for a planar rigid body. We use the LMP for a particle system and take the limit m i 0, N giving us the LMP. The principle of linear momentum for a planar rigid body Ṗ = Ma C = F ext (IV.4) P..... linear momentum of M.... mass of a C.... acceleration of the center of mass of F ext... resultant external force acting on Note: If F ext = 0 then linear momentum is conserved, P = const. IV.2 Angular momentum principle Figure IV.2: The rigid body is again divided into infinitesimal areas of mass m i. The center of mass CM is also depicted in the figure. Further, denotes an arbitrary reference point. We again consider a planar rigid body of mass M that is divided in N areas of mass m i at r i, see Figure IV.2. The center of mass is located at r C and an arbitrary point is located at r. Remember is totally arbitrary and may also move. The vector pointing from to the mass element m i is denoted by ϱ i = r i r. To find the angular momentum w.r.t. a point we use the formula for a system of N

84 76 IV.3. Moment of inertia and angular momentum transfer formula particles with mass m i and take the limit N and m i 0, N N H = lim ϱ i N (P ) = lim ϱ i N ( m iv i ) = ϱ v dm. m i 0 i=1 m i 0 i=1 Equivalently, we can derive the angular momentum principle, see result below. (IV.5) The principle of angular momentum for a planar rigid body Ḣ + v P = M ext (IV.6) H.... angular momentum of w.r.t. a point v..... velocity of point P M ext linear momentum of body... resultant torque w.r.t. point Note: If M ext v = 0 or = C or v v C, = 0 and one of the three following conditions is fulfilled, i.e. then angular momentum is conserved, H = const. Recall that we can compute the resultant external torque w.r.t. a point at r as M ext = where F ext k M free j K k=1 r k F ext k + J j=1 M free j. is an external force acting on the body at r k and r k is r k = r k r. denotes a free moment that is exerted onto the body. Note: As we have only considered particles without physical dimension so far, a free moment was not defined. However, for a rigid body the resultant torque comes from the torque produced by all external forces and free moments. IV.3 Moment of inertia and angular momentum transfer formula We will now go back to formula (IV.5) for calculating the angular momentum w.r.t. to some point, H = ϱ v dm, and see how we can compute this in practice since the above formula is rather cumbersome.

85 Chapter IV. Kinetics of planar rigid bodies 77 Case 1: Point is fixed or coincident with the center of mass C We can first replace v using the velocity transfer formula w.r.t. the point v = v + ω ϱ, where ϱ = r r, and then insert it into the general formula for computing the angular momentum w.r.t., H = ϱ (v + ω ϱ ) dm = ϱ dm v + ϱ (ω ϱ ) dm. Since either v = 0 or C the first term of above equation vanishes, i.e. ϱ dm v = 0. Remember how we defined C: The center of mass C is the point for which the mass moment ϱ C dm is equivalent to zero ϱ C dm = 0. Therefore, we only need to evaluate the second term H = ϱ (ω ϱ ) dm, which we do as follows: H = (ϱ x i + ϱ y (ω j) k ( ϱ x i + ϱ y j)) dm, = (ϱ x i + ϱ y (ω j) ( ϱ x j ϱ y i)) dm, = ω ( (ϱ x ) 2 + (ϱ y )2) dm k. (IV.7) We now define the above integral to be the mass moment of inertia w.r.t. the point. Definition. The mass moment of inertia w.r.t. a point at r for a planar rigid body is defined as 2 ( I := ϱ dm = (ϱ x ) 2 + (ϱ y )2) dm (IV.8) where ϱ = r r.

86 78 IV.3. Moment of inertia and angular momentum transfer formula Note: If C then we also call I the centroidal moment of inertia. With (IV.8) the angular momentum (IV.7) becomes H = I ω k. The angular momentum of a planar rigid body w.r.t. a fixed point, respectively C H = I ω k (IV.9) H... angular momentum of w.r.t. I.... mass moment of inertia of w.r.t. ω..... angular velocity of Case 2: Point is a general point We now go back to the general case and we will find a relation between the angular momenta H, H A for two general points and A. This allows us to easily compute the angular momentum w.r.t. any desired point. We first consider the angular momentum H w.r.t. an arbitrary point at r, i.e. H = ϱ v dm, and replace ϱ = r C + ϱ C as well as v with the use of the velocity transfer formula, v = v C + ω ϱ C. C denotes the center of mass at r C, r C = r C r and ϱ C = r r C. From that we get H = (r C + ϱ C ) (v C + ω ϱ C ) dm, = r C v C dm + r C ω ϱ C dm + ϱ C v dm. (IV.10) Note that we did not replace v in the last term. From above equation we can simplify a few things. The first integral P = v C dm

87 Chapter IV. Kinetics of planar rigid bodies 79 is the linear momentum of. The second integral is equal to 0 by definition of the center of mass as discussed before, ϱ C dm = 0, and the last integral is actually nothing else than the angular momentum w.r.t. C H C = ϱ C v dm. We already saw how to compute this. Therefore (IV.10) becomes H = H C + P r C. (IV.11) We can do the same for some other point A at r A and compute the angular momentum w.r.t. A, H A = H C + P r CA with r CA = r A r C. When we subtract (IV.12) from (IV.11) we get or H H A = P (r C r CA ) H = H A + P r A (IV.12) where r A = r C r CA = r r A. This is called the angular momentum transfer formula with which we can relate the angular momenta of some planar rigid body w.r.t. two arbitrary points A, with each other. The angular momentum transfer formula for a planar rigid body H = H A + P r A (IV.13) H H A... angular momentum of w.r.t. some arbitrary point... angular momentum of w.r.t. some arbitrary point A P..... linear momentum of r A... vector pointing from point A to point Practical computation of the moment of inertia For common bodies we can find the moment of inertia tabulated for a uniform mass distribution ρ = const. and about the center of mass, see e.g. Figure IV.3. Note that the moment of inertia is defined w.r.t. some point as well as some axis, which is in our case of planar rigid bodies the axis k, i.e. the axis perpendicular to the planar body. What is also most useful is the fact that integration is a linear operation and we can apply the superposition principle for moments of inertia w.r.t. to the same point.

88 80 IV.3. Moment of inertia and angular momentum transfer formula Figure IV.3: The above table summarizes the moments of inertia for the most common 2D rigid bodies. Superposition of moments of inertia One can compose the moment of inertia I of a planar rigid body consisting of several bodies i, i = 1,..., n with moments of inertia I i about some common point by the superposition principle, i.e. I = n i=1 I i (IV.14) Example IV.1: Moment of inertia of a Mercedes logo Consider the Mercedes logo consisting of a ring and 3 equally lengthened bars depicted in Figure IV.4. We know the moments of inertia I bar about the center for a bar and I ring for the ring, also about. We now want to know what the moment of inertia I logo for the whole structure is. To obtain I logo we can simply use the superposition principle to get I logo = 3I bar + I ring.

89 Chapter IV. Kinetics of planar rigid bodies 81 Figure IV.4: The moment of inertia of the Mercedes logo may be partitioned into the moment of inertia of bars and a ring. Example IV.2: Moment of inertia for a hollow cylinder Consider a hollow cylinder of mass M, inner radius R 1 and outer radius R 2. What is the moment of inertia I about its center? Hint: The moment of inertia of a cylinder of radius R i and mass M i about its center is I i = 1 2 M iri 2. We use the superposition principle by subtracting the moment of inertia of a cylinder of radius R 1 from the moment of inertia from a cylinder of radius R 2, I = 1 2 M 2R M 1R 2 1. We now say that M = M 2 M 1 and after some simplifications we find I to be I = 1 2 M(R2 1 + R 2 2). Finally we can also relate the moment of inertia of an arbitrary point A with the centroidal moment of inertia via Steiner s theorem. We will not derive the theorem but only state the result, see below and Figure IV.5. Parallel axis theorem (Steiner s theorem) I A = I C + M r CA 2 (IV.15) I A..... moment of inertia about some point A I C..... moment of inertia about the center of mass C M..... mass of the body r CA... distance between center of mass C and point A Note: Steiner s theorem is not valid for any two arbitrary points! I C specifically denotes the moment of inertia about the center of mass and cannot be replaced with any other moment of inertia. When one wants to know the moment of inertia I A about some arbitrary point A and knows I about Steiner s theorem must be applied twice. Note: From Steiner s theorem it follows that the moment of inertia is minimal w.r.t. the center of mass.

90 82 IV.4. Work-energy principle Figure IV.5: Steiner s theorem may be used to relate the moment of inertia of the same rigid body about the center of mass C and any point A. IV.4 Work-energy principle Figure IV.6: The rigid body is divided into small areas to derive the kinetic energy and the work-energy principle. To come up with the work-energy principle for a 2D rigid body, we first define its kinetic energy. We divide the planar rigid body of mass M into small areas of mass m i at r i, each with velocity v i, see Figure IV.6. When the partitions become sufficiently small, they can be regarded as particles. We can then use the formula for kinetic energy of a particle system and then take the limit m i 0, N to obtain T = 1 2 lim N m i 0 N m i v i 2 = 1 2 i=1 v 2 dm. (IV.16) This is the general form to compute the kinetic energy of a planar rigid body. However, there are two ways to simplify the equation above. We will derive one case, where we relate the general velocity v to the velocity of the center of mass v C via the velocity transfer formula. With ϱ C = r r C and the angular velocity ω the velocity transfer formula states v = v C + ω ϱ C.

91 Chapter IV. Kinetics of planar rigid bodies 83 Inserting it into (IV.16) we obtain T = 1 (v 2 C + ω ϱ C ) (v C + ω ϱ C ) dm, = 1 2 v C 2 dm + v C (ω ϱ C ) dm ω ϱ C 2 dm. (IV.17) We now evaluate each integral individually. The first integral yields 1 v 2 C 2 dm = 1 2 v C 2 dm = 1 2 v C 2 M since dm = M. We find the second integral to be v C (ω ϱ C ) dm = v C ω ϱ C dm = 0 per definition of the center of mass C, i.e. ϱ C dm = 0. The third term evaluates to ω ϱ 2 C dm = 2 ω2 = 1 2 ω2 = 1 2 ω2 k ϱ C 2 dm k 2 ϱc 2 sin 2 π 2 dm ϱ C 2 dm. Above we used that ω = ω k ϱ C and that k = 1. In the integral we find the definition of the centroidal moment of inertia, i.e. 2 I C = ϱ C dm, therefore the last integral becomes ω ϱ 2 C dm = 2 ω2 I C.

92 84 IV.4. Work-energy principle We can insert all these results into (IV.17) and obtain T = 1 2 M v C I Cω 2. We see that the kinetic energy of a 2D rigid body consists of two parts, a translational part 1 2 M v C 2 and a rotational part 1 2 I Cω 2. The second way to obtain a simpler equation from the general result (IV.16) is to use the velocity transfer formula to relate any velocity v with the instantaneous center of rotation for which v 0 and the velocity transfer formula then states v = ω ϱ where ϱ = r r. With this choice we get T = 1 2 I ω 2 where I denotes the moment of inertia about the instantaneous center of rotation. The derivation works very similarly. The kinetic energy T of a planar rigid body T = 1 2 M v C I Cω 2 (IV.18) or T = 1 2 I ω 2 (IV.19) M... mass of v C... velocity of the center of mass I C... centroidal moment of inertia ω.... angular velocity of I... moment of inertia about the center of rotation (!) To derive the work-energy principle we look at a rigid-body system of particles and then approximate the planar rigid body in the same manner as before, i.e. divide it into N areas of m i and then take the limit m i 0, N. The work-energy principle for a 2D rigid body is then stated below. Work-energy principle for a planar rigid body W ext 12 = T 2 T 1 (IV.20) W ext work done by the external forces between states 1 and 2 T i..... kinetic energy of the system associated with state i

93 Chapter IV. Kinetics of planar rigid bodies 85 Note: As discussed before, the hypothesis of rigid body greatly simplifies the work-energy principle. If the body is not rigid, the internal forces do work and the integrals become more complicated. As for other types of systems we can define the potential energy and the work-energy principle for conservative systems. The derivation are analogous to the previous chapters and we will simply state the results. Definition. The potential energy V (r) of a planar rigid body is defined as the sum of the potentials V k of all (external) potential forces F k and moments M k acting on the system, k = 1,..., K, K V (r) := V k. k=1 (IV.21) Work-energy principle for a conservative system The total mechanical energy E := T + V of a conservative system is conserved along all motions, T 1 + V 1 = T 2 + V 2 (IV.22) or equivalently, d dt E(t) = 0. E... total mechanical energy T i... kinetic energy of the planar rigid body at state i V i... potential energy of the planar rigid body at state i IV.5 Degrees of freedom In Section II.3 we discussed the degrees of freedom for a system of particles. Recall that the degrees of freedom for a system are defined as the number of independent generalized coordinates that are necessary to describe the complete motion of the system. For one particle, in general, we need two coordinates to describe its motion in 2D and when we have holonomic constraints on the system this number reduces, see Section II.2 on the definition of a holonomic constraint. To derive the degrees of freedom for a planar rigid body we look at a rigid-body system of N particles in 2D. We then compute the DOF for the system and finally take the limit N. Recall that a rigid-body system of particles was defined as a system of particles where any

94 86 IV.5. Degrees of freedom (a) (b) Figure IV.7: Determination of the degrees of freedom for a planar rigid body. (a) The procedure consists of connecting one particle to the others with enough rigid links to make there is no relative motion between the particles. (b) The most common choice for the DOF of a unconstrained planar rigid body is the position of the center of mass (x C, y C ) and the orientation ϕ of the body. pair of particles are at a constant distance from each other, see Figure IV.7 (a). This can be described by the holonomic constraint r i (t) r j (t) = const i, j, that we already introduced when speaking of a massless rod. Let us now imagine to build a rigid body by adding one particle at a time and connect it with the others with enough rigid links such that any relative motion is prevented. We need therefore one link to connect the second particle to the first, two additional ones to connect the third particle, etc... This can be summarized in the below table: N = 1 : DOF = 2 N = 2 : DOF = = 3, N = 3 : DOF = = 3, N = 4 : DOF = = 3, N = 5 : DOF = = 3, This procedure shows that: DOF = 3. for an unconstrained planar rigid body. This result was expected since we know from our previous experience that we can describe, amongst other ways, the motion of a planar rigid body with the position (x C, y C ) of the center of mass and some rotation angle ϕ, see Figure IV.7 (b). Note: We could also choose three independent position coordinates to describe the motion but we could not use three angles as they are never all independent from each other. Now for L unconstrained planar rigid bodies the degrees of freedom are DOF = 3L

95 Chapter IV. Kinetics of planar rigid bodies 87 and for L planar rigid bodies that are subjected to K independent holonomic constraints f j (r 1 (t),..., r L (t), t) = 0, j = 1,..., K we obtain DOF = 3L K for the degrees of freedom. The degrees of freedom for a system of planar rigid bodies DOF = 3L K (IV.23) DOF... degrees of freedom L number of planar rigid bodies in the system K..... number of independent holonomic constraints on the system Note: For a system of DOF degrees of freedom we need DOF independent 2 nd order ODEs to fully describe the motion of the system. This set of equation is called the equations of motion of the system. Example IV.3: Possible holonomic constraints for a planar rigid body We have already seen holonomic constraints before in our discussion of particle systems, see Example II.1. Recall that a holonomic constraint is a scalar equation that only depends on the current position of the considered system and time. We show here some other very typical and often used holonomic constraints. (a) (b) Figure IV.8: Typical holonomic constraints. (a) A slider whose motion is constrained in the vertial direction and in the rotation. (b) Rolling without slipping constraint.

96 88 IV.5. Degrees of freedom (a ) Slider Consider a planar rigid body that can slide between two bars, see Figure IV.8 (a). The body cannot move up and down in y-direction nor can the body be rotated. This corresponds to two independent holonomic constraints, i.e. and y C const. = 0 ϕ = 0, where C denotes the center of mass. Therefore, the degrees of freedom reduce to DOF = = 1. This result was also intuitively visible since the slider can only move in x-direction but no other motion is possible. (b ) Rolling without slipping in 2D A disk of radius R rolls without slipping along i as shown in Figure IV.8 (b). We find the motion to be constrained by the following equations. First, the center of mass C cannot move in j-direction, or ẏ C = 0 y C (t) R = 0 where y C denotes the position of the center of mass in j-direction. The second constraint states that the contact point A is the instantaneous center of rotation and therefore v A = 0. Via the velocity transfer formula v C = ω r AC we can find the velocity of the center of mass in i-direction, i.e. or ẋ C = ϕr ẋ C + ϕr = 0. Unlike the first constraint this is a non-holonomic constraint but we can actually integrate it to x C + ϕr = const. We call this type of constraints integrable non-holonomic constraints since they are integrable and only depend on the current position and time as for a true holonomic constraint. We count them as holonomic constraints since such constraints also reduce the degrees of freedom. We then have two constraints and the degrees of freedom become DOF = = 1

97 Chapter IV. Kinetics of planar rigid bodies 89 Example IV.4: Possible non-holonomic constraints for planar rigid bodies We will now state three examples for purely non-holonomic constraints that are not integrable and therefore do not reduce the degrees of freedom. In other words, non-holonomic constraints do not prevent a certain configuration to be attained but dictate how the system could evolve to reach such configuration. (a) (b) (c) Figure IV.9: Examples of non-holonomic constraints are shown. (a) The constraint that governs the steering of a car is depicted here. (b) An airplane pursuing another airplane. (c) The horizontal motion of the slider is imposed. (a ) Steering/propulsion constraint We consider a distinguished point A in the line of symmetry of a body, see Figure IV.9 (a). This constraint states that the velocity of A must also fall on this line of symmetry. The fact that a car does not skid sideways is represented by this constraint. The point A is the middle point of the rear axis. As stated before the velocity v A = ẋ A i + ẏ A j of A must point "forward", i.e. along the axis of symmetry. We further denote e as the unit vector aligned with the rear axis, e = sin ϕ i + cos ϕ j, where ϕ denotes the angle between i and the axis of symmetry. The steering constraint states that v A e = 0, since both vectors must be orthogonal to each other. From that we obtain (ẋ A i + ẏ A j) ( sin ϕ i + cos ϕ j) = 0, ẋ A ẏa(t) tan ϕ(t) = 0 This constraint is non-holonomic and not integrable as a function of the current position only. This is actually a good result since we want to go with a car wherever

98 90 IV.5. Degrees of freedom we want: A holonomic constraint would define a region where we are not allowed to go. Think about parking your car. You cannot drive into the parking spot whatever way you want to. This is exactly what a non-holonomic constraint does: telling us the way to do it but not forbid it. (b ) Pursuit constraint An airplane is moving with velocity v T in positive i-direction. A second airplane with velocity v C = ẋ C i + ẏ C j tries to chase the first airplane as shown in Figure IV.9 (b). It does so by always flying towards the current direction in which it sees the first airplane. The position of the first airplane is r T = v t t i + y t j and the position of the chasing airplane is r C = x C i + y C j. We denote the angle between the directions in which the airplanes are moving as ϕ and use the following equation to describe the constraint on the motion of the chasing airplane, tan ϕ = ẏc = y C y T, ẋ C v T t x C ( vt t x C (t) ) ẏ C (t) + ( y C (t) y T )ẋc (t) = 0 Again this is a purely non-holonomic constraint. (c ) Slider with imposed motion We look at the same slider as in the previous example but this time the motion ẋ C in x-direction is imposed to be ẋ C = v 0, see Figure IV.9 (c). We have already seen that the "normal" slider is subjected to two constraints, i.e. y C const. = 0 and ϕ C = 0. We now imposed a third constraint ẋ C = v 0. When we integrate it we get x C (t) = x c (t 0 ) + v 0 (t t 0 ), x c (t) v 0 t x c (t 0 ) + v 0 t 0 = 0 From above equation we can see that the constraint is integrable but depends on the previous position of the slider and therefore strictly speaking does not reduce the DOF. If we still count it as holonomic constraint we get DOF = 3 3 = 0. Another possibility is to count it as half a constraint: DOF = = 1 2.

99 Chapter IV. Kinetics of planar rigid bodies 91 Note: From above examples we can see that counting DOF in the presence of non-holonomic constraint can be very subtle. We will not pursue this topic further, though. IV.6 Example problems In the remaining section of this chapter we will apply the derived principles for a planar rigid body to concrete problems. Example IV.5: Falling stick A stick of mass m, length L and center of mass C at r C is falling to the ground due to gravity, see Figure IV.10 (a). The point denotes the contact point with the ground. We will assume that the stick is thin and therefore remains in touch with the ground independent of the angle ϕ, y = 0. Further, we assume a smooth surface and therefore there is no friction between the ground and the stick. Note that the stick can slide along the surface. (a ) Draw a free-body diagram and decide on the necessary generalized coordinates to describe the problem. (b ) Find the equation of motion by applying the appropriate principles. (a) (b) Figure IV.10: A falling stick of mass m and length L on a smooth surface. (a) The necessary quantities are defined. (b) The free-body diagram of the system is depicted. Note that there is no friction force since the surface is assumed to be smooth. (a ) FD and generalized coordinates From the free-body diagram in Figure IV.10 (b) we see that the only forces acting on the body are the normal force N attacking at and the gravitational force mg. Per assumption there is no friction force. The system is subjected to one constraint, i.e. y = 0

100 92 IV.6. Example problems and DOF = 2. We choose the angle ϕ and the horizontal position x C of the center of mass as the generalized coordinates to describe the system. (b ) Equation of motion i Linear momentum principle The linear momentum principle for the system states d dt P = ma C = mg + N. For the x-component of (IV.24) we get (IV.24) mẍ C = 0 since there are no forces acting in x-direction. Integrating this twice yields x C (t) = x C (0) + v C (0)t. (IV.25) When evaluating (IV.24) in y-direction we see that this includes the unknown force N. Instead we can apply the angular momentum principle about, which does not include N. ii Angular momentum principle about When using the angular momentum principle w.r.t. we avoid having the unknown force N in our equations, see Figure IV.10. The angular momentum principle about states Ḣ + v P = M. (IV.26) In the following we will compute all necessary terms. We start with the linear momentum P = mṙ C. From the geometry we obtain that r C = x C i + l 2 sin ϕ j, which we then differentiate in time, ṙ C = ẋ C i + l 2 ϕ cos ϕ j, and insert into the linear momentum, P = m(ẋ C i + l 2 ϕ cos ϕ j). We can also find r from the geometry and then differentiate in time to get v : v = d dt (x C l 2 cos ϕ) i = (ẋ C + l 2 ϕ sin ϕ) i. To compute H we first compute H C = I C ϕ k

101 Chapter IV. Kinetics of planar rigid bodies 93 and then use the angular momentum transfer formula We obtain: H = H C + P r C. H = I C ϕ k + m(ẋ C i + l 2 ϕ cos ϕ j) ( l 2 cos ϕ i l sin ϕ j), ( 2 ( = I C + m l2 ) 4 cos2 ϕ ϕ m l ) sin ϕ k. 2ẋC We now need to differentiate in time, ( ( Ḣ = I C + m l2 ) 4 cos2 ϕ 1 4 ml2 sin(2ϕ) ϕ 2 m l sin ϕ m l ) ϕ cos ϕ, ( 2ẍC 2ẋC ( = I C + m l2 ) 4 cos2 ϕ 1 4 ml2 sin(2ϕ) ϕ 2 m l ) ϕ, cos ϕ, 2ẋC where we used that ẍ C = 0, which we know from the linear momentum principle (IV.24). We also need to compute the vector product (v P ): v P = m(ẋ C + l 2 ϕ sin ϕ) i (ẋ C i + l 2 ϕ cos ϕ j), = m l 2 (ẋ C + l 2 ϕ sin ϕ) ϕ cos ϕ k. The torque w.r.t. is M = mg l 2 cos ϕ k. We can now insert all the terms into (IV.26) and obtain the equation of motion (I C + m l2 4 cos2 ϕ) ϕ 1 8 ml2 ϕ 2 sin(2ϕ) + 1 mgl cos ϕ = 0. 2 We have not computed the centroidal moment of inertia yet I C. To compute it we use 2 I C = ϱ C dm with dm = ρ dx dy = m w x l dx dy. w x is the width of the rod and l the length, see Figure IV.11. Since we assumed the rod to be very thin (w x l) we can say that ϱ C = y 2, where y denotes the distance away from the center of mass in the direction of l. So we obtain for I C I C = l 2 l 2 wx 2 wx 2 y 2 m dx dy. w x l

102 94 IV.6. Example problems Figure IV.11: The stick is assumed to be thin, i.e. w x l. We can then use the fact that integration about y is symmetric to simplify the integral to l 2 I C = 2 0 wx 2 wx 2 y 2 m w x l dx dy. Finally we can evaluate the integral: (( I C = 2m l 3 ) 2) w x l w x 0 = ml2. We could have also found the result in tables. Eventually, we plug I C into our equation of motion to get the final result, ( ) ml2 3 + cos2 ϕ ϕ 1 8 ml2 ϕ 2 sin(2ϕ) + 1 mgl cos ϕ = 0 2 (IV.27) Together with (IV.25) these is our set of 2 nd order ODEs, which describe the motion of the system. (IV.25) was easily solved but (IV.27) is highly nonlinear and can only be solved numerically. Note: The two equations of motion are fully decoupled from each other, i.e. there is one separate equation for each degree of freedom, ϕ and x C. This is usually not the case. Example IV.6: Falling hinged stick We again consider a falling (thin) stick of mass m and length l, see Figure IV.12 (a). ut this time the contact point is hinged and cannot slide. This system only has one degree of freedom and we choose the angle ϕ as the generalized coordinate. (a ) Which is the equation of motion for the system? (b ) Find the acceleration of the point A at the top of the stick. Then take its vertical component for the moment the stick is released from rest and compare it to the gravitational acceleration g. Discuss the result.

103 Chapter IV. Kinetics of planar rigid bodies 95 (a) (b) Figure IV.12: A falling hinged stick of mass m and length L is considered. (a) The stick is hinged at point and subjected to gravity. (b) The forces acting on the body are shown in the free-body diagram. (a ) Equation of motion To find the equation of motion we apply the angular momentum principle about the hinge point, Ḣ + v P = M. Since the reaction forces of the hinge do not produce a torque about we only have to include the torque coming from the gravitational forces, M = mg l 2 cos ϕ k. For this problem we can compute the angular momentum directly because the point is fixed: H = I ϕ k with I = 1 3 ml2 from tables. Also note that v = 0 and therefore (v P = 0). Inserting into the AMP we find the equation of motion: 1 3 ml2 ϕ = mg l 2 cos ϕ ϕ g l cos ϕ = 0 (IV.28) (b ) Acceleration of point A

104 96 IV.6. Example problems To find the acceleration a A of the top point A we differentiate the vector r A twice in time, i.e. a A = d2 dt 2 r A = d2 (l cos ϕ i + l sin ϕ i), dt2 = l d ( ϕ sin ϕ i + ϕ cos ϕ j), dt = l( ϕ 2 cos ϕ ϕ sin ϕ) i + l( ϕ 2 sin ϕ + ϕ cos ϕ) j. We now take the vertical component a vert = a A j of the acceleration, a A j = l( ϕ 2 sin ϕ + ϕ cos ϕ), and replace ϕ with (IV.28), a A j = 3 2 g cos2 ϕ l ϕ 2 sin ϕ. Next, we want to know what the vertical acceleration of point A is in the moment we release the stick from rest, i.e. ϕ(0) = 0, a vert (0) = 3 2 g cos2 ϕ. Noticeable, a vert (0) > g when cos 2 ϕ > 2 3, i.e. for ϕ > 35. The end of the stick has an acceleration larger than the gravitational acceleration. How can this be explained? In the hinge at there are reaction forces in both x and y-direction, see Figure IV.12 (b). These forces rotate the rigid body and add an extra component to the vertical acceleration of point A. Example IV.7: Sweet spot of baseball bats (center of percussion) An experienced baseball player is able to hit to ball on a distinguished spot of the bat, called the center of percussion. When the ball hits the bat at this spot the player does not feel an impact force in direction parallel to the motion of the ball. The same is also true for racket sports like tennis or sword fighting. How can we find this sweet spot of a bat? (a ) Modelling and free-body diagram For our computation we will model the situation when a player hits the ball as follows: The bat is assumed to be a 2D planar rigid body and the wrist of the player holding the bat is modeled as a smooth joint fixed in point, see Figure IV.13. The end of the bat is hinged in this point. The distance between the center of mass C and the point is r C. The unknown sweet spot is denoted by P and the distance between and P is r P. We will model the ball hitting the bat as an impulsive force F acting over the time interval [t, t + ], t + t. F acts at the sweet spot P. The reaction force in is denoted by R = R x i + R y j.

105 Chapter IV. Kinetics of planar rigid bodies 97 Figure IV.13: The baseball bat is modeled as a planar rigid body. The center of percussion is denoted as P and the respective forces that are acting on the body are shown as well. Note that F is impulsive. (b ) Rotation of bat after impact First, we need to find the angular velocity ω + after the impact. We will apply the angular momentum principle w.r.t. the point, Ḣ = M. (IV.29) Note that v = 0 and therefore the term v P disappears. The only force producing a torque about is the impact force F, i.e. M = F r P k = F r P k. Next, we integrate (IV.29) over the interval [t, t + ]: H (t + ) H (t ) = t + t M dt = t + t F dt r P k. (IV.30) Although the time interval is infinitesimally small, t + t, the integral of F does not vanish because F is impulsive, i.e. not bounded. We denote the integral I F = t + t F dt. Since is fixed we can compute the angular momentum using the moment of inertia I about, H = I ω k.

106 98 IV.6. Example problems Also, we assume that the angular velocity ω before the impact is equal to zero, ω = 0, and hence H (t ) = 0. The angular momentum H (t + ) is equal to H (t + ) = I ω + k. Plugging everything into (IV.30) we obtain I ω + k = I F r P k, which can be solved for ω + : ω + = I F r P I (IV.31) (c ) Sweet spot We defined our sweet spot as the point of impact for which R x = 0. To find this spot we need to relate the reaction forces R at with r P so we can find at what distance r P from the sweet spot lies. For that we apply the angular momentum principle w.r.t. the center of mass C and again integrate over the interval [t, t + ], H C (t + ) H C (t ) = t + We find the torque M C about C to be M C = F (r P r C ) k + R x r C k. t M C dt. (IV.32) Note that in general both forces F and R x are impulsive and therefore we cannot neglect them, although we are integrating over an infinitesimally small interval. However, we are interested in finding the sweet spot for which R x = 0. So we get t + t M C dt = t + t F (r P r C ) k dt = I F (r P r C ) k. The angular momentum before the impact is zero, H C (t ) since ω = 0 and the angular momentum after the impact can be computed using the centroidal moment of inertia, H C (t + ) = I C ω + k = I C I F r P I k. We here used (IV.31) to replace ω +. Plugging everything into (IV.32) yields I C I F r P I k = I F (r P r C ) k, which we solve for r P, r P = I I I C r C. Using the parallel axis theorem we relate I and I C to each other, I = I C + mr 2 C.

107 Chapter IV. Kinetics of planar rigid bodies 99 Finally we obtain r P = I m r C (IV.33) This is the location of the sweet spot or also called center of percussion in mechanical terminology. When a rigid body is hit at this point, no parallel reaction force is experienced at the fixed axis of rotation. Thin rod When we assume the bat is a thin rod with I = 1 3 ml2 and r C = 1 2L we get r P = 2 3 L. Hammer A hammer, modeled as point mass of mass m at L, with I = ml 2 and r C = L has its center of percussion at r P = L. Example IV.8: Chimney demolition Consider an old chimney that is scheduled to be demolished. Explosives in the bottom part of the chimney are causing it to bend over and fall to the ground, see Figure IV.14. During the fall the structure experiences positive normal stress ("pulling") that causes the structure to break. In the following, we are interested in finding the location of the point with the largest stress, i.e. the location of the largest bending moment. This will be the point where the chimney is first going to break. (a ) Modelling and free-body diagram We will model the falling chimney as a falling hinged stick (rigid body), see Example (IV.6), of mass m and length L with constant cross section. This assumption is valid until the first break in the chimney but that is sufficient for our case since we are specifically interested in the location of the first break. The system then has one degree of freedom and we choose the angle θ between the vertical and the actual position of the chimney as our generalized coordinate. At the smooth joint the tangential force L and the normal force N are acting, see Figure IV.15. Further, we denote the center of mass with C. Gravity acts upon the chimney (otherwise there is no fall).

108 100 IV.6. Example problems Figure IV.14: A chimney is being demolished by initiating an explosion. Figure IV.15: The chimney is modeled as a falling hinged stick that is hinged in a smooth joint at. The free-body diagram is shown above.

109 Chapter IV. Kinetics of planar rigid bodies 101 (b ) Equation of motion We find the equation of motion to be the same as the equation of motion (IV.28) for Example IV.6. We only need to replace ϕ = π 2 θ, to obtain θ 3 g 2 l sin θ = 0 (IV.34) (c ) Normal reaction force N We will first compute the normal reaction N acting at normal to the chimney. When applying the angular momentum principle about the center of mass C the only force producing a torque is N. This makes it very convenient for us to compute N. The AMP about C states Ḣ C = M C (IV.35) with and H C = I C θ k M C = N l 2 k. The centroidal moment of inertia is I C = 1 12 ml2. y inserting this values into (IV.35) we obtain I C θ k = N l 2 k. We next replace θ with (IV.34) and solve for N : N = 2I C l θ = 1 mg sin θ. 4 (IV.36) (d ) Distribution of bending moment T A (x) To find the bending moment at any location x of the chimney, we first isolate a subchimney of varying length x, see Figure IV.16. Since we cut free a part of the chimney we have to introduce the forces N A (x) and L A (x) as well as the desired bending moment T A (x), which are exerted on the portion of the chimney we are considering by the rest of it. The cutting point is denoted by A. Since we are only interested in T A we apply the angular momentum principle about A to avoid involving the unknown forces N A, L A, L : Ḣ A + v A P = M A. (IV.37) Since we rotate around, v A v C and therefore v A P = 0. We find H A by applying the angular momentum transfer formula between A and C: H A (x) = H C (x) + P (x) r CA (x) (IV.38)

110 102 IV.6. Example problems (a) (b) Figure IV.16: We are interested in finding the bending moment of the chimney. (a) A subchimney of length x is cut free. (b) There is bending moment T A acting on the subchimney when it is cut free. Note that since we are actually regarding a portion of the chimney of length x all quantities above are dependent on the actual length x. The partial mass m x can be expressed as m x = x l m. Similarly, we find 1 H C = I C θ k = 12 m xx 2 1 θ k = 12 mx3 θ k l and P r CA = m x v C r CA k = ( m x l ) ( x 2 θ) x 2 k. Inserting all the terms into (IV.38) we get H A = 1 m 6 l x3 θ k. and Ḣ A = 1 m 6 l x3 θ k, by taking the time derivative. We replace θ using the equation of motion (IV.34) and get Ḣ A = 1 mg 4 l 2 x3 sin θ. For the moment about A we obtain ( ( M A = T A N x + m x ) g x ) l 2 sin θ k

111 Chapter IV. Kinetics of planar rigid bodies 103 Combining all above results together with (IV.36) and put it into (IV.37) we get 1 mg ( 4 l 2 x3 sin θ = T A N x + m x ) g x l 2 sin θ and we can then solve it for T A (x): T A (x) = mg 4l 2 sin θx(x l)2. (IV.39) We can see that for x [0, l] T A < 0, since for the considered interval θ [0, π 2 ], sin θ > 0, see Figure IV.17. This result is also visible from the acceleration θ of the chimney: θ = 3 g 2 l sin θ. ecause θ is inversely proportional to the length of the chimney a longer beam accelerates slower than a shorter one. In other words, the lower portion of the chimney on its own would be rotating faster than the whole chimney but is kept back by the upper part resulting in a negative bending moment. You can also see this in Figure IV.14 where the lower part of the chimney starts rotating more quickly after the first break. Since T A (0) = T A (L) = 0 and in between T A < 0, there must be a minimum for x [0, l]. We find it by differentiating w.r.t. x, d dx T A(x) = mg sin θ(x l)(3x l), 4l2 and set the derivative to zero. This gives us the location of the maximum bending moment (maximum absolute value), i.e. x = l 3 (IV.40) This is the location at which the chimney is going to break first. Note that in practice when then chimney has a non-constant cross section the location varies. Figure IV.17: The distribution of the bending moment is qualitatively illustrated. The minimum is at x = l 3.

112 104 IV.6. Example problems Example IV.9: Dynamics of a towed wheel Consider a wheel that is towed with constant velocity u = const. as illustrated in Figure IV.18. The wheel is linked to an inclined arm in the geometric center of the wheel. The arm is hinged in the pivot axis at. The overall center of mass C is located at a horizontal distance a away from. The distance between the center of mass and the point of contact G is b. We want to discuss the stability of such a wheel when it is towed but also when it is pushed. (a ) Model the given situation in 2D, find the degrees of freedom and then draw an appropriate free-body diagram. (b ) Find the equation of motion. (c ) Discuss the stability of the dynamics. First, consider the special case where the arm is not inclined (a=b=0). For the general case, linearize the equation of motion and then discuss the stability based on the eigenvalues of the problem. (a) (b) Figure IV.18: A wheel towed with constant velocity u. (a) Side view of the wheel with relevant geometric quantities. (b) The top view illustrates in which direction the wheel is towed. We will model the system as 2D from this point of view. (a ) Modelling and free-body diagram i Two-dimensional model The two-dimensional model we use corresponds to the top view of the three-dimensional model, see Figure IV.18 (b). We will therefore not consider the kinematics of the wheel but simply of G and C. The constant towing velocity u is assumed to be in horizontal direction, i.e. negative i direction. We denote the angle between the towing direction and

113 Chapter IV. Kinetics of planar rigid bodies 105 the orientation of the wheel as θ. ii Degrees of freedom Since we have one rigid body the unconstrained system has three degrees of freedom. Now is not allowed to move in j direction, v j = 0. This corresponds to a non-holonomic constraint that can be integrated to a holonomic one. The towing velocity u is imposed, which is also non-holonomic constraint. However, this one when integrated still depends on the initial position, i.e. x (t) = x (t 0 ) + u(t t 0 ). From our discussion about non-holonomic constraints in Example IV.4 we know this is a debatable case. In this case we regard this constraint as holonomic and therefore the degrees of freedom are reduced to one, DOF = 3 2 = 1. We now choose the angle θ as the generalized coordinate to describe the system. iii Free-body diagram The forces acting on the body are illustrated in Figure IV.19 (a). The towing force F T acts at and the rolling friction force F r acts in the direction of the wheel. Since the wheel does not purely roll in the direction where it points, there is also a third force present, i.e. the lateral friction force F L resisting the lateral sliding. Note that this force only exists when a pneumatic tire is put on the wheel, otherwise the wheel can laterally slide without resistance. Further, we will use the unit vectors e = cos θ i sin θ j pointing in the direction of the wheel and n = sin θ i + cos θ j pointing in the direction perpendicular to the wheel. (a) (b) Figure IV.19: The free-body diagram of the system is depicted here (a) as well as the model for the lateral friction force (b).

114 106 IV.6. Example problems iv Analysis of the lateral friction force F L Let us take a closer at the motion of the wheel and discuss how we can describe the lateral friction force F L, see Figure IV.19 (b). The velocity v G of the wheel consists of a rolling component v roll in the direction of the wheel and one lateral component v lat normal to it, i.e. v G = v roll e + v lat n. The slip angle α is the angle between the rolling direction and the actual direction of motion, tan α = v roll v lat. Experiments have shown that the magnitude F L of the lateral friction force can be described as the product between some cornering coefficient C α and the slip angle α, F L = C α α (IV.41) Note that without lateral friction steering a car would be impossible. We will now find a relation for α in terms of the geometry and the generalized coordinate θ to express the lateral friction force F L accordingly. For α we need to find v lat and v roll. Via the angular velocity transfer formula we can find v G from v = u, which is given, v G = v + ω r G, where the angular velocity is ω = θ k, the velocity of point is v = u i and the vector pointing from to G is r G = (a + b)(cos θ i + sin θ j). When we now project v G onto n, and e respectively, and insert the calculated terms we get the lateral velocity v lat and the rolling velocity v roll : v lat = v G n = u sin θ + θ(a + b), v roll = v G e = u cos θ. We can now express α by its definition as ( sin α = sin tan 1 v ) roll = v lat v lat v G = v lat. vlat 2 + v2 roll Plugging in all necessary expressions we obtain α = sin 1 u sin θ + θ(a + b). u 2 + 2u θ(a + b) sin θ + θ 2 (a + b) 2

115 Chapter IV. Kinetics of planar rigid bodies 107 With this we can express the lateral friction force as desired, F L = C α sin 1 u sin θ + θ(a + b) n. u 2 + 2u θ(a + b) sin θ + θ 2 (a + b) 2 (IV.42) (b ) Equation of motion For finding the equation of motion we apply the angular momentum principle w.r.t.. This choice ensures that we exclude the unknown forces, F T and F r since they do not produce a torque about. The AMP about states Ḣ + v P = M. (IV.43) Next, we need to find all necessary quantities to apply the AMP. We already know the velocity of, i.e. v = u i. Via the velocity transfer formula we also find the velocity v C of the center of mass, v C = v + ω r C, = u i + θ k (a cos θ i + a sin θ j), = (u + a θ sin θ) i + a θ cos θ j, the linear momentum P is then: ( P = mv C = m (u + a θ sin θ) i + a θ ) cos θ j. We can now compute the vector product v P using the above equations: v P = mua θ k. To compute the angular momentum about, we first compute the angular momentum H C about the center of mass C, then use the angular momentum transfer formula. The angular momentum about C can be computed with the centroidal moment of inertia I C, i.e. H C = I C θ k. The angular momentum about is given by ( H = H C + P r C = (I C + ma 2 ) θ ) + mua sin θ k with the vector r C = ( a cos θ k a sin θ j) pointing from C to. We also need to differentiate the angular momentum about in time, ( Ḣ = (I C + ma 2 ) θ + mua θ ) cos θ k.

116 108 IV.6. Example problems The last quantity that needs to be computed is the torque about, i.e. M = r G F L, = (a + b)(cos θ i + sin θ j) F L, = C α sin 1 u sin θ + θ(a + b) (a + b) k. u 2 + 2u θ(a + b) sin θ + θ 2 (a + b) 2 We now substitute all the terms into (IV.43) to obtain the equation of motion: (I C + ma 2 ) θ + C α sin 1 u sin θ + θ(a + b) (a + b) = 0 u 2 + 2u θ(a + b) sin θ + θ 2 (a + b) 2 (IV.44) This is a 2 nd order nonlinear ODE in θ for which we need the initial conditions θ(0), θ(0) to be able to solve it. The equilibrium for the towing is θ(t) 0. (c ) Stability i Non-inclined wheel In the case there is no inclination in the arm, i.e. a = b = 0 we find the following simplified equation of motion from (IV.44): θ = 0. We can directly integrate this twice with the initial conditions θ(0), θ(0) and obtain θ(t) = θ(0) + θ(t)t. From this result we see that for an initial perturbation from the equilibrium position, i.e. θ(0) 0, θ(0) 0, the wheel never reaches the equilibrium. The system becomes unstable. Therefore, only the inclined arm creates a possibility to stabilize the towing. We now ask ourselves for what specific configurations the system is stable. ii Stability in the general case To investigate the stability in the general case, we will linearize the equation of motion (IV.44) around the equilibrium θ = 0, which is the desired final configuration when towing. For that we Taylor-expand it around θ = 0, θ = 0 and only keep the linear terms. The result of the Taylor-expansion is (I C + ma 2 ) θ + C α ( (uθ + θ(a + b) + O(θ 3 ) )( 1 u 2 + O(θ2, θθ) )) (a + b) = 0 and after we drop the terms of higher order and set θ to zero we obtain (I C + ma 2 (a + b) 2 ) θ + C α θ + C α sign(u)(a + b)θ = 0 u (IV.45) which is the linearized 2 nd order ODE. Note that we substituted equation above. The general solution of this can be written as θ(t) = K 1 e λ 1t + K 2 e λ 2t u u 2 = sign(u) in the (IV.46)

117 Chapter IV. Kinetics of planar rigid bodies 109 for K j, λ j C, j = 1, 2. λ j denotes the jth eigenvalue of the linearized equation of motion or the jth distinct root of the characteristic polynomial of the equation, i.e. (I C + ma 2 )λ 2 (a + b) 2 + C α λ + C α sign(u)(a + b) = 0. u When we solve the characteristic polynomial we obtain (a + b) 2 C α ± C u 2 (a + b) 4 α u 2 4(I C + ma 2 )C α sign(u)(a + b) λ 1,2 = 2(I C + ma 2. (IV.47) ) To continue we need to differentiate two cases, the case when sign(u) > 0 and when sign(u) < 0. Case 1: sign(u) > 0, towing In this case the real part of the eigenvalues are negative, i.e. R(λ i ) < 0. We can now express the general solution (IV.46) such that we split λ j into its real and imaginary part: θ(t) = K 1 e λ1t + K 2 e λ2t, 2 = K j e R(λj)t e I(λj)t, = j=1 2 K j e R(λ j)t ( cos(i(λ j )t) + i sin(i(λ j )t) ). j=1 We can see that the term e R(λ j)t t 0 approaches 0 for the considered case. The equilibrium of the towed wheel is asymptotically stable. To design robust wheels we want the oscillations to die out fast to reach the equilibrium more quickly. The larger R(λ j ) is, the faster the rate of decay e R(λj)t, ( e R(λj)t (a + b) 2 ) t = exp C α 2 u (I C + ma 2. ) For faster decay the following things can be done: Select a tire with a large C α A larger inclination (a+b) of the wheel helps also. Make u smaller. Make the inertia term I C + ma 2 small. ut also be careful about I(λ j ) since it determines the frequency of the oscillation. We do not want to hit the resonance frequency of other parts of the vehicle. Case 2: sign(u) < 0, pushing

118 110 IV.6. Example problems In this case both eigenvalues are real, i.e. λ j R, j = 1, 2, whereas one of them is positive and one is negative, λ 1 < 0 < λ 2. The general solution then becomes θ(t) = K 1 e λ 1t + K 2 e λ 2t and the term with the positive eigenvalue λ 2 becomes unstable, i.e. e R(λ 2)t t. Note that this is all within our linearized observations that is only a valid assumption when we are close to the equilibrium around which we linearized. Of course for an unstable solution that approaches infinity this assumption is not given anymore. In reality, the wheel would turn 180 and be towed again, therefore returning to the stable solution. Think, e.g. about a cart in the supermarket.

119 Chapter V Kinematics of three dimensional rigid bodies In this chapter, we introduce the concept of a rigid body in three dimensions and discuss how we can describe the kinematics of such a body. Section V.1 repeats the basic concepts that should be known from Chapter III about the kinematics of planar rigid bodies. Afterwards, we introduce how rotations in three dimensions can be described in Section V.2. In Section V.3 we will determine the angular velocity of a body in three dimensions and find the velocity transfer formula for a rigid body. V.1 asic concepts Consider a three dimensional rigid body as sketched in Figure V.1 described in the inertial frame [ i, j, k ]. Remember that we defined a rigid body as a body for which the distance r A between any two points A, remains constant for all times, r A = const. A,, as in Section III.1. Figure V.1: A three dimensional rigid body. 111

120 112 V.1. asic concepts V.1.1 Degrees of freedom To determine the degrees of freedom for a three dimensional rigid body, we consider a triangle AC consisting of the three non-collinear points A,, C, as shown in Figure V.2. We observe that the position of is fully determined by these three points. Figure V.2: A three-dimensional rigid body. We now use the formula (II.8) to obtain DOF = dim n k, with which we can determine the degrees of freedom of a particle system. For our three points (n = 3) with k = 3 independent holonomic constraints (rigid links) we find the degrees of freedom for an unconstrained rigid body in three dimensions to be DOF = = 6. The degrees of freedom for a system of rigid bodies DOF = 6L k (V.1) DOF... degrees of freedom L number of three-dimensional rigid bodies in the system k number of independent holonomic constraints on the system V.1.2 Relative velocities We denote the position of two points A, by r A and r respectively, and the vector pointing from A to by r A = r r A (see Figure V.3 (a)). Let us write r as r = r A + r A. Differentiating in time gives d dt r = ṙ = ṙ A + ṙ A,

121 Chapter V. Kinematics of three dimensional rigid bodies 113 or equivalently, v = v A + ṙ A (V.2) From the definition r A r A = const. of a rigid body we know that ṙ A r A = 0. Therefore, the relative velocity ṙ A of w.r.t. A can be interpreted such that the point constantly rotates around A, generally with varying rotation axis and speed, at a constant distance r A from A. Alternatively, we can say that the point rotates around the 3D joint A that moves with velocity v A as suggested in Figure V.3 (b). Note: The current discussion is similar to that in Section III.2 for planar rigid bodies. Now in order to describe the rotation of around A we need to understand first the geometry of three dimensional rotations. (a) (b) Figure V.3: There are various ways to describe the relative motion between the points A, of the same rigid body. (a) Two points are indicated and the according position vectors are shown. (b) One may partition the velocity of into a rotation with relative velocity ṙ A around A plus the velocity v A of A. V.2 Three dimensional rotation In this section we will discuss mathematical view on three-dimensional rotations. Let us first define what a rotation is, see below. Definition. A three-dimensional rotation is a linear, orientation preserving transformation that also preserves length.

122 114 V.2. Three dimensional rotation Figure V.4: A vector r is shown before and after it is rotated by the transformation R. From the above definition we may state the transformation as r R r with r R 3 and R : R 3 R 3 (see Figure V.4 for an example). Since the transformation R is length preserving we must have R r = r, r R 3, or, equivalently, r 2 = R r 2. This last equation can also be expressed in terms of the Euclidean inner product (dot product) as r, r = R r, R r. A property of the inner product is that R r, R r = r, R T R r, and therefore r, r = r, R T R r. For the last equation to be true for all r R 3, we must have R T R = I, leading to the following two properties of the transformation R : R T = R 1, det ( R T R ) = det ( R T ) det ( R ) = det ( R ) 2 = 1. (V.3) (V.4) Further note what happens to two orthogonal vectors r s R 3 \ 0 after they are transformed, R r, R s = r, R T R s. (V.5)

123 Chapter V. Kinematics of three dimensional rigid bodies 115 y (V.3), this is equivalent to r, R T R s = r, s, and since the two vectors r, s are orthogonal to each other, we must have r, s = 0. Thus, by equation (V.5), we conclude that two initially orthogonal vectors remain orthogonal under any length-preserving linear map, such as a rotation: R r R s. Such a map (i.e. linear and length-preserving) is called an orthogonal transformation R and is denoted as R O(3), where O(3) refers to the group of orthogonal transformations. Any transformation R O(3) maps a cube ( e 1, e 2, e 3 ) based at the origin into such a cube as well, i.e rotates the cube (see Figure V.5 for an illustration). Note, however, that the orientation ("handidness") of ( e 1, e 2, e 3 ) is not necessarily preserved by R O(3). Figure V.5: A transformation R O(3) maps a cube at the origin to a rotated cube that is also located at the origin. Note that the orientation, however, is not necessarily preserved. Example V.1: Reflection w.r.t. (y, z) - plane Such a transformation R O(3) is not orientation-preserving. A reflection w.r.t. (y, z) - plane maps the three unit vectors [ i, j, k ] as follows: R i = i, R j = j, R k = k. We see that the transformation does not preserve orientation when we consider the determinant of [ i, j, k ] before and after the rotation: det ( i j k ) = 1, det ( R i R j R k ) = 1.

124 116 V.2. Three dimensional rotation In general, for three linearly independent vectors a, b, c R 3, we find the determinant after the transformation to be det ( R a R b R c ) = det ( R ( a b c )) = det ( R ) det ( a b c ). For the orientation of [a, b, c] to be preserved, we need sign ( det ( a b c )) = sign ( det ( R a R b R c )) and therefore det ( R ) > 0. Since for a transformation R O(3), its determinant by (V.4) is det ( R ) = ±1, it follows that for a A O(3) that also preserves orientation, we must have det ( A ) = 1. We call this group of transformations within the orthogonal group O(3), the special orthogonal group SO(3), more specifically as { SO(3) = A R 3 3 : A T A = A A T = I ; det ( A ) } = 1. (V.6) Any transformation A SO(3) describes a rotation in 3D. We state a few properties of A, SO(3) in the following: A, A SO(3). A 1 = A T SO(3). A A 1 = A A T = A T A = I. However, 3D rotations do not commute in general, i.e. A, SO(3) : A A. Note: On the other hand, rotations in two dimensions (i.e. members of the group SO(2)) do commute. Rotations in three-dimensional space Any rotation of a vector r R 3 about a point can be represented as a linear operation r R r, (V.7) where R SO(3). SO(3) denotes the special orthogonal group, i.e. the group of rotations, in three dimension that is defined as { SO(3) = A R 3 3 : A T A = A A T = I ; det ( A ) } = 1. (V.8)

125 Chapter V. Kinematics of three dimensional rigid bodies 117 Example V.2: Plane rotated twice Consider the plane A in the (y, z) - plane, as shown in Figure V.6. We now rotate the plane A first around the x-axis by 90 and then by 90 around the y-axis. This does not yield the same result as when we first rotate 90 around the y-axis and then 90 around the x-axis, i.e. R 90 y R x 90 R 90 x R 90 y. Figure V.6: A plane is shown that is first rotated around the x-axis and then around the y-axis, and then the other way around. The resulting planes have a different orientations, showing that rotations in 3D are not commutative. Matrix representation of a vector rotation To find the matrix representation of a given rotation in a given basis [ i, j, k ], we apply the rotation to every basis vector to obtain the representation of R in [ i, j, k ] in the form [ R ] [ i, j, k] = [ R i R j R k ]. In Figure V.7, the procedure described above is applied to a rotation around the positive basis vector j. The resulting transformation matrix can be found below.

126 118 V.2. Three dimensional rotation Figure V.7: To find the matrix representation of R the transformation R is applied to all basis vectors. Here e 1 = i, e 2 = j, e 3 = k. Rotation around the basis vectors To rotate a vector by an angle φ around one of the three coordinate axes [ i, j, k ] in the positive direction, we use the following matrices representing the corresponding rotation in the frame [ i, j, k ] : R φ x = 0 cos φ sin φ, 0 sin φ cos φ cos φ 0 sin φ R φ y = 0 1 0, sin φ 0 cos φ cos φ sin φ 0 R φ z = sin φ cos φ (V.9) (V.10) (V.11) Note: We can decompose any rotation R SO(3) into a sequence of the three rotations around each basis vector of a given basis [ i, j, k ]. Representation of rotated vectors A vector r R 3 represented in the [ i, j, k ] frame as r = n r i e i, i=1 that is transformed under R SO(3), can be represented after the transformation as 3 r R r = R ij r j e i. i,j=1

127 Chapter V. Kinematics of three dimensional rigid bodies 119 V.3 Angular velocity of time-dependent rotations Figure V.8: A rotating vector may be described by a time-dependent rotation R (t). The trajectory between 0 and the current time t can then be written using the parameter s [0, t]. Consider a vector r(t) rotating about a point 0 in three-dimensional space, as illustrated in Figure V.8. We represent the vector in the coordinate frame [ i, j, k ]. With the initial position r(0) of the vector, we can represent the rotated vector r(t) as r(t) = R(t)r(0), where R (t) SO(3) rotates the vector r(0) to its current orientation r(t) for any time t R. In the following, we are interested in computing the velocity ṙ(t) of a vector rotating in space. First, we take the time derivative of r(t), i.e. d r(t) = ṙ(t) = Ṙ (t)r(0). dt We can replace r(0) with and obtain r(0) = R 1 r(t) = R T r(t) ṙ(t) = Ṙ (t)r T (t)r(t). Since R 1 = R T, we have R R T = I. When differentiating this in time, we obtain Ṙ R T + R Ṙ T = 0, Ṙ R T = R Ṙ T. (V.12)

128 120 V.3. Angular velocity of time-dependent rotations Since for any two matrices A, (A) T = T A T we find for R Ṙ T that R Ṙ T = (Ṙ R T ) T. Together with (V.12), we then obtain Ṙ R T = (Ṙ R T ) T, which shows that Ṙ R T is a skew-symmetric matrix. We define the angular velocity matrix as Ω (t) := Ṙ R T R 3 3, (V.13) a skew-symmetric matrix (Ω = Ω T ) defined for any rotation R (t) SO(3). We now have an expression for the velocity of r(t) in terms of Ω : Velocity of a rotating vector A vector r(t) rotating around a point as r(t) = R(t)r(0) with R(t) SO(3) has the velocity ṙ(t) = Ω(t)r(t), (V.14) where Ω (t) := Ṙ R T R 3 3, Ω = Ω T. In general we can write a skew-symmetric matrix Ω R 3 3 as 0 ω 3 ω 2 Ω = ω 3 0 ω 1. ω 2 ω 1 0 We observe that 0 ω 3 ω 2 r 1 ω 3 r 2 + ω 2 r 3 i j k Ω r = ω 3 0 ω 1 r 2 = ω 3 r 1 ω 1 r 3 = ω 1 ω 2 ω 3 ω 2 ω 1 0 r 3 ω 2 r 1 + ω 1 r 2 r 1 r 2 r 3. Therefore, with the vector ω = ω 1 i + ω 2 j + ω 3 k, we can write Ω r = ω r. Note: This result holds true for any skew-symmetric matrix. With the help of the last equation we can rewrite (V.14) as ṙ(t) = Ω r = ω r. Here ω is referred to as the angular velocity vector.

129 Chapter V. Kinematics of three dimensional rigid bodies 121 Angular velocity vector of a rotation The velocity of a vector r(t) = R(t)r(0) rotating around a point can be described as ṙ(t) = ω(t) r(t), (V.15) where ω denotes the corresponding angular velocity vector of the rotation family R (t) SO(3). Angular velocity of a three-dimensional rigid body For now, let us come back to our initial question of how the velocities v A, v of the points A, of a rigid body are related to each other. From (V.2) we know that v = v A + ṙ A. We also know that ṙ A describes the relative rotation of about A. With (V.15) at hand, we can express the relative velocity ṙ A as ṙ A = ω A r A, where ω A is the angular velocity obtained relative to the point A. We therefore have v = v A + ω A r A. (V.16) Let us now examine how the angular velocity vector ω A depends on the point A. Let us consider three points A,, C of a rigid body as shown in Figure V.9. Figure V.9: Three arbitrary points A,, C of the same rigid body are chosen. Using (V.16), we express the velocity of point C as v C = v A + ω A r AC or, alternatively, as v C = v + ω r C.

130 122 V.3. Angular velocity of time-dependent rotations Subtracting the above two equations from each other gives v v A = ω A r AC ω r C. We also know from (V.15) that v v A = ω A r A. (Note: When we consider the rotation of any two points, C around the point A the rotation of these points is described by the same rotation family, i.e. the same ω A.) Setting the last two expressions equal, we obtain ω A r A = ω A r AC ω r C, or, equivalently, ω A (r A r AC ) = ω r C. (V.17) Since r A r AC = r C we can also rewrite (V.17) as or (ω A ω ) r C = 0, (ω A ω ) e C = 0, (V.18) where e C = r C denotes the unit vector pointing from to C. We observe that r C in (V.18), the term (ω A ω ) does not depend on our choice of C. Therefore, by varying C, we may let e C point in any direction without influencing the term (ω A ω ). This implies that (V.18) must be true for unit vector e C = e R 3, i.e. (ω A ω ) e = 0, e R 3. This may only hold true if ω A = ω. Since our choice of the points A, was also arbitrary, we conclude that ω ω A ω, i.e. there exists a unique angular velocity ω for the body. We can now restate (V.16) with this unique angular velocity ω to obtain the following:

131 Chapter V. Kinematics of three dimensional rigid bodies 123 The velocity transfer formula for a rigid body v = v A + ω r A (V.19) v.... velocity of some point v A.... velocity of some other point A ω..... unique angular velocity of r A... vector pointing from A to Note: Unlike rotations, angular velocities are additive and commutative. We can therefore find the angular velocity ω by first identifying partial angular velocities ω i w.r.t. distinguished axes and then (vectorially) add them up, i.e. ω = i ω i. It is important, however, that ω i must be expressed w.r.t. the same basis, i.e. coordinate frame. For more information on how vectors can be transformed from one basis to another, see Section VI.4. Note: Another useful concecpt is the instantaneous axis of rotation. If there exist two points A, with v A = v = 0 then the axis r A defines the instantaneous axis of rotation. We now show that the angular velocity ω is always parallel to this axis. To see this, consider the velocity transfer formula between those two points A, : v = v A + ω r A. Since v A = v = 0, it follows that ω r A = 0, which is only true if ω r A. (V.20) Example V.3: Rolling cone Consider a rigid cone that rolls without slipping on the (x, y) - plane with its top point hinged in O = (0, 0, R) as sketched in Figure V.10. The cone has radius R at its basis, along which we select a point C, as shown. The angular velocity about the z-axis of the cone is given as φ. Compute the instantaneous velocity v C of point C using the velocity transfer formula. (a ) Angular velocity We want to use the velocity transfer formula to infer the the velocity v C from

132 124 V.3. Angular velocity of time-dependent rotations Figure V.10: A cone rolls without slipping in the shown configuration. the velocity v O = 0. We first need to identify the angular velocity ω. We denote the point of contact with the ground by A. Since the cone rolls without slipping we must have v A = 0. Similarly, the hinge point O has zero velocity, v O = 0. When we apply the velocity transfer formula between O and A, we obtain v A = v O + ω r OA = ω r OA = 0. From the equation above we obtain ω r OA. Therefore, r OA defines the instantaneous axis of rotation (see also (V.20)). To find ω we now examine the geometry of the problem more carefully (see Figure V.11). The angle α denotes half the angle of the cone; φ is the angle between the x-axis and the projection of the axis r AO onto the (x, y)-plane. We denote the unit vector pointing from the origin of the coordinate system to A by e r = cos φ i + sin φ j, and the component of ω in the radial direction e r by ω r. We can now express the angular velocity ω as ω = ω r e r + ω z k, where ω z = φ is given. Since ω r OA, we find that tan α = ω z ω r. Therefore, the angular velocity of the cone can be written as ω = φ ( ) cos φ i + sin φ j + φ k (V.21) tan α

133 Chapter V. Kinematics of three dimensional rigid bodies 125 Figure V.11: The detailed geometry of the rolling cone. Note that we decomposed the angular velocity ω into the sum of ω r and ω z. (b ) Instantaneous velocity of point C The velocity transfer formula between A and C gives v C = v A + ω r AC = ω r AC. Recall that v A = 0. The geometry in Figure V.11 gives r AC = R e φ + R k, where e φ = sin φ i + cos φ j. Thus, we obtain the velocity v C in the form ( v C = φ tan α ( ( v C = R φ cos φ sin φ ) tan α ) ( ) ( ) cos φ i + sin φ j + φ k R sin φ i R cos φ j + R k ( i + sin φ + cos φ ) tan α ) j + 1 tan α k

134 126 V.3. Angular velocity of time-dependent rotations

135 Chapter VI Kinetics of three-dimensional rigid bodies This chapter covers the kinetics of a three-dimensional rigid body. First, we state the momentum principles and the work-energy principle, see Section VI.1-VI.3. The derivation is analogous to the derivation of same the principles for planar rigid bodies. Section VI.4 then introduces the concept of rotating frames and illustrates their use in applying the momentum principles. As an application, in Section VI.5, we discuss the Euler equations for a spinning top, i.e., a three-dimensional rigid body hinged in a point. VI.1 Linear momentum principle Figure VI.1: A three-dimensional rigid body subjected to forces F ext and moments M ext. To derive the LMP and AMP, is partitioned into infinitesimal small pieces of mass m i. denotes an arbitrary reference point and C = CM denotes the center of mass. 127

136 128 VI.1. Linear momentum principle Consider a rigid body in three dimensions as shown in Figure VI.1. We use the inertial frame [ i, j, k ] to describe the motion of. We may split the body into infinitesimal mass elements dm, whose centers of masses are located are labeled by the position vector r. We can then approximate the system as a system of N particles with mass m 0. In Cartesian coordinates, dm becomes dm = ρ dx dy dz, where ρ = ρ(x, y, z) is the volume density of ([ρ] = kg /m 3 ). The mass M of is M = dm. (VI.1) (VI.2) As in the case of system of particles we define the center of mass C to be the point for which the mass moment vanishes, i.e., ϱ C dm = 0 (VI.3) holds with ϱ C = r r C. This definition gives us the following formula for the location of the center of mass C. Location of the center of mass of a rigid body r C = 1 M r dm = 1 M r ρ dv (VI.4) r C... vector pointing to the center of mass from the origin r..... general position vector in the body dm... infinitesimal mass element ρ..... volume density of ([ρ] = kg /m 3 ) dv... infinitesimal volume element Similar to the derivation in Section IV.1 in 2D, we obtain the linear momentum and the linear momentum principle in the following form: Linear momentum and linear momentum principle of a rigid body The linear momentum P of a rigid body is P := Mv C (VI.5) and the linear momentum principle for states Ṗ = Ma C = F ext. (VI.6)

137 Chapter VI. Kinetics of three-dimensional rigid bodies 129 P... linear momentum of M... mass of v C... velocity of the center of mass of a C... acceleration of the center of mass of Note: If F ext = 0, then linear momentum is conserved, P = const. (conservation of linear momentum). VI.2 Angular momentum principle The angular momentum and the angular momentum principle for a three dimensional rigid body as shown in Figure VI.1, can be derived analogously to the two-dimensional case. Refer to Section IV.2 and Section IV.3 for more information regarding the derivation. For the results in 3D, see below: Angular momentum and angular momentum transfer formula for a rigid body The angular momentum H of a rigid body w.r.t. some point is H = ϱ v dm, (VI.7) where ϱ = r r and v = ṙ. To relate the angular momentum of two points A,, one can use the angular momentum transfer formula, i.e. H = H A + P r A (VI.8) H H A... angular momentum of w.r.t. some arbitrary point... angular momentum of w.r.t. some arbitrary point A P..... linear momentum of r A... vector pointing from point A to point The principle of angular momentum for a three-dimensional rigid body Ḣ + v P = M ext (VI.9) H.... angular momentum of w.r.t. a point v..... velocity of point P M ext linear momentum of body... resultant torque w.r.t. point

138 130 VI.2. Angular momentum principle Note: If M ext v = 0 or = C or v v C, = 0 and one of the following three conditions is fulfilled then the angular momentum of the body is conserved, i.e. H = const. (conservation of angular momentum). VI.2.1 Moment of inertia tensor For a two-dimensional body, we found the mass-moment of inertia to be a scalar with which we are able to compute the angular momentum. In three dimensions, this quantity becomes the moment of inertia tensor I R 3 3. It again provides a helpful for computing the angular momentum. Calculation of the angular momentum for fixed or C As in 2D, using the general formula (VI.7) to compute the angular momentum is cumbersome. However, when is fixed or coincides with the center of mass C, we can simplify the equation. We begin with the general formula H = ϱ v dm, where ϱ = r r, as illustrated in Figure VI.1. We then use the velocity transfer formula between and a general point labeled by ϱ to express v at that point as H = ϱ v dm = ϱ (v + ω ϱ ) dm. We then split H into two parts: H = ϱ dm v + ϱ (ω ϱ ) dm. The first integral vanishes for our choice of : ϱ dm v = 0. This holds either because is fixed (v = 0), or because C, and hence, by definition, ϱ dm = 0. The expression for the angular momentum then reduces to H = ϱ (ω ϱ ) dm.

139 Chapter VI. Kinetics of three-dimensional rigid bodies 131 To simplify the calculations further, we shift the origin of the coordinate system to coincides with the point. We can then write ϱ as x ϱ = y. z Further recall that the cross product among three vectors a, b and c satisfies a (b c) = (a c)b (a b)c. We use this identity to conclude ϱ (ω ϱ ) = (ϱ ϱ )ω (ϱ ω)ϱ. Substituting x ϱ = y z and ω 1 ω = ω 2 ω 3 into the above expression gives ϱ (ω ϱ ) = ( x 2 + y 2 + z 2) ω 1 x ω 2 (xω 1 + yω 2 + zω 3 ) y, ω 3 z (y 2 + z 2 )ω 1 yxω 2 zxω 3 = (x + z 2 )ω 2 xyω 1 zyω 3, (x 2 + y 2 )ω 3 xzω 1 yzω 2 y 2 + z 2 xy xz ω 1 = xy x 2 + z 2 yz ω 2. xz yz x 2 + y 2 ω 3 With this result, we may rewrite the angular momentum H as H = ϱ (ω ϱ ) dm = I ω. (VI.10) Here I is the moment of inertia tensor of the body w.r.t. the point. It is defined as I xx I xy I xz I := I xy I yy I yz, I xz I yz I zz (VI.11)

140 132 VI.2. Angular momentum principle with I xx = (y 2 + z 2 ) dm, I xy = xy dm, I xz = xz dm, I yy = (x 2 + z 2 ) dm, I yz = yz dm, I zz = (x 2 + y 2 ) dm. The angular momentum of a rigid body w.r.t. a fixed point or w.r.t. its center of mass C H = I ω (VI.12) H... angular momentum of w.r.t. I... moment of inertia tensor of w.r.t. ω..... angular velocity of Principal axes of a body Consider a basis [u 1, u 2, u 3 ] for which the inertia tensor becomes diagonal, i.e. I = diag ( I 1 I 2 I 3 ). (VI.13) We call the axes [u 1, u 2, u 3 ] the principal axes of the body and I 1, I 2, I 3 are the principal moments of inertia. We may find such orthonormal basis [u 1, u 2, u 3 ] by solving the eigenvalue problem (I λi )u = 0, (VI.14) where we obtained the eigenvalues λ i = I i from det ( I λi ) = 0. (VI.15) We indicated with I the identity matrix. Therefore the principal axes correspond to the (orthonormal!) eigenbasis of I and the principal moments of inertia to the eigenvalues. The eigenbasis is orthonormal, because I is symmetric. When we use a basis [ i, j, k ] for ] representing the inertia tensor [I, we may find the eigenbasis [u 1, u 2, u 3 ] by solving xyz the eigenvalue problem stated above. We may then transform any vector [r] xyz, represented in the [ i, j, k ] basis, to its representation [r] 123 in the eigenbasis [u 1, u 2, u 3 ] via and [r] xyz = U [r] 123 [r] 123 = U T [r] xyz where U = ( u 1 u 2 u 3 ) SO(3).

141 Chapter VI. Kinetics of three-dimensional rigid bodies 133 Note: Consider a coordinate frame where, e.g., the x-axis corresponds to an axis of symmetry of, as for the body shown in Figure VI.2. We then find I xz = I xy = 0 due to the symmetry. When there are two axes of symmetry for, two of the principal axes will coincide with the two axes of symmetry. Aligning the x and y axes with the two directions gives I xz = I xy = I yz = 0, i.e., all off-diagonal terms of I principal form. become 0 and the moment of inertia tensor is in its Figure VI.2: An axis of symmetry (say, x) is always a principal axis. Practical computation of the moment of inertia tensor For common bodies and constant density ρ = const. the moment of inertia tensors are tabulated in Figure VI.3. In all cases the tensor is given in its principal basis. As for two dimensional bodies we can relate the moment of inertia tensor w.r.t. some point O to the tensor w.r.t. the center of mass C (see below and Figure VI.4). Parallel axis theorem (Steiner s theorem) or I O = I C + M ( r CO I r CO r T ) CO b 2 + c 2 ab ac I O = I C + M ab a 2 + c 2 bc ac bc a 2 + b 2 (VI.16) (VI.17) I O I C... moment of inertia tensor about some point O... moment of inertia tensor about the center of mass C M.... mass of the body r CO... vector pointing from center of mass C to point O, r CO = (a, b, c) Note: Recall that I C must be the centroidal moment of inertia tensor otherwise the above formula is not valid.

142 134 VI.3. Work-energy principle Figure VI.3: A list of moment of inertia tensors for various rigid bodies with constant density. The moment of inertia tensor is additive as was the moment of inertia of a planar rigid body. This can be used for rigid body systems consisting of several bodies for which the moment of inertia tensor is either tabulated or much easier to calculate, see below. Superposition of moments of inertia tensors One can calculate the moment of inertia tensor I of a rigid body consisting of several component-bodies i, i = 1,..., n with moments of inertia tensors I i (about some common point and w.r.t. the same coordinate system) by the superposition principle, i.e. I = n i=1 I i (VI.18) VI.3 Work-energy principle To establish a work-energy principle for 3D rigid bodies we first compute the kinetic energy of the system. In a manner analogous to two-dimensional rigid bodies we have a general

143 Chapter VI. Kinetics of three-dimensional rigid bodies 135 Figure VI.4: When changing the reference point of the moment of inertia tensor from any point O to the center of mass C = CM or vice versa, Steiner s theorem can be applied. formula for the kinetic energy T : T = 1 v 2 dm. 2 We now replace v with (VI.19) v = v + ω ϱ, where is either fixed or coincides with the center of mass CM. Further, let ϱ = r r. We then obtain T = 1 2 v 2 + ω ϱ dm = 1 2 v 2 dm + v ω ϱ dm For our choice of, the second term always vanishes: v ω ϱ dm = 0. ω ϱ 2 dm. This is because when is fixed, then v = 0. On the other hand when CM, then ϱ dm = 0 holds by the definition of the center of mass. We will now take a closer look at the integrand of the third term: 2 ( ) ) ω ϱ = ω ϱ (ω ϱ.

144 136 VI.3. Work-energy principle We recall the following identity valid for three vectors a, b, c: (a b) c = (b c) a. When we apply this identity for a = ω, b = ϱ and c = ) ) (ω ϱ (ω ϱ = ( ) ϱ (ω ϱ ) ω. Plugging this back into the third term of the kinetic energy gives ω ϱ 2 dm = ϱ 2 (ω ϱ ) dm ω. (ω ϱ ), we obtain that y (VI.10), this evaluates to 1 ϱ 2 (ω ϱ ) dm ω = 1 2 (I ω) ω = 1 2 (I ω) T ω = 1 2 ωt I ω. Note that we switched from dot product notation (" ") to matrix multiplication and used the fact that I = I T. The last term to be evaluated is the first integral in the expression for T, which simply yields 1 v 2 2 dm = 1 dm v 2 2 = 1 2 M v 2. Putting everything together, we obtain T = 1 2 M v ωt I ω for the kinetic energy T when is either fixed (v = 0) or coincides with the center of mass CM. The kinetic energy T of a rigid body T = 1 2 M v C ωt I C ω (VI.20) or T = 1 2 ωt I ω (VI.21) M... mass of v C... velocity of the center of mass I C... centroidal moment of inertia tensor ω.... angular velocity of I... moment of inertia tensor about some fixed point with v = 0

145 Chapter VI. Kinetics of three-dimensional rigid bodies 137 As in 2D, we may now also find the work-energy principle from the work-energy principle for a system of N particles with mass m after taking the limit N, m 0: Work-energy principle for a three-dimensional rigid body W ext 12 = T 2 T 1 (VI.22) W ext work done by the external forces between states 1 and 2 T i..... kinetic energy of the system associated with state i In case the system is conservative (i.e. all external forces are potential or do no work): Work-energy principle for a conservative system The total mechanical energy E := T + V of a conservative system is conserved along all motions, T 1 + V 1 = T 2 + V 2 (VI.23) or equivalently, d dt E(t) = 0. E... total mechanical energy T i... kinetic energy of the rigid body at state i V i... potential energy of the rigid body at state i Note: Recall that the potential of a system is defined as the sum of all potentials coming from the potential forces and torques, i.e. V (r) := K V k. k=1 VI.4 Rotating frames In many three dimensional applications it is convenient to express mechanical quantities in a rotating frame attached to the body. Especially, the moment of inertia tensor is timedependent in most cases when not expressed in the frame of the principal axes. Therefore, in problems concerning rotations we tend to express quantities of interest in the frame of the principal axes. As a consequence, there are two main challenges to be considered. First, we need to be able to transform vectors from their representation in one frame into their

146 138 VI.4. Rotating frames representation in another frame (see Section VI.4.1). Second, the momentum principles only hold true in their classic form in inertial frames. To still be able to use the momentum principles when the quantities are expressed in a rotating frame we need to differentiate these quantities in time relative to an inertial frame (see Section VI.4.2). Figure VI.5: A vector u may be represented ] using an arbitrary frame, e.g. the inertial frame [e 1, e 2, e 3 ] or a moving frame [f 1, f 2, f 3. VI.4.1 Rotation transformation Consider a frame [e 1, e 2, e 3 ] and an arbitrary ] vector u expressed in this frame, i.e. [u] e. Next, we introduce another frame [f 1, f 2, f 3 that is rotated around a certain axis w.r.t. the first frame [e 1, e 2, e 3 ], as shown in Figure VI.5. We may now find for any rotation a transformation matrix R e f = R fe SO(3) that transforms a vector from its representation [u] e in the e-frame into its representation [u] f in the f-frame. We obtain [u] f = R fe [u] e. To transform the vector back, we can use the transformation matrix R ef = R 1 fe = R T fe, [u] e = R ef [u] f = R T fe [u] f. Specifically, we now state the transformation for the three cases when the coordinate frames are rotated w.r.t. a basis axis. When the coordinate frame f is obtained by rotating the e-frame around the e 1 -axis by an angle α in the positive direction (right-hand rule), the transformation matrix R fe = R x SO(3) becomes R x = 0 cos α sin α, 0 sin α cos α (VI.24)

147 Chapter VI. Kinetics of three-dimensional rigid bodies 139 and hence [u] f = R x [u] e. Note that in this case, e 1 = f 1. We can similarly identify the transformation matrices for the case when the f-frame is obtained by rotating the e-frame about its y- or z-axis in the positive direction, by an angle β or γ, respectively: cos β 0 sin β R y = 0 1 0, sin β 0 cos β cos γ sin γ 0 R z = sin γ cos γ (VI.25) (VI.26) Note: We may put together any transformation matrix between two frames by applying the above three rotation matrices R x, R y, R z in a sequence, choosing the angles α, β and γ appropriately. Example VI.1: ] Consider a frame [f 1, f 2, f 3 that is rotated relative to an inertial frame [e 1, e 2, e 3 ] by an angle β in the y-direction, i.e., about the axis e 2 f 2. Another frame ] [g 1, g 2, g 3 is rotated w.r.t. the f-frame about their common f 3 g 3 axis. We may then find the transformation matrix R ge to transform a vector represented in the e-frame into its representation in the g-frame by first applying the transformation from the e-frame to the f-frame, and then to the g-frame as follows: [u] g = R gf [u] f = R gf R fe [u] e = R ge [u] e. Since R gf = R z and R fe = R y we obtain R ge = R z R y. In case the rotation about an axis is negative (by the right hand rule), we simply construct the rotation matrix as the transpose of the rotation matrix for the respective axis in the positive direction For example, for negative rotations about the x-axis, we have R x = R T x. VI.4.2 Differentiation of quantities in rotating frames ] Let us consider a vector quantity u represented in some rotating frame [f 1, f 2, f 3, as illustrated in Figure VI.5. The angular velocity of the frame f w.r.t. a known inertial frame is Ω. What can we do when we want to express the momentum principle w.r.t. a rotating frame f? The key is that when differentiating a quantity we need to carry out this differentiation in an inertial frame relative to which f rotates with Ω. The vector u is represented as u = 3 u i f i i=1

148 140 VI.4. Rotating frames in the rotating frame f, with u i denoting the coordinates of u w.r.t. f i. Differentiating u in time yields ( 3 ) u = d 3 3 u i f dt i = u i f i + u i ḟ i. i=1 i=1 i=1 From (V.15), we know that the velocity ḟ i angular velocity Ω as follows: of a rotating vector can be calculated using its ḟ i = Ω f i. Therefore, the derivative u can be expressed as u = 3 3 u i f i + u i Ω f i. i=1 i=1 We rewrite this result as u = ů + Ω u, denoting the time derivative of u in the moving frame by ů = 3 u i f i. i=1 Differentiation of quantities in moving frames u = ů + Ω u (VI.27) u... absolute time derivative of u in an inertial frame ů... time derivative of u in the moving frame Ω... angular velocity of the rotating frame relative to the inertial frame All quantities must be expressed in the moving frame! Note: For a body-attached frame, the angular velocity Ω of the frame is just the angular velocity ω of the body: Ω = ω. Example VI.2: Eccentric and skewed disk on a rotating shaft Consider a disk of mass m, height h, radius R and center of mass at C, as illustrated in Figure VI.6. The disk is eccentrically mounted on a massless shaft at the point D, which coincides with the origin of the chosen inertial coordinate system

149 Chapter VI. Kinetics of three-dimensional rigid bodies 141 [ i, j, k ]. The eccentricity e measures the projection of the distance between D and C, onto the [ i, j ] -plane. Moreover, the disk is skewed at an angle δ relative to the horizontal. The distance between D and the floating bearing of the shaft is l 1, respectively the distance between D and the fixed bearing A of the shaft is l 2. The magnitutde of the angular velocity of the shaft is denoted by ϕ. Gravity acts in the negative k - direction. What are the reaction forces at A and? Figure VI.6: A skewed disk mounted eccentrically on a vertically rotating shaft. (a ) Degrees of freedom The system (shaft and disk) has in its unconstrained state, six degrees of freedom. The fixed bearing represents three constraints and the floating bearing adds an additional two constraints. We then obtain that the constrained system has one degree of freedom: DOF = = 1.

150 142 VI.4. Rotating frames We choose ϕ, i.e., the current rotation angle of the shaft, as a generalized coordinate to describe the motion. (b ) Free-body diagram When considering the free-body diagram in Figure VI.7, we see that the following forces act on the system. At both bearings A and, normal forces N A = N A e r and N = N e r arise. They result from the frictionless point contact with the bearing and point in the direction of C in the [ i, j ] -projection. This is the direction of e r = cos ϕ i + sin ϕ j. Since A is a fixed joint, there is also a vertical reaction force T A = T A k exerted in the direction of the shaft. At C, the gravitational force mg acts upon the disk. Figure VI.7: The free-body diagram of the skewed and eccentric rotating shaft. (c ) Linear momentum principle in the inertial frame [ i, j, k ] Since we are interested in finding the reaction forces, we first apply the linear momentum principle Ṗ = F ext.

151 Chapter VI. Kinetics of three-dimensional rigid bodies 143 To determine the necessary ingredients, we will make use of the unit vector e r, as defined above that points from the shaft to C in the [ i, j ] -plane. We find the resultant external force to be F ext = (N A + N ) e r + (T A mg) k, or, equivalently, F ext = (N A + N ) cos ϕ i + (N A + N ) sin ϕ j + (T A mg) k. We also need the acceleration of the center of mass C. We find the position of C from the geometry to be: r C = e ( e r sin δ k) = e ( cos ϕ i + sin ϕ j sin δ k ). Differentiating this twice in time yields ṙ c = e ϕ ( sin ϕ i + cos ϕ j ), r C = e ϕ ( sin ϕ i + cos ϕ j ) + e ϕ 2 ( cos ϕ i sin ϕ j ) = e ( ϕ sin ϕ + ϕ 2 cos ϕ ) i + e ( ϕ cos ϕ ϕ 2 sin ϕ ) j. When we now plug everything into the linear momentum principle and evaluate it for each direction individually, we obtain the following three equations: em ( ϕ sin ϕ + ϕ 2 cos ϕ ) = (N A + N ) cos ϕ, em ( ϕ cos ϕ ϕ 2 sin ϕ ) = (N A + N ) sin ϕ, Combining (VI.28) and (VI.29), we obtain (VI.28) (VI.29) 0 = T A mg. (VI.30) em ϕ 2 = N A + N, (VI.31) and from (VI.30), we find that T A = mg. (VI.32) From (VI.31), we see that the force (N A + N ) points away from the center of mass C. (N A + N ) acts as a centripetal force, keeping C on its circular orbit. (d ) Angular momentum principle in principal frame (non-inertial!) Just from the linear momentum principle, we do not have enough equations to determine the reaction forces N A, N. Therefore, we also need to apply the angular momentum principle. ] We choose to apply it w.r.t. C and use the principal frame [f 1, f 2, f 3, shown in Figure VI.8. This choice was made because in this configuration we can easily determine the moment of inertia tensor I C. It is diagonal, time-independent and readily available from the table in Figure VI.3. Note: In the following any column vector or matrix representation of a quantity is expressed in the principal frame!

152 144 VI.4. Rotating frames (a) (b) Figure VI.8: The angular momentum principle is applied in the principal frame [f 1, f 2, f 3 ]. (a) The orientation of the principal frame is shown. (b) The principal frame coincides with the axes of symmetry of the disk. Since v C P, the AMP about C states that Ḣ C = M ext C. Since C is the center of mass of the disk, the angular momentum w.r.t. C can be computed as H C = I C ω. Since the shaft is the axis of rotation, the angular velocity ω points along this axis and it has the magnitude ϕ: ω = ϕ k. In the principal frame f ω can be expressed as ϕ sin δ ω = ϕ sin δf 1 + ϕ cos δf 3 = 0, ϕ cos δ where the coordinates of the last vector are determined w.r.t. the [f 1, f 2, f 3 ] -basis.

153 Chapter VI. Kinetics of three-dimensional rigid bodies 145 Note: We can transform any vector u represented in the inertial e-frame as [u] e into its representation [u] f in the principal frame f by the transformation matrix R fe = R y R z. We decomposed the transformation into two parts to describe the rotation of f. First, we rotate around the positive z-axis by φ. Then, from the frame arising from the first rotation, we rotate around the new positive y-axis by δ. Therefore, cos δ 0 sin δ R fe = sin δ 0 cos δ It is then straightforward to check that ϕ sin δ [ω] f = R fe [ω] e = 0 ϕ cos δ. cos ϕ sin ϕ 0 sin ϕ cos ϕ We now come back to computing the angular momentum H C. From the lookup table in Figure VI.3, we find the moment of inertia tensor about C in the principal frame: I α 0 0 I C = 0 I β I γ with I α = I β = 1 ( ) 4 m R 2 + h2, 3 I γ = 1 2 mr2. Finally, we obtain I α 0 0 ϕ sin δ I α ϕ sin δ H C = I C ω = 0 I β 0 0 = I γ ϕ cos δ I γ ϕ cos δ To differentiate H C we apply (VI.27) since H C is expressed in the moving f-frame. We obtain Ḣ C = H C + Ω H C, where Ω denotes the angular velocity of the body-attached frame, and hence Ω = ω. Substitution into the formula for ḢC gives I α ϕ sin δ ϕ sin δ I α ϕ sin δ Ḣ C = I γ ϕ cos δ ϕ cos δ I γ ϕ cos δ I α ϕ sin δ = 1 2 ϕ2 sin(2δ)(i γ I α ) I γ ϕ cos δ

154 146 VI.5. Euler equations for spinning tops For the torque M ext C M ext C = about C, we find 0 N (l 1 + e sin δ) N A (l 2 l sin δ) + T A e cos δ 0 Plugging everything into the AMP yields I α ϕ sin δ ϕ2 sin(2δ)(i γ I α ) = N (l 1 + e sin δ) N A (l 2 l sin δ) + T A e cos δ. I γ ϕ cos δ 0 From the first and third component of this equation we obtain ϕ = 0 and hence ϕ = const. = ω 0. We could have anticipated this result since the angular momentum in the z-direction is conserved about C. The second component of the AMP yields 1 2 ϕ2 sin(2δ)(i γ I α ) = N (l 1 + e sin δ) N A (l 2 l sin δ) + T A e cos δ. (VI.33) When we combine (VI.33) with (VI.31), we can solve this for N A, N to finally obtain ( 1 8 N = m R h2) sin(2δ)ω0 2 + e ( g cos δ + ω0 2(l 2 sin δ) ), l 1 + l ( N A = m R h2) sin(2δ)ω0 2 + e ( g cos δ ω0 2(l 1 + sin δ) ). l 1 + l 2 Note that the first term with "sin(2δ)" denotes the effect of the skewedness and the second term with "e" represents the effect of the eccentricity. Note: For e = δ = 0, i.e. a unskewed disk centrally placed, both reaction forces disappear: N A = N = 0. However, when e, δ 0 then N A, N ω0 2 grow (!) quadratically with increasing angular velocity. In practice, this is very important for high-speed turbo machineries. They must be carefully balanced.. VI.5 Euler equations for spinning tops Consider a rigid body that is hinged at the origin O of the inertial coordinate system [x, y, z], and rotates freely, as illustrated in Figure VI.9. The system may be subjected to forces and torques. The center of mass of the body is located at C and the axis r OC is a principal axis of the body. For such systems, we can simplify the general angular momentum principle to Euler s equations for spinning tops. We will derive these equations by applying the angular momentum principle about C in the principal frame of the body.

155 Chapter VI. Kinetics of three-dimensional rigid bodies 147 Figure VI.9: A spinning top hinged at the origin O. The degrees of freedom are indicated by three angles (Euler angles). Degrees of freedom The fixed joint at O corresponds to three independent holonomic constraints. Therefore, the system has three degrees of freedom: DOF = 6 3 = 3. Euler s idea was to describe such system using three angles as generalized coordinates. Those three angles are constructed by a "3-1-3" rotation sequence from the inertial frame to the principal frame of the body, see Figure VI The first rotation is a rotation by ϕ about the axis z z 1 to the coordinate system [x 1, y 1, z 1 ]. The corresponding transformation is given by cos ϕ sin ϕ 0 [u] 1 = sin ϕ cos ϕ 0 [u] xyz We call this angle the precession angle, whose angular velocity component is ω ϕ = ϕ e ϕ = ϕ e z = ϕ e z1. 2. Next we rotate by the nutation angle ν about the axis x 1 x 2 to the coordinate frame [x 2, y 2, z 2 ] such that z 2 is aligned with the principal axis r OC of the body. The

156 148 VI.5. Euler equations for spinning tops second transformation is given by [u] 2 = 0 cos ν sin ν [u] 1, 0 sin ν cos ν with the associated angular velocity vector ω ν = ν e ν = ν e x1 = ν e x2. Note that the angular velocity of the frame is now Ω 2 = ω ν + ω ϕ. 3. Finally, we rotate about z 2 z 3 by the angle ψ such that the resulting frame [x 3, y 3, z 3 ] corresponds to a principal frame of the spinning top. The transformation is cos ψ sin ψ 0 [u] 3 = sin ψ cos ψ 0 [u] 2, or, equivalently, cos ψ sin ψ cos ϕ sin ϕ 0 [u] 3 = sin ψ cos ψ 0 0 cos ν sin ν sin ϕ cos ϕ 0 [u] xyz, sin ν cos ν when expressed relative to the original inertial frame [x, y, z]. The angular velocity of this last transformation is ω ψ = ψ e ψ = ψ e z2 = ψ e z3, the angle ψ is called spin. The set (ϕ, ν, ψ) of generalized coordinates fully describes the motion of the spinning top. They are commonly referred to as Euler angles. Other variants of Euler angles also exist, but the present one is the most common choice. Note: The spin ψ comes from the fast spinning of the top around its principal axis. The precession ϕ describes the change of orientation of the spinning axis. The top precesses around the inertial axis z. The nutation ν describes small oscillations of the spinning axis during the precession. Free-body diagram The spinning top may also be subjected to a resultant force F ext and a resultant torque M ext C, leading to a reaction force N of general orientation at the joint at O (see Figure VI.10). Angular velocity The total angular velocity of the body is ω = ω ϕ + ω ν + ω ψ,

157 Chapter VI. Kinetics of three-dimensional rigid bodies 149 Figure VI.10: The free-body diagram of the spinning top. which is also the angular velocity of the principal frame. We transform each component of the angular velocity to the principal frame [x 3, y 3, z 3 ] using the above transformation and then add them up. We obtain ω 1 ϕ sin ν sin ψ + ν cos ψ ω = ω 2 = ϕ sin ν cos ψ ν sin ψ. (VI.34) ω 3 ϕ cos ν + ψ Angular momentum about C Since we are in the principal frame, the moment of inertia tensor is diagonal and timeindependent. We then find the angular momentum to be I ω 1 I 1 ω 1 H C = I C ω = 0 I 2 0 ω 2 = I 2 ω 2. (VI.35) 0 0 I 3 ω 3 I 3 ω 3 Also note that the angular velocity Ω 3 = ω of the principal frame [x 3, y 3, z 3 ] coincides with the angular velocity of the body. Angular momentum principle w.r.t. C in the principal frame The angular momentum principle about C states Ḣ C = M ext C + r 0C N. Note that v c P = 0. The arising torques are M 1 M C = M 2 = M ext C + r OC N. M 3

158 150 VI.5. Euler equations for spinning tops We also need to compute the time derivative of the angular momentum: Ḣ C = H C + ω H C, I 1 ω 1 ω 1 I 1 ω 1 = I 2 ω 2 + ω 2 I 2 ω 2. I 3 ω 3 ω 3 I 3 ω 3 Plugging everything into the angular momentum principle, we obtain the following set of Euler s equations for a spinning top in its principal frame: Euler s equations of motion for a spinning top I 1 ω 1 + (I 3 I 2 )ω 3 ω 2 = M 1, I 2 ω 2 + (I 1 I 3 )ω 1 ω 3 = M 2, I 3 ω 3 + (I 2 I 1 )ω 2 ω 1 = M 3. (VI.36) I i.... principal moments of inertia ω i... i th component of the angular velocity expressed in the principal frame M i... i th component of the resultant torque expressed in the principal frame Example VI.3: Axisymmetric gyroscope Consider an axisymmetric gyroscope hinged at its tip at O, as shown in Figure VI.11. We are going to use the notation and coordinate frames we have just introduced for a general spinning top (see above). In this case, since the body is axisymmetric, the frame [x 2, y 2, z 2 ] is already a principal frame although it is not attached to the body! We can, therefore, also apply the angular momentum principle with all quantities expressed in the [x 2, y 2, z 2 ]-frame. We are interested in applying the AMP to such a gyroscope when it is in steady precession, i.e. ν = ν 0 = const., ϕ = ϕ 0 = const. and ψ = ψ 0 = const. (a ) Angular velocity The angular velocity of the [x 2, y 2, z 2 ]-frame is ν 0 Ω = ϕ e ϕ + ν e ν = ϕ sin ν = ϕ 0 sin ν 0, ϕ cos ν ϕ 0 cos ν 0 and the angular velocity of the body is ν 0 ω = ϕ e ϕ + ν e ν + ψ e ψ = ϕ sin ν ϕ cos ν + ψ = ϕ 0 sin ν 0 ϕ 0 cos ν 0 + ψ 0.

159 Chapter VI. Kinetics of three-dimensional rigid bodies 151 Figure VI.11: An axisymmetric gyroscope. Note that the frame [x 2, y 2, z 2 ] is already a principal frame since the gyroscope is axisymmetric and the third transformation to Euler s angles is not necessary. There are still three degrees of freedom needed to describe the motion. All quantities are expressed w.r.t. the [x 2, y 2, z 2 ]-frame. We used the fact that we consider steady precession, therefore ν = 0. Note that Ω ω since the frame is not attached to the body. (b ) Angular momentum prinicple w.r.t. C in the [x 2, y 2, z 2 ]-frame The angular momentum principle about C states Ḣ C = M C. We find the angular momentum to be I 1 ω 1 H C = I 2 ω 2, I 3 ω 3 as for the general spinning top but with I 1 = I 2. its time derivative in the absolute frame is Ḣ C = H C + Ω H C. In case of steady precession, the time derivative in the moving frame is zero: H C = 0. Therefore, we obtain Ḣ C = Ω H C,

160 152 VI.5. Euler equations for spinning tops which, after substitution into the angular momentum principle, yields Ω H C = M C. (VI.37) Evaluating the left-hand side of (VI.37) for steady precession gives I 3 Ω 2 ω 3 I 1 Ω 3 ω 2 Ω H C = I 1 Ω 3 ω 1 I 3 Ω 1 ω 3 = I 3 ϕ 0 sin ν 0 ( ϕ 0 cos ν 0 + ψ 0 ) I 1 ϕ I 1 Ω 1 ω 2 I 1 Ω 2 ω 1 0 sin(2ν) 2 Note that ω 1 = Ω 1 = 0 in our case. In steady precession, the external torque is parallel to the nutation axis x 1 x 2 and its magnitude balances (I 3 Ω 2 ω 3 I 1 Ω 3 ω 2 ). Consider, for example, an axisymmetric gyroscope that spins with ψ 0 and is slightly tilted (see Figure VI.12). Gravity will create a reaction force N, which in turn creates a torque M C about C. However, instead of making the gyroscope fall, this torque causes the gyro to develop a ϕ 0 0 and therefore a Ω 0 to maintain (VI.37). The gyroscope reacts to N by rotating about N! An equilibrium ν = ν 0 is reached when the external torque precisely balances the gyro-action torque (Ω H C ). Then the gyroscope steadily precesses around N. Note: To reach equilibrium, we should have that ψ 0 ϕ 0.. Figure VI.12: When an axisymmetric gyroscope spins fast its angular momentum H C is almost parallel to ω ψ.

161 Chapter VII Vibrations In this chapter, we only consider two-dimensional rigid bodies for simplicity but all results extend to three-dimensional rigid bodies. We will study what happens to a system that oscillates around one of its stable equilibria. This is a central topic to dynamics since in practice most systems operate in equilibrium and this type of analysis allows us to see how a system responds to small perturbations around this equilibrium position. We call the resulting perturbed motion vibration. Definition. Vibrations are small oscillations around a stable equilibrium. Since we only consider small oscillations around a stable equilibrium, we can approximate the motion by linearizing the equation of motion around the equilibrium. This step allows us to use a set of analytical tools only valid for linear systems that greatly simplify our analysis. Note: Our analysis is only valid when the equilibrium is stable, since then small oscillations remain small. For unstable equilibria, the oscillation becomes large even for arbitrary small initial perturbations. We start out with one of the most common example of vibrations, i.e. a forced damped pendulum. On this example we illustrate the derivation of the linearized equations of motin starting from the momentum principles. Subsequently, we analyze the general equation for vibrations with one degree of freedom (see Section VII.1). Afterwards, we undertake the analysis of multi-degree-of-freedom vibrations in Section VII.2. Example VII.1: Forced-damped pendulum Consider a pendulum of mass m connected to a massless bar of length l, hinged in a smooth joint at O, see Figure VII.1. The smooth joint moves according to x 0 = a sin ωt in the i-direction. We describe the motion with the generalized coordinate ϕ denoting the deflection from the horizontal position. Gravity acts upon the system, as does a linear, angular damper, exerting a damping torque M d = c ϕ. (a ) Find the equation of motion using the angular momentum principle about O. (b ) Linearize the equation of motion, first for a = 0 and then for a

162 154 Figure VII.1: A damped and forced pendulum. (a ) Equation of motion We use the angular momentum principle Ḣ O + v O P = M O w.r.t. the joint O to find the equation of motion. The time derivative of the angular momentum H O is Ḣ O = ml 2 ϕ, and the torque about O is M O = ( mgl sin ϕ c ϕ) k. The velocity of O is given as v O = ẋ O (t) i = a ω cos ωt i with which we can compute v and P = mv via the velocity transfer formula. When we plug all ingredients into the AMP, we find the equation of motion: ml 2 ϕ malω 2 sin ωt cos ϕ = mgl sin ϕ c ϕ (VII.1) This equation of motion is a 2 nd order, nonlinear, non-autonomous ODE. (b ) Linearization of the equation of motion i Unforced vibration, a = 0 To linearize the equation of motion, we first need to find a stable equilibrium around which to linearize. The two equilibria are ϕ 0 = 0 and ϕ 0 = π, for which

163 Chapter VII. Vibrations 155 ϕ(t) ϕ 0. We now introduce a new variable x = ϕ ϕ 0 that describes the perturbation from the equilibrium. With x and a = 0, we find (VII.1) to be ml 2 ẍ + cẋ + mgl sin(x + ϕ 0 ) = 0. After linearizing the equation (i.e. Taylor-expanding it to first order at x = 0, ϕ = ϕ 0 and neglecting higher order terms in the expansion), we obtain Mẍ + Cẋ + Kx = 0, (VII.2) where we introduced the parameters M = ml 2, C = c and K = mgl cos ϕ 0. This is the equation for free vibration at equilibrium. It is a 2 nd order homogeneous, linear ODE. As stated before, for a stable equilibrium and an initially small perturbation ( x(0), ẋ(0) small), the linearized model remains valid for all times since the perturbation from the equilibrium will remain small. Note that in this case, as expected, only ϕ 0 = 0 is a stable equilibrium. For this choice of ϕ 0, K = mgl and therefore the quantities M, C, K > 0 are all positive. ii Forced vibration, 0 < a 1 For small a 1, the linear approximation is expected to remain valid. We again linearize (VII.1) but this time we add the force arising from the motion of the joint to our linearization. We obtain Mẍ + Cẋ + Kx = F 0 sin ωt, (VII.3) with F 0 = malω 2 cos ϕ 0. We call such an externally excited vibration a forced vibration. The above equation is a 2 nd order inhomogeneous, linear, non-autonomous ODE. VII.1 One-DOF vibrations We now consider a general system of one degree of freedom, oscillating around its stable equilibrium. As illustrated in the previous example, we can express the linearized equation of motion of such systems with a sinusoidal inhomogeneous term as Mẍ + Cẋ + Kx = F 0 sin ωt. (VII.4) The variable x denotes the perturbation from the stable equilibrium. One can, e.g., think of this system as a cart attached to a wall via a spring and a damper (see Figure VII.2). It is excited by the external force F 0 sin ωt, which causes oscillations around the equilibrium position. We now introduce the characteristic damping δ := C 2M, the natural frequency ω 0 := K M

164 156 VII.1. One-DOF vibrations Figure VII.2: A cart of mass m is shown that is attached to a spring as well as a damper and is excited by an external force. The system is already linear and hence does not have to be linearized. and the normalized force amplitude f 0 = F 0 M. We use these parameters to rewrite the equation of one DOF vibrations as follows: The general equation for vibrations with one degree of freedom ẍ + 2δẋ + ω 2 0x = f 0 sin ωt (VII.5) x(t)... deflection from the stable equilibrium δ..... characteristic damping [ 1 /s] ω 0 f natural frequency [ 1 /s].... amplitude of normalized excitation force ω..... frequency of excitation Sometimes, we also use the parameter D, called Lehr s damping, defined as D := δ ω 0 = C 2 KM (VII.6) VII.1.1 Free vibrations Let us first consider the case where f 0 = 0, i.e. becomes ẍ + 2δẋ + ω0x 2 = 0. we have a free vibration. Then (VII.5) (VII.7) This equation can also be converted into a 2 2 system of 1 st order linear, homogeneous ODEs. For this purpose, we introduce the state vector x R 2, defined as ( ) ( ) x1 x x = =. ẋ x 2

165 Chapter VII. Vibrations 157 We call the space of possible x values the phase space. Hence, we can represent a motion (x 1, x 2 ) in the phase space as a trajectory. In the phase space, (VII.7) is represented by a system of ODEs ( ) 0 1 ẋ = A x = x. (VII.8) 2δ ω 2 0 We can find the general solution of (VII.7) via the ansatz x(t) = Ae λt with A, λ C, A 0. Plugging this ansatz into (VII.7) we obtain Ae λt (λ 2 + 2δλ + ω 2 0) = 0. To solve this equation, we need to find the roots λ 1,2 of the characteristic polynomial λ 2 + 2δλ + ω 2 0, given that A = 0 is of no interest. When solving for these roots, we obtain λ 1,2 = δ ± δ 2 ω0 2 (VII.9) for the roots of the characteristic polynomial of the matrix A. We now need to distinguish two cases: Case 1: distinct roots, λ 1 λ 2, λ j C For two distinct roots λ 1 λ 2, λ j C, we find the general solution of (VII.7) to be x(t) = A 1 e λ 1t + A 2 e λ 2t, with A j C. Using the initial conditions x(0) = x 0, ẋ(0) = v 0, we can find A 1, A 2. The solution can also be represented in phase space as x(t) = A 1 s 1 e λ 1t + A 2 s 2 e λ 2t, where s j = (1, λ j ). The initial conditions in the phase space take the form ( ) x0 x 0 = R 2. v 0 Note: Although A j, λ j C, the solution x(t) is always real. General solution for free vibrations (one DOF) with distinct roots λ 1 λ 2 The solution in the time domain is x(t) = A 1 e λ 1t + A 2 e λ 2t, (VII.10) while in the phase space it is of the form x(t) = A 1 s 1 e λ 1t + A 2 s 2 e λ 2t, (VII.11) where λ 1,2 = δ ± δ 2 ω 2 0 and s j = (1, λ j ). The coefficients A 1, A 2 C can be determined from the initial conditions (x 0, v 0 ). From these general solutions further sub-cases arise.

166 158 VII.1. One-DOF vibrations Overdamped vibrations Overdamped vibrations arise when δ > ω 0 and hence λ 1,2 = δ ± δ 2 ω 2 0 are real, distinct and negative: λ 1 < λ 2 < 0, λ 1, λ 2 R. From (VII.10) we find the solution in the time domain in the form ( ( ) ) ( ( ) ) x(t) = A 1 exp δ δ 2 ω0 2 t + A 2 exp δ + δ 2 ω0 2 t, (VII.12) and from (VII.11) we obtain the solution in the phase space as ( ( ) ) ( ( ) ) x(t) = A 1 s 1 exp δ δ 2 ω0 2 t + A 2 s 2 exp δ + δ 2 ω0 2 t. (VII.13) Note that A 1, A 2 R and ( ) ( ) 1 s 1 = δ 1 δ 2 ω0 2, s 2 = δ + δ 2 ω0 2. This case is called overdamped because damping is large enough to prevent oscillations around the equilibrium. The solution decays exponentially from the initial position towards the equilibrium position, as shown in Figure VII.3. A nonzero initial velocity v 0 0 will cause the system to either overshoot in the beginning (v 0 > 0), or undershot before approaching the equilibrium (v 0 < 0). In the phase space, solutions decay to an invariant line (blue), then creep towards the equilibrium in a close vicinity of this line. There is another invariant line (red), along which solutions decay to the origin at a fast rate, withou any creeping. Note that two solutions cannot ever cross due to the uniqueness. As a consequence, the invariant lines separate the phase space into four regions. This type of equilibrium is called a node. (a) (b) Figure VII.3: The motions in an overdamped vibration are qualitatively shown in the time domain and the phase space. (a) The solution is exponentially decaying towards the equilibrium. Depending on the initial velocity v 0, there is an overshot or undershot. (b) In the phase space, solutions converge towards the equilibrium along an invariant line. The equilibrium is called a node.

167 Chapter VII. Vibrations 159 Underdamped vibrations Underdamped vibrations occure when the damping is weak enough to allow for oscillations around the equilibrium. In this case, we have δ < ω 0. The roots of the characteristic polynomial then satisfy λ 1,2 = δ ± i ω0 2 δ2 C. Substitution into (VII.10) then gives x(t) = e δt ( A 1 e i ω 2 0 δ2t + A 2 e i ω 2 0 δ2) for the solution in the time domain. ecause x(t) R, the constants A 1, A 2 must be complex conjugate, i.e., A 2 = A 1. With the help of the two real constants C 1, C 2 R, we can therefore write: A 1 = 1 2 (C 1 ic 2 ), A 2 = 1 2 (C 1 + ic 2 ). Using Euler s formula, e ix = cos x + i sin x, we obtain the solution in the time domain as ) )) x(t) = e (C δt 1 cos ( ω 20 δ2 t + C 2 sin ( ω 20 δ2. (VII.14) This equation describes an oscillation around the stable equilibrium with damped natural frequency ω d = ω0 2 δ2, and with an amplitude decaying as e δt (see Figure VII.4). For the solution in the phase space, we use (VII.19) and the same real constants C 1, C 2 R as above to obtain ( x(t) = C 1 C 1 δ + C 2 ω d ) ( e δt cos(ω d t) + C 2 C 2 δ C 1 ω d ) e δt sin(ω d t). (VII.15) (a) (b) Figure VII.4: An underdamped vibration is qualitatively shown in the time domain and the phase space. (a) The solution in the time domain oscillates around the stable equilibrium with (pseudo-) period T d and exponentially decaying amplitude e δt. (b) In the phase space, there are no invariant lines anymore. The solutions converge as non-intersecting spirals towards the equilibrium.

168 160 VII.1. One-DOF vibrations The phase portrait in this case is clearly distinguished from the previously discussed phase portraits. There is no invariant line anymore. Solutions form spirals converging towards a stable focus-like equilibrium, as shown in Figure VII.4. Remember that solutions are unique and therefore they must not intersect in the phase portrait. (a) (b) Figure VII.5: Undamped vibration in the time domain and the phase space. (a) The solution in the time domain oscillates around the stable equilibrium with period T and constant amplitude C 0. (b) The solution in the phase space is described by elliptical trajectories encircling the equilibrium. Undamped vibrations Undamped vibrations arise in the presence of zero damping: δ = 0. The roots of the characteristic polynomial are then purely imaginary: λ 1,2 = ±iω 0. We can use the solutions from the overdamped case and simply set δ = 0, ω d = ω 0. In the time domain, we then obtain x(t) = C 1 cos(ω 0 t) + C 2 sin(ω 0 t) from (VII.14). We can also rewrite this solution as ( x(t) = C1 2 + C2 2 C 1 cos(ω C C2 2 0 t) + C 2 sin(ω C C2 2 0 t) ). We define now the constant C 0 and the angle ϕ 0 via C 0 = C1 2 + C2 2, sin ϕ 0 = C 1, cos ϕ C C2 2 0 = C 2. C C2 2 With these constants, we can express the solution as x(t) = C 0 sin(ω 0 t + ϕ 0 ). (VII.16) The constants C 0, ϕ 0 can always be determined from C 1 and C 2, which in term are obtained from the initial conditions (x 0, v 0 ). We see that the solutions describe a sinusoidal oscillation with phase shift ϕ 0 and natural frequency ω 0 (see Figure VII.5). The solution

169 Chapter VII. Vibrations 161 will never converge to the equilibrium but constantly oscillate around it. To identify the solution in the phase space, we set δ = 0 and ω d = ω 0 in (VII.15) to obtain x(t) = ( C1 C 2 ω 0 ) ( ) C2 cos(ω 0 t) + sin(ω C 1 ω 0 t). 0 (VII.17) This defines elliptic trajectories that encircle the center-type equilibrium (see Figure VII.5). No solution converges to the equilibrium. Note: The phase portrait is structurally unstable. The addition of the slightest amount of damping (δ > 0) causes all solutions to converge to the origin (underdamped case). Case 2: repeated roots, λ 1 = λ 2 = λ R For this case, the discriminant of (VII.9) vanishes, which is only possible when δ = ω 0. In this case the solution takes the following form: General solution for free vibrations (one DOF) with a repeated root λ = λ 1 = λ 2 The solution in the time domain is x(t) = A 1 e λt + A 2 te λt, (VII.18) and in the phase space, it is x(t) = (A 1 s 1 + A 2 s 1 )e λ 1t + A 2 s 2 te λ 2t, (VII.19) where λ 1,2 = δ = ω 0, s j = (1, λ j ) and s 1 = (0, 1). The coefficients A 1, A 2 R can be found via the initial conditions (x 0, v 0 ). Note that this only occurs when δ = ω 0. Critically damped vibrations The case of repeated roots is called critically damped. The solution in the time domain looks qualitatively the same as for the case of overdamped systems. In the phase space, however, the phase portrait is somewhat different, compared to the case of overdamped vibrations. The two invariant lines from the case of overdamped vibrations have merged into a single invariant line in the direction ( ) 1 s 1 =. ω 0 This type of equilibrium is called a degenerate (stable) node, see Figure VII.6 (b). VII.1.2 Forced vibrations In this section, we consider oscillations under nonzero external forcing (f 0 0). general equation for a one DOF vibration system (VII.5) is The ẍ + 2δẋ + ω 2 0x = f 0 sin ωt. Recall that we can express the solution of an inhomogeneous linear ODE as the superposition of the general solution of the associated homogeneous ODE and one particular

170 162 VII.1. One-DOF vibrations (a) (b) Figure VII.6: The solution to a critically damped vibration is qualitatively shown in the time domain and the phase space. (a) The solution is exponentially decaying towards the equilibrium. Depending on the initial velocity v 0, there is either an overshot or an undershot. (b) In the phase space, the solutions converges towards equilibrium. There is only one invariant line. The equilibrium is a degenerate node. solution of the inhomogeneous ODE: x(t) = x h (t) + x p (t), where x h (t) denotes the general solution of the homogeneous ODE and x p (t) is a particular solution of the inhomogeneous ODE. General solution of a forced vibration with one degree of freedom The general solution x(t) to a forced vibration with one degree of freedom can be written as x(t) = x h (t) + x p (t), (VII.20) where x h (t) denotes the general solution to the associated free vibration, i.e. f 0 = 0, and x p (t) denotes a particular solution to the forced vibration. We already found the general solution for the homogeneous case, see (VII.10) and (VII.18). We are now interested in finding the particular solution of (VII.5). Note: For damped systems (δ 0), the homogeneous solution x h (t) converges to 0, x h t 0, and, therefore, we can approximate a general solution x(t) for longer times as x(t) x p (t). The ansatz we choose to find the particular solution x p (t) is x p (t) = A sin(ωt ϕ), (VII.21)

171 Chapter VII. Vibrations 163 where ω is the frequency of the external forcing. Substituting this ansatz into (VII.5), we obtain A( ω 2 + ω 2 0) sin(ωt ϕ) + 2δAω cos(ωt ϕ) = f 0 sin(ωt). With trigonometric identities and sin(ωt ϕ) = sin(ωt) cos ϕ cos(ωt) sin ϕ, cos(ωt ϕ) = cos(ωt) cos ϕ + sin(ωt) sin ϕ, we obtain ) sin(ωt) (A(ω 0 2 ω 2 ) cos ϕ + 2δAω sin ϕ f 0 ( ) = cos(ωt) A(ω0 2 ω 2 ) sin ϕ + 2δAω cos ϕ. This equation only holds true for all times t if both sides are equal to zero, i.e., A(ω 2 0 ω 2 ) cos ϕ + 2δAω sin ϕ = f 0, A(ω 2 0 ω 2 ) sin ϕ + 2δAω cos ϕ = 0. Combining the above equations gives A(ω 2 0 ω 2 ) = f 0 cos ϕ, 2δAω = f 0 sin ϕ. Solving these last two equations for the amplitude A of the particular solution, we obtain A = f 0 (ω 2 0 ω 2 ) 2 + 4δ 2 ω 2. (VII.22) Similarly, we find that ϕ = tan 1 ( 2δω ω 2 0 ω2 ). (VII.23) To analyze the response to forcing in more detail we introduce the magnification factor V = A (f 0 /ω 2 0 ), the ratio between the amplitude A generated by the forcing f 0 sin(ωt), and the amplitude of the static response under constant forcing f 0. We find the static response amplitude (f 0 /ω0 2 ) from the forced vibration equation (VII.5), looking for a steady state solution x (i.e. setting ẍ = ẋ = 0): ẍ + 2δẋ + ω0x 2 = f 0 x = f 0 ω0 2. (VII.24)

172 164 VII.1. One-DOF vibrations With the frequency ratio η := ω, ω 0 and Lehr s damping D := δ, ω 0 the amplification factor V can be written as V = 1 (1 η 2 ) 2 + (2Dη) 2. (VII.25) We can also rewrite the phase shift ϕ of the particular solution in terms of η and D to obtain ( ) 2Dη ϕ = tan 1 1 η 2. (VII.26) Finally, we state the particular solution (VII.21) in terms of the amplification factor V : The steady-state solution (particular solution) for a one-dof forced vibration x p (t) = F 0 V (η, D) sin (ωt ϕ(η, D)) K (VII.27) x p (t)..... steady state solution F amplitude of external periodic force K stiffness factor from (VII.4) V (η, D)... amplification factor ω frequency of exciting force ϕ(η, D)... phase shift of steady state solution We now further analyze the amplification factor V (η, D) and the phase shift ϕ(η, D) (see Figure VII.7). For each damping D, there exists a maximum of V. For D = 0 (no damping), V diverges when the excitation frequency ω approaches the natural frequency ω 0 : V (η, D = 0) η 1. This phenomenon is commonly referred to as resonance, arising when a system is excited precisely at its natural frequency ω 0. The less damping there is, the closer the maximum amplification is to the natural frequency ω 0, and the larger is the maximum of the amplification factor V. For the phase shift ϕ(η, D), we see that η < 1 yields ϕ < π 2. In this case, we say that the

173 Chapter VII. Vibrations 165 system oscillates in phase. For η > 1 the phase shift is larger than π 2, i.e. ϕ > π 2. Such oscillations are called out of phase. In this case, the exciting frequency is larger than the natural frequency and the system lacks the ability to keep up with the oscillation. When D 0, the phase shift jumps at η = 1 from 0 to π 2. Furthermore, we can interpret the general solution x(t) = x h (t) + x p (t) as one that invariably settles on the particular solution x p (t), while the damped oscillations x h (t) of the homogeneous system die out, as shown in Figure VII.8. Figure VII.7: The amplification factor V (η, D) and the phase shift ϕ(η, D), plotted as a function of the frequency ratio η, for various damping values D. The red arrows indicate the curve of the maximum amplification for select values of D. Note that, for D = 0, the phase shift jumps at the natural frequency ω 0.

174 166 VII.2. Multi-DOF vibrations Figure VII.8: The general solution, indicated in red, consists of the homogeneous solution and the particular solution, indicated in green. For nonzero damping, the homogeneous solution dies out and the solution approaches the particular solution. VII.2 Multi-DOF vibrations Figure VII.9: Two carts of mass m 1 and m 2 are attached to each other and each to a wall via respective spring-damper elements. Note that this system is already linear. In this section, we consider vibrations of mechanical systems with multiple degrees of freedom, such as the system shown in Figure VII.9. This system has n = 2 degrees of freedom and oscillates around its equilibrium. From our previous discussion of multi-dof systems, we know that we need n independent generalized coordinates to describe such an oscillatory system. In general, let r(t) R n be the vector of generalized coordinates for an n-degree-of-freedom system, and let r 0 R n be a stable equilibrium (constant solution of the equations of motion) of the same system. We denote the deviation from the equilibrium r 0 by x(t) = r(t) r 0 (t). For small deviations x(t) (i.e. for vibrations), we can linearize the system of equations of motion. The resulting linear system is of the general form M ẍ + C ẋ + K x = F (t),

175 Chapter VII. Vibrations 167 with the positive definite mass matrix M = M T R n n, the positive semidefinite damping matrix C = C T R n n, and the positive semidefinite stiffness matrix K = K T R n n. F (t) R n denotes the vector of the external forces acting on the system. Note: A positive definite matrix A R n n is defined as a matrix whose associated quadratic form x T A x satisfies x T A x > 0, x 0. Similarly, a positive semidefinite matrix R n n is a matrix whose associated quadratic form satisfies x T x 0, x 0. Equation for vibrations with n degrees of freedom M ẍ + C ẋ + K x = F (t) (VII.28) x(t)... deviation from the stable equilibrium, x(t) R n M.... positive definite mass matrix, M = M T R n n C..... positive semidefinite damping matrix, C = C T R n n K.... positive semidefinite stiffness matrix, K = K T R n n F (t)... external force acting on the system, F (t) R n In the most general case, the linearization of an arbitrary mechanical system around its equilibrium yields M ẍ + (C + G )ẋ + (K + N )x = F (t), where we call the skew-symmetric matrix G = G T the gyro matrix and the likewise skewsymmetric matrix N = N T the follower force matrix. We will not treat this general case here. VII.2.1 Free, undamped vibrations Let us first consider the case of a free and undamped vibration. Equation (VII.28) then becomes M ẍ + K x = 0. (VII.29)

176 168 VII.2. Multi-DOF vibrations We choose the ansatz x(t) = ue λt (VII.30) with u C n \ {0}, λ C, to find the solution of (VII.29). When we substitute this ansatz into (VII.29), we obtain (λ 2 M + K )ue λt = 0, (λ 2 M + K )u = 0. (VII.31) Note: Equation (VII.31) is referred to as the generalized eigenvalue problem for the matrices M and K. Indeed, rewriting the equation as K u = λ 2 M u, multiplying with M 1 from the left and defining := M 1 K and Λ = λ 2 lead to the classic eigenvalue problem u = Λu. ack to our equations, we now multiply (VII.31) with u T from the left to obtain u T M uλ 2 + u T K u = 0. Substituting u = v + iw, where v, w R n, we obtain (v T M v + w t M w)λ 2 + (v T K v + w T K w) = 0. (VII.32) We now introduce the two parameters α, β as and α = v T M v + w t M w, β = v T K v + w T K w. Since M is positive definite, it follows that α > 0. Similarly, because K is positive semidefinite β 0. The eigenvalues λ of (VII.31) can therefore be written, using (VII.32), as or λ 2 = β α = ω2 λ = ±iω (VII.33) for some ω 0. To find the eigenvalues λ, we use the fact the matrix (λ 2 M + K ) must be singular according to (VII.31). This implies that det ( λ 2 M + K ) = 0. (VII.34) Equation (VII.32) is called the characteristic equation of (VII.29), an n th -order polynomial in λ 2. The roots of this polynomial define the 2n eigenvalues λ j of (VII.29).

177 Chapter VII. Vibrations 169 Note: To find the corresponding eigenvector u j of λ j, one needs to solve (VII.31) with λ j substituted. Since the eigenvalue problem is quadratic in λ, the eigenvectors u j and u j+1 of the eigenvalue pair λ j,j+1 = ±iω coincide and are purely real: u j = u j+1 R n. From (VII.33) we see that the eigenvalues only arise in pure imaginary pairs or zero pairs. We can now find two independent solutions to (VII.29) for each eigenvalue pair. We differentiate betweem two cases: ω = 0 and ω > 0. Case 1: purely imaginary eigenvalue pairs, λ j,j+1 = ±iω j, ω j > 0, u R n In this case, we find the solution to be x k (t) = u j (t)(p j e iω jt + q j e iωt ) with some constants p j, q j C. This follows by superposing the two independent solutions x j (t) = p j u j (t)e iω jt, x j+1 (t) = q j u j (t)e iω jt, obtained from the general ansatz (VII.30) after substituting the eigenvalue λ j and λ j+1, and the corresponding eigenvectors u j = u j+1. From the undamped case of a one-dof system, we already know that q j = p j must hold since x k (t) R. We can therefore write the solution as x k (t) = u j (t) ( c j cos(ω j t) + d j sin(ωt) ), where c j, d j R. Equivalently, we have x k (t) = u j A j sin(ω j t + ϕ j ), (VII.35) with amplitude A j R and phase shift ϕ j [0, 2π). oth of these constants can be determined from the initial conditions. This case corresponds to a undamped oscillation. Case 2: zero eigenvalue pairs, λ j = λ j+1 = 0, u R n One solution for this case can be found directly from the ansatz (VII.30): x j = p j e 0 t u j = p j u j. When plugging this into (VII.29), we obtain or K x = 00, K u j = 0, (VII.36) This implies that u j is an eigenvector of K with eigenvalue 0. To find a second independent solution for the eigenvalue pair, λ j = λ j+1 = 0, we use the ansatz x j+1 (t) = q j (ũ j + u j t).

178 170 VII.2. Multi-DOF vibrations When we plug this into the vibration equation (VII.29), we obtain K (ũ j + u j t) = 0. y (VII.36), this is the same as K ũ j = 0. We are free to choose any ũ j to satisy the above equation. y (VII.36), we can choose ũ j = u j. So the general solution for the eigenvalue pair, λ j = λ j+1 = 0, becomes x k (t) = p j u j + q j u j (1 + t), or, with two different constants j, C j R, x k (t) = u j ( j + C j t). (VII.37) Physically, this corresponds to a rigid-body translation with uniform velocity. General solution Finally, we can find the general solution of the full linearized system as the superposition of all individual solution x k, k = 1,..., n: x(t) = n x k (t). k=1 Depending on whether the k th eigenvalue pair is purely imaginary or zero, we plug in (VII.35) or (VII.37), as we summarize: General solution to a free, undamped vibration of n degrees of freedom The general solution x(t) to (VII.29) is x(t) = u 1 A 1 sin(ω 1 t + ϕ 1 ) + u 3 A 3 sin(ω 3 t + ϕ 3 ) u j ( j + C j t) + u j+2 ( j+2 + C j+2 t) +... (VII.38) with j = 1,..., 2n. The constants (A j, ϕ j ) and ( j, C j ) can be determined using the given initial conditions. To find the eigenvector u j, commonly called normal mode or mode shape, the eigenvalue problem (λ 2 jm + K )u j = 0 (VII.39) for each eigenvalue pair λ j,j+1 must be solved. Remember that λ 2 j = λ2 j+1 = ω2 j and u j = u j+1. The eigenvalues, also called modal frequencies, are determined by the characteristic equation det ( λ 2 M + K ) = 0. (VII.40)

179 Chapter VII. Vibrations 171 Note that each purely imaginary eigenvalue pair λ j,j+1 = ±iω j adds an oscillatory term x k = u j A j sin(ω j t + ϕ j ) to the general solution x(t), whereas a rigid-body translation (λ j,j+1 = 0) adds a term x k = u j ( j + C j t). M... mass matrix K... stiffness matrix Example VII.2: Two masses connected via a spring Consider a system of two masses m 1 and m 2, connected to each other via a spring with spring constant k, as illustrated in Figure VII.10. oth masses can slide on the ground without friction. The system has two degrees of freedom (n = 2). We choose the position x 1 of the center of mass m 1 and the position x 2 of the center of mass m 2 as generalized coordinates. For x 1 = x 2 = 0, the system is in unstretched, stable equilibrium. Therefore, x j denotes the deviation of m j from the equilibrium. (a ) From the equation of motion, find the linearized vibration equation. (b ) What are the corresponding eigenvalues and mode shapes of the vibration? (c ) State the general solution for this vibration and interpret it physically. Figure VII.10: Two masses, m 1 and m 2, are attached to each other via a spring but can otherwise move freely. (a ) Equation of vibrations

180 172 VII.2. Multi-DOF vibrations Applying the linear momentum principle to each mass individually, we obtain m 1 ẍ 1 = k(x 2 x 1 ), m 2 ẍ 2 = k(x 2 x 1 ). These equations are already linear and describe the deviation from the equilibrium (0, 0). With the state vector ( ) x1 x =, x 2 we bring the equations of motion to matrix form. We obtain ( ) ) ( ) ( ) m1 0 (ẍ1 k k x1 + = 0. 0 m 2 ẍ 2 k k x 2 Note: With the mass matrix, ( ) m1 0 M = 0 m 2 the associated quadratic form satisfies x T M x = m 1 x m 2 x 2 2 > 0, therefore the mass matrix M is positive definite, as assumed earlier. Similarly, the quadratic form of ( ) k k K = k k satisfies x T K x = k(x 1 x 2 ) 2 0, and hence the stiffness matrix K is positive semidefinite, as we generally assumed. (b ) Eigenvalues and mode shapes i Eigenvalues To find the eigenvalues of this vibration problem, we need to solve the characteristic equation det ( λ 2 M + K ) = 0. Substitution into the characteristic equation gives ( λ det 2 ) m 1 + k k k λ 2 = 0, m 2 + k or, equivalently, λ 2( m 1 m 2 λ 2 + k(m 1 + m 2 ) ) = 0. As roots to this equation, we obtain the two eigenvalue pairs k(m 1 + m 2 ) λ 1,2 = ±i = ±iω 1, λ 3,4 = 0, m 1 m 2 i.e., one purely imaginary pair and one zero pair.

181 Chapter VII. Vibrations 173 VII.2.2 ii Mode shapes We find the mode shapes u 1 and u 3 by solving (λ 2 jm + K )u j = 0 for each eigenvalue pair. Substituting λ 1,2 yields ( ) ( ) ω 2 1 m 1 + k k u11 k ω1 2m = k u 21 Solving this equation for the mode shape u 1, we obtain ( ) ( ) u11 m2 u 1 = =. u 21 m 1 Similarly, we find the mode shape u 3 for the eigenvalue pair λ 3,4 in the form ( ) 1 u 3 =. 1 (c ) General solution The general solution from (VII.38) is x(t) = u 1 A 1 sin(ω 1 t + ϕ 1 ) + u 3 ( 3 + C 3 t), which is of the specific form ( ) m2 x(t) = A m 1 sin k(m 1 + m 2 ) t + ϕ m 1 m 2 ( ) 1 ( C 3 t), with the eigenvalues and mode shapes substituted. The fist term in this solution corresponds to an oscillation around the overall center of mass due to the spring force between the masses. The second term describes the rigid-body translation of the overall center of mass. Note that for this system, the total linear momentum is conserved (no external forces act on the system in the direction of the motion). Modal decomposition of free, undamped vibrations We again consider the free, undamped vibration equation M ẍ + K x = 0, with M, K R n n and x R n. From (VII.39), we observe that ω 2 j M u j + K u j = 0, ω 2 k M u k + K u k = 0 (VII.41) (VII.42) for two distinct eigenvalue pairs with ω j, ω k with k j. Combining these equations as u T k (VII.41) ut j (VII.42),

182 174 VII.2. Multi-DOF vibrations we obtain (ω 2 k ω2 j )u T j M u k = 0. Since (ωk 2 ω2 j ) 0 for k j, it follows that u T j M u k = 0 (VII.43) Let us also combine (VII.41) and (VII.42) in this way: u T k (VII.41) + ut j (VII.41). This gives us (ωk 2 + ω2 j )u T j M u k + 2u j K u k = 0, or, equivalently, u j K u k = 0, (VII.44) according to (VII.43). These results will imply that we can decompose (VII.29) into n independent vibration equations instead of with an n-dimensional coupled system. To obtain this decomposition, we first define the (regular) modal matrix U as U = [ u 1 u 3... u 2n 1 ] R n n, uj = 1, from the corresponding mode shapes u j, normalizing them to u j = 1. From (VII.43), we observe that U T M U = diag ( m 1 m 2... m n ), a diagonal matrix with entries m j. Similarly, U T K U = diag ( k 1 k 2... k n ) is also a diagonal matrix. We now introduce the modal coordinates y by letting n x(t) = U y(t) = y k u 2k 1. k=1 (VII.45) Since U is non-singular we can also write y(t) = U 1 x(t). When we now substitute x(t) = U y(t) into (VII.29) and then multiply by U T from the left, we obtain U T M U ÿ + U T K U y = 0.

183 Chapter VII. Vibrations 175 Since both (U T M U ) and (U T K U ) are diagonal matrices, we obtain n independent 2 nd order linear, homogeneous ODEs for the modal coordinates y: m j ÿ j + k j y j = 0, j = 1,..., n. We say that system is now decoupled into its n independent vibration modes. Note that each of these modes can be excited separately, and we can interpret the general solution (VII.45) as a weighted superposition of its individual modes, as shown, e.g., in Figure VII.11. Modal decomposition of free, undamped vibrations A free, undamped vibratory system of n degrees of freedom, M ẍ + K x = 0, can be decoupled into its normal modes using the modal coordinates y(t) defined as x(t) = U y(t), (VII.46) where each coordinate represents one normal mode. The matrix U = ( u 1 u 3... u 2n 1 ) R n n (VII.47) is the corresponding modal matrix of the system with u j = 1, j = 1, 3,..., 2n 1 denoting the normalized eigenvectors of the system. In the modal coordinates, the equations of motion become U T M U ÿ + U T K U y = 0, (VII.48) with U T M U = diag ( ) m 1 m 2... m n, (VII.49) U T K U = diag ( ) k 1 k 2... k n, (VII.50) or, equivalently, m j ÿ j + k j y j = 0, j = 1,..., n. (VII.51) Note: The eigenfrequency/natural frequency of each mode, i.e. the eigenfrequency of the corresponding eigenvalue pair, is easily computed from the modal decomposition as k j ω j = 0, j = 1,..., n. m j VII.2.3 Multi-DOF forced, damped vibrations Finally, we consider at the general equations of forced and damped vibrations for a system of n degrees of freedom: M ẍ + C ẋ + K x = F (t). (VII.52)

184 176 VII.2. Multi-DOF vibrations Figure VII.11: The frame of a car is excited by the external force p(t), as shown. The solution to this problem can be decomposed into the individual modes of the system, which are graphically displayed for the given situation (a two-mode model is assumed for the car). We assume that the damping matrix C is a linear combination of the mass matrix M and the stiffness matrix K : C = am + bk. (VII.53) We will show that, for this type of damping (structural damping), the system can still be decoupled into one-dof oscillations using the modal coordinates y(t) = U 1 x(t) of the corresponding free, undamped system. As before, we substitute the modal coordinates into our vibration equation (VII.52), then multiply by U T from the left to obtain U T M U ÿ + U T C U ẏ + U T K U y = U T F (t). From (VII.49), we already know that U T M U = diag ( m 1 m 2... m n ) and from (VII.50), we have U T K U = diag ( k 1 k 2... k n ). y the structural damping hypothesis (VII.53) we have U T C U = U T (am + bk )U = a U T M U + b U T K U, or, equivalently, U T C U = a diag ( m 1 m 2... m n ) + b diag ( k1 k 2... k n ).