Rigid Body Dynamics: Kinematics and Kinetics. Rigid Body Dynamics K. Craig 1

Rigid Body Dynamics: Kinematics and Kinetics Rigid Body Dynamics K. Craig 1

Topics Introduction to Dynamics Basic Concepts Problem Solving Procedure Kinematics of a Rigid Body Essential Example Problem Kinetics of a Rigid Body Supplement: Rigid Body Plane Kinetics Essential Example Problem Rigid Body Dynamics K. Craig 2

Introduction Dynamics The branch of mechanics that deals with the motion of bodies under the action of forces. Newtonian Dynamics This is the study of the motion of objects that travel with speeds significantly less than the speed of light. Here we deal with the motion of objects on a macroscopic scale. Relativistic Dynamics This is the study of motion of objects that travel with speeds at or near the speed of light. Here we deal with the motion of objects on a microscopic or submicroscopic scale. Rigid Body Dynamics K. Craig 3

Newtonian Dynamics Kinematics This is the study of the geometry of motion. It describes the motion of bodies without reference to the forces which either cause the motion or are generated as a result of the motion. It is used to relate position, velocity, acceleration, and time without reference to the cause of the motion. Kinetics This is the study of the relation existing between the forces acting on a body, the mass distribution of the body, and the motion of the body. It is used to predict the motion caused by given forces or to determine the forces required to produce a given motion. Rigid Body Dynamics K. Craig 4

Basic Concepts Space Space is the geometric region occupied by bodies. Position in space is determined relative to some geometric reference system by means of linear and angular measurements. The basic frame of reference (perspective from which observations are made) for the laws of Newtonian mechanics is the primary inertial system which is an imaginary set of rectangular axes assumed to have no translation or rotation in space. Rigid Body Dynamics K. Craig 5

Measurements show that the laws of Newtonian mechanics are valid for this reference system as long as any velocities involved are negligible compared with the speed of light (186,000 miles per second). Measurements made with respect to this reference system are said to absolute. A reference frame attached to the surface of the earth has a somewhat complicated motion in the primary system, and a correction to the basic equations of mechanics must be applied for measurements made relative to the earth s reference frame. Rigid Body Dynamics K. Craig 6

In the calculation of rocket- and space-flight trajectories, the absolute motion of the earth becomes an important parameter. For most engineering problems of machines and structures which remain on the earth s surface, the corrections are extremely small and may be neglected. For these problems, the laws of Newtonian mechanics may be applied directly for measurements made relative to the earth, and, in a practical sense, such measurements will be referred to as absolute. Time Time is the measure of the succession of events and is considered an absolute quantity in Newtonian mechanics. Rigid Body Dynamics K. Craig 7

Mass Mass is the quantitative measure of the inertia or resistance to change in motion of a body. It is also the property which gives rise to gravitational attraction and acceleration. In Newtonian mechanics, mass is constant. Newton s Law of Universal Gravitation The force of attraction between two bodies of mass M and m, respectively, separated by a distance r, is given by: GMm F= ê r G = 6.673 10 r M r m 3 11 2 2 Rigid Body Dynamics K. Craig 8 ê r m kg s

Mass Moment of Inertia The mass moment of inertia of a rigid body is a constant property of a body and is a measure of the radial distribution of the body s mass with respect to an axis through some point. It represents the body s resistance to change in angular motion about the axis through the point. Force Force is the vector action of one body on another. There are two types of forces in Newtonian mechanics: Direct contact forces between two bodies. Forces which act at a distance without physical contact, of which there are only two: gravitational and electromagnetic. Rigid Body Dynamics K. Craig 9

Particle A particle is a body of negligible dimensions. Also, when the dimensions of a body are irrelevant to the description of its motion or the action of the forces acting on it, the body may be treated as a particle. It can also be defined as a rigid body that does not rotate. Rigid Body A rigid body is a body whose changes in shape are negligible compared with the overall dimensions of the body or with the changes in position of the body as a whole. Coordinate A coordinate is a quantity which specifies position. Any convenient measure of displacement can be used as a coordinate. Rigid Body Dynamics K. Craig 10

Degrees of Freedom This is the number of independent coordinates needed to completely describe the motion of a mechanical system. This is a characteristic of the system itself and does not depend upon the set of coordinates chosen. Constraint A constraint is a limitation to motion. If the number of coordinates is greater than the number of degrees of freedom, there must be enough equations of constraint to make up the difference. Generalized Coordinates These are a set of coordinates which describe general motion and recognize constraint. Rigid Body Dynamics K. Craig 11

A set of coordinates is called independent when all but one of the coordinates are fixed, there still remains a range of values for that one coordinate which corresponds to a range of admissible configurations. A set of coordinates is called complete if their values corresponding to an arbitrary admissible configuration of the system are sufficient to locate all parts of the system. Hence, generalized coordinates are complete and independent. Newton s Laws of Motion (for a particle) 1 st Law: Every particle continues in its state of rest or of uniform motion in a straight line unless compelled to change that state by forces acting on it. That is, the velocity of a particle can only be changed by the application of a force. Rigid Body Dynamics K. Craig 12

2 nd Law: The time rate of change of the linear momentum of a particle is proportional to the resultant force (sum of all forces) acting upon it and occurs in the direction in which the resultant force acts. 3 rd Law: To every action there is an equal and opposite reaction, i.e., the mutual forces of two bodies acting upon each other are equal in magnitude, opposite in direction, and collinear. These laws have been verified by countless physical measurements. The first two laws hold for measurements made in an absolute frame of reference, but are subject to some correction when the motion is measured relative to a reference system having acceleration. Rigid Body Dynamics K. Craig 13

Units SI Units The primary dimensions are: mass, M, length, L, and time, T. The units are: mass (kg), length (m), and time (sec). This is an absolute set of units based on mass, which is invariant. Force, F, has dimensions of ML/T 2 with the unit newton (N). 1 N = 1 kg m 2 s Rigid Body Dynamics K. Craig 14

US Customary Units The primary dimensions are: force, F, length, L, and time, T. The units are: force (lb), length (ft), and time (sec) This is a relative set of units dependent upon the local force of gravitational attraction. Mass, M, has dimensions FT 2 /L with the unit slug. 2 lb s 1 slug = 1 ft Rigid Body Dynamics K. Craig 15

When close to the surface of the earth, g = 9.81 m/s 2 in SI units, and g = 32.2 ft/s 2 in US Customary units. Some useful conversions: 1 ft = 0.3048 m 1 lb = 4.448 N 1 slug = 14.59 kg Weight = mg = magnitude of the gravitational force acting on mass m near the surface of the earth. Rigid Body Dynamics K. Craig 16

Scalar A scalar is any quantity that is expressible as a real number. Vector A vector is any quantity that has both magnitude and direction. Because the study of Newtonian mechanics focuses on the motion of objects in threedimensional space, we are interested in threedimensional vectors. A unit vector has a magnitude of unity. Rigid Body Dynamics K. Craig 17

There are three types of vectors: Free Vector: no specified line of action or point of application Sliding Vector: specified line of action but no specified point of application Bound Vector: specified line of action and specified point of application. A bound vector is unique, i.e., only one vector can have a specified direction, magnitude, line of action, and origin. Note that vector algebra is valid only for free vectors. Consequently, the result of any algebraic operation on vectors, regardless of the type of vector, results in a free vector. Rigid Body Dynamics K. Craig 18

Matrices An array of numbers arranged in rows and columns is called a matrix. A m n matrix has m rows and n columns. Our use of matrices will initially be restricted to coordinate transformations and later to the concept of the inertia matrix. ˆi ˆ 1 1 0 0 i ˆj ˆ 1 0 cos sin = α α j kˆ 0 sinα cosα kˆ 1 ˆi 1 1 1 = ˆi ˆj = cosα ˆj+ sinα kˆ kˆ = sinα ˆj+ cosα kˆ Rigid Body Dynamics K. Craig 19

Notation and Reference Frames A reference frame is a perspective from which observations are made regarding the motion of a system. A moving body, such as an automobile or airplane, frequently provides a useful reference frame for our observations of motion. Even when we are not moving, it is often easier to describe the motion of a point by reference to a moving object. This is the case for many common machines, such as linkages. An engineer needs to be able to correlate observations of position, velocity, and acceleration of points on moving bodies, as well as the angular velocities and angular accelerations of those moving bodies, from both fixed and moving reference frames. Rigid Body Dynamics K. Craig 20

Reference Frames and Notation P a???? Meaningless! R1 ω???? R Proper Notation 2 ω???? R R1 ω R P y 1 y a R 1 R2 α ω z z 1 O ˆi ˆ 1 1 0 0 i ˆj ˆ 1 0 cos sin = α α j kˆ 0 sinα cosα kˆ 1 ˆi ˆ 2 cos φ sin φ 0 i 1 ˆj = sinφ cosφ 0 ˆj kˆ 0 0 1 kˆ 2 1 2 1 y 1 x 2 Reference Frames: φ x 1 R ground: xyz R 1 shaft: x 1 y 1 z 1 R 2 disk: x 2 y 2 z 2 O Rigid Body Dynamics K. Craig 21 y 2

φ î î ĵ ĵ = unit vector in x direction 1 1 = unit vector in y direction 1 1 = unit vector in x direction 2 2 = unit vector in y direction 2 2 ˆi = cosφ ˆi + sinφ ˆj 2 1 1 ˆj = sinφ ˆi + cosφˆj 2 1 1 ˆi ˆ 2 cos φ sin φ i 1 = ˆ j sin cos φ φ ˆj 2 1 ˆi = cosφ ˆi sinφ ˆj 1 2 2 ˆj = sinφ ˆi + cosφ ˆj 1 2 2 ˆi ˆ 1 cos φ sin φ i 2 = ˆ j sin cos φ φ ˆj 1 2 Rigid Body Dynamics K. Craig 22

Procedure for the Solution of Engineering Problems GIVEN State briefly and concisely (in your own words) the information given. FIND State the information that you have to find. DIAGRAM A drawing showing all quantities involved should be included. BASIC LAWS Give appropriate mathematical formulation of the basic laws that you consider necessary to solve the problem. ASSUMPTIONS List the simplifying assumptions that you feel are appropriate in the problem. Rigid Body Dynamics K. Craig 23

ANALYSIS Carry through the analysis to the point where it is appropriate to substitute numerical values. NUMBERS Substitute numerical values (using a consistent set of units) to obtain a numerical answer. The significant figures in the answer should be consistent with the given data. CHECK Check the answer and the assumptions made in the solution to make sure they are reasonable. Check the units, if appropriate. LABEL Label the answer (e.g., underline it or enclose it in a box). Rigid Body Dynamics K. Craig 24

Kinematics of a Rigid Body Angular Velocity of a Rigid Body Differentiation of a Vector in Two Reference Frames Addition Theorem for Angular Velocities Angular Acceleration of a Rigid Body Reference Frame Transformations Velocity and Acceleration of a Point Coriolis Acceleration and Centripetal Acceleration Essential Example Problem Rigid Body Dynamics K. Craig 25

Angular Velocity of a Rigid Body R is the ground reference frame with coordinate axes xyz fixed in R R 1 reference frame is a rigid body moving in reference frame R with coordinate axes x 1 y 1 z 1 fixed in R 1 β is any vector fixed in R 1 ˆˆ i form a right-handed set of 1 j1 kˆ 1 mutually perpendicular unit vectors fixed in R 1 Angular velocity is the time rate of change of orientation of the body. It is not in general equal to the derivative of any single vector. y y 1 R 1 z 1 A z O R x R d dt β x 1 β R = ω R1 β Defining equation for ω Rigid Body Dynamics K. Craig 26

Simple Angular Velocity of a Rigid Body When a rigid body R 1 moves in a reference frame R in such a way that there exists throughout some time interval a unit vector whose orientation in both R 1 and R is independent of the time, then rigid body R 1 is said to have simple angular velocity in R throughout this time interval. For example: R R1 R R1 ω = ω kˆ =θ kˆ θ= angular speed of R in R Here R 1 has simple angular velocity in R (ω 1 ) and R 2 has simple angular velocity in R 1 (ω 2 ). R 2 does not have simple angular velocity in R. Rigid Body Dynamics K. Craig 27 1

Differentiation of a Vector in Two Reference Frames If R and R 1 are any two reference frames, the first time derivatives of any vector β (not fixed in either R or R 1 ) in R and in R 1 are related to each other as follows: R dβ dt R1dβ = + ω β dt R R1 ( ) Rigid Body Dynamics K. Craig 28 y y 1 R 1 z 1 A z O R x x 1 β

Addition Theorem for Angular Velocities Consider multiple reference frames: R 1, R 2,, R N The following relation applies, whether the angular velocities are simple or not: ω = ω + ω + + ω R R R R R R R R N 1 1 2 N 1 N There exists at any one instant only one Also R RN RN ω = ω R ω R N This addition theorem is very powerful as it allows one to develop an expression for a complicated angular velocity by using intermediate reference frames, real or fictitious, that have simple-angularvelocity relations between each of them. Rigid Body Dynamics K. Craig 29 R

R ω = ω + ω R R R R R 2 1 1 2 simple angular velocity NOT simple angular velocity Rigid Body Dynamics K. Craig 30

Angular Acceleration The angular acceleration of reference frame R 1 in reference frame R is given by: There is no addition theorem for angular accelerations. When R 1 has simple angular velocity in R, e.g., kˆ = kˆ 1 R R1 R R1 ω = ω ˆk R R1 R R1 α = α ˆk ω=θ α=ω=θ R R R R R R d ω d 1 1 1 R1 α = = Rigid Body Dynamics K. Craig 31 y dt y 1 R 1 z 1 A z O R x ω dt x 1 R

Reference Frame Transformations φ 1 2 R 1 x 1 y 1 z 1 R 2 x 2 y 2 z 2 R1 R2 R1 R2 R1 R2 1 2 R ω = ω kˆ = ω R ω =φ kˆ ˆi 1 cosφ sinφ 0 ˆi 2 ˆ j 1 sin cos 0 ˆ = φ φ j2 ˆ ˆ k 1 0 0 1 k 2 ˆi ˆ 2 cosφ sinφ 0 i 1 ˆ j ˆ 2 = sinφ cosφ 0 j1 kˆ 0 0 1 kˆ 2 1 Define: V= V ˆi + V ˆj = V ˆi + V ˆj x 1 y 1 x 2 y 2 1 1 2 2 R 1 dv What Is? dt Rigid Body Dynamics K. Craig 32

R dv dt 1 1 R ˆ d ˆ ˆ ˆ ˆ i1 and ˆj 1 = Vi x1 1+ Vj y1 1 = V xi 1 1+ V yj 1 1 dt are fixed in R R1 R1 dv d = V ˆ ˆ x i 2 2 + Vy2j 2 dt dt R2 d ˆ ˆ R1 R 2 Vx i 2 2 + Vy2j = 2 + ( ω V) dt = V ˆi + V ˆj + ( φ kˆ V) One Approach Another Approach Are the two approaches equivalent? ˆi and ˆj 2 2 are fixed in R 2 x 2 y2 2 1 2 d = Vx cos V 1 y sin ˆi 1 2 dt φ+ φ + d V ˆ x sin V 1 y cos j 1 2 dt φ+ φ + φ k ˆ (V ˆi + V ˆj ) Rigid Body Dynamics 1 x1 1 y1 1 K. Craig 33 1

= V x cosφ V 1 x φsin φ+ V 1 y sin φ+ V 1 yφ ˆ cosφ 1 i2 + V x sin φ V 1 x φcosφ+ V ˆ 1 y cosφ V 1 yφ sin φ j 1 2 + φv ˆ ˆ x j 1 1 φ Vyi 1 1 = V ˆ ˆ ˆ ˆ x cosφi 1 2 sinφ j 2 + V y sinφ i 1 2 + cosφ j 2 + φv ˆ ˆ ˆ x sinφi 1 2 cosφ j 2 +φv y cosφ i 1 2 sin φ ˆj 2 + φv ˆ ˆ x j 1 1 φ Vy i 1 1 = V ˆi + V ˆj φ V ˆj +φ V ˆi +φv ˆj φv ˆi x 1 y 1 x 1 y 1 x 1 y 1 1 1 1 1 1 1 = Vi ˆ + Vj ˆ Same Result! x 1 y 1 1 1 Rigid Body Dynamics K. Craig 34

R Velocity and Acceleration of a Point The solution of nearly every problem in dynamics requires the formulation of expressions for velocities and accelerations of points of a system under consideration. ( R ) 1 R 1 v = v + ω r + v P R A a a R R 1 AP R 1 P + α r + a R R1 R 1 P + 2 ω v R P R A R AP P ( R R AP r ) R R1 1 = + ω ω Rigid Body Dynamics K. Craig 35 z Reference Frames R - Ground xyz R 1 - Body x 1 y 1 z 1 y O y 1 Relative Acceleration Centripetal Acceleration Tangential Acceleration Coriolis Acceleration R 1 z 1 R A x P x 1

Derivation r = r + r OP OA AP d y v = (r ) dt OA r R R d OA d AP = (r ) + (r ) dt dt z O R R1 R A d AP R R1 AP = v + (r ) + ( ω r ) dt R A R1P R R1 AP = v + v + ( ω r ) R 1 R 1 v = v + ω r + v R R P OP ( ) Rigid Body Dynamics K. Craig 36 y 1 R 1 R P R A R AP P A z 1 x r OP x 1 r AP P

d d ( R = = + ω ) + dt dt R R R P R P R A R AP P 1 R1 a ( v ) v r v R R d ( v ) dt d dt R d dt = a R A R A d r ( r ) ( r ) dt = α + ω + ω R R AP R AP R AP ( ) R1 R1 R1 ω = α + ω d dt R R1 AP R R1 R1 P R R1 ( r ) [ v ( r AP )] R1 R1 P R1 P R R1 R1 P R1 P R R1 R1 P ( v ) = ( v ) + ( ω v ) = a + ( ω v ) ( R = + ω ω ) R R 1 AP R 1 P R R1 R 1 P + α r + a + 2 ω v R1 1 a a r R P R A R R AP Rigid Body Dynamics K. Craig 37

Anatomy of Coriolis and Centripetal Acceleration Situation: A turntable, with its center pivot O fixed to ground, is rotating clockwise at a constant angular rate. An ant is at point P on the turntable walking at a constant speed v, relative to the turntable, towards some food at point 2. What is the absolute acceleration of the ant? R P a Rigid Body Dynamics K. Craig 38

= + ω ω R R 1 OP R 1 P R R1 R 1 P + α r + a + 2 ω v R R1ˆ R R1ˆ ˆ R R1 = k ˆ ˆ ω 1 ( ω k1 rj 1) + 2 ω k1 vj 1 ω r ˆj 2ωvˆi R1 R1 a a ( r ) R P R O R R OP 2 = 1 + 1 = Centripetal Acceleration + Coriolis Acceleration Rigid Body Dynamics K. Craig 39

The approximate acceleration of the ant with respect to the R reference frame is the difference between its velocity at points 2 and 1 divided by t. Then we take the limit as t 0. The result is: R P a Acceleration in the y direction: radial [ ] [ ] Δ v = 2 + 4 v= vcos θ ω (r+δr)sinθ v = ω Δ ω Δ Δv 2 = = ω Δt 2 2 2 v r t v( t) v radial aradial lim r Δ t 0 Centripetal Acceleration due to term 4 v has no effect on a radial depends on ant s position cosθ 1 sin θ θ Rigid Body Dynamics K. Craig 40

Acceleration in the x direction: tangential [ ] [ ] Δ v = 1 + 3 rω= vsin θ+ω (r+δr)cosθ rω [ ] = vωδ t+ω r+ωvδt rω Δv tangential atangential = lim =ω v+ω v= 2ωv Δ t 0 Coriolis Acceleration Δt cosθ 1 sin θ θ independent of ant s position effect of ω on v (term 1 ) is always exactly the same as the effect of v on ω (term 3 ). effect of ω changing the orientation of v is exactly the same as the effect of v carrying rω to a different radius, changing its magnitude. Rigid Body Dynamics K. Craig 41

Rigid Body 3D Kinematics Example Rigid Body Dynamics K. Craig 42

Rigid Body Kinematics Essential Example Given: Find: R R R R1 ω = = R ω = 1 = 1 2 P a 5i ˆ constant 4k ˆ constant R r = 0.06 m R 1 R 2 O θ = 30º y 2 ˆi ˆ 1 1 0 0 i ˆj = 0 cosα sinα ˆj kˆ 0 sinα cosα kˆ 1 1 y 1 x 2 y 1 α y Reference Frames: R ground: xyz R 1 shaft: x 1 y 1 z 1 R 2 disk: x 2 y 2 z 2 O φ x 1 z O = θ ˆ + θ ( ) ( ) OP r rcos i1 rsin j1 ˆ z 1 Rigid Body Dynamics K. Craig 43

R R R 2 2 R P RO R R ( ) 2 R R 2 OP R R 2 OP a = a + ω ω r + α r R 2 P R R2 R 2 P + a + 2 ω v O a = 0 Point O at end of rotating shaft is fixed in R P a = 0 Point P is fixed in R P 2 (disk) v = 0 R R ( ) ( 5i 4k ) 1 R2 R R1 R1 R2 ω = ω + ω = + R R R2 R R d ω d 2 α = = 5i ˆ+ 4kˆ dt dt dk = + = ω dt = 4 5i ˆ kˆ = 20j ˆ R 1 0 4 4 kˆ 1 1 1 ( R R ) 1 1 = 20 ˆjcosα+ ksin ˆ α ( ) Rigid Body Dynamics K. Craig 44 ˆ 1 ˆ

After Substitution and Simplification: ˆ ˆ ˆ ( ) ( ) ( ) R P a = 16rcosθ i1+ 41rsinθ j1+ 40rcosθ k1 Alternate Solution: R R = + ω ( ω ) + α R 1 P R R1 R 1 P + a + 2 ω v R1 R1 R1 a a r r R P R O R R OP R OP R O a = 0 R1 ω = = 5i ˆ constant R R R d ω 0 dt 1 R1 α = = ( ) ( ) OP r = rcosθ i1+ rsinθ j1 ˆ ˆ Rigid Body Dynamics K. Craig 45

R R R R ( ) 1 P R 1 O R 1 R2 R 1 R 2 OP R 1 R 2 OP a = a + ω ω r + α r 1 O a = 0 R ω = 4kˆ 1 R1 R 1 R2 R1 R d ω d α = = 4kˆ 1 0 dt dt = ( ) 1 P 1 O 1 2 OP v = v + ω r 1 O OP v = 0 r = ( rcosθ ) ˆi + ( rsinθ) ˆj 1 2 1 2 R R R R R After Substitution and Simplification: 1 1 (P is fixed in R 2 ) ˆ ˆ ˆ ( ) ( ) ( ) R P a = 16rcosθ i1+ 41rsinθ j1+ 40rcosθ k1 Same Result Rigid Body Dynamics K. Craig 46

Kinetics of a Rigid Body Rigid Body Degrees of Freedom Linear Momentum Angular Momentum Mass Moments of Inertia & Parallel Axis Theorem Principal Axes and Planes of Symmetry Translation Theorem for Angular Momentum Equations of Motion Euler s Equations Kinetic Energy and Work-Energy Principle Impulse-Momentum Principle Essential Example Problem Rigid Body Dynamics K. Craig 47

Rigid Body Degrees of Freedom If a system of particles becomes a continuum and the measured distances between points in the system remains constant, the system is said to be a rigid body. The same laws of motion that influence a system of particles must also govern the motion of a rigid body. The difference is that with a continuum present, the summation of physical quantities for discrete particles now becomes an integration over the whole volume. An unconstrained rigid body has 6 degrees of freedom (3 translational and 3 rotational) and 6 equations of motion are needed to specify its motion. Rigid Body Dynamics K. Craig 48

Linear Momentum of a Rigid Body C mass center O reference point in body B xyz body-fixed axes XYZ ground axes L= linear momentum of rigid body = = + ω B B R O R B center of mass = v dm+ ω rdm location total mass B B R O R 1 B = m v + ( ω mr) m= dm r = rdm m B B RO R B RC = m v + ( ω r) = m v R O R B L vdm v ( r) dm R C m v Rigid Body Dynamics K. Craig 49 L =

Angular Momentum of a Rigid Body C mass center O reference point in body B xyz body-fixed axes XYZ ground axes Angular Momentum of B O about point O H = (r v)dm H O B = + ω B R O R B r v ( r) dm v r dm r ( r) dm r ( r) dm R O R B R B = + ω = ω B B B R Rigid Body Dynamics K. Craig 50 B O v = 0 if point O is fixed in R r dm = 0 if point O coincides with C

H R ω R B HO = r ( ω r)dm B R B H = ρ ( ω ρ)dm B HO = H ˆ ˆ ˆ xi+ Hyj+ Hzk ω = ω ˆi+ ω ˆj+ ω kˆ R B R B R B R B x y z Here we assume either point O is fixed in R or coincident with point C. O B Independent of the orientation of the xyz body-fixed axes, but their components are not. Point O is fixed in R Point C is the mass center of B Z Rigid Body Dynamics K. Craig 51 Y z Ground R y X r O dm r x ρ C Rigid Body B

We can show by integration that: = + 2 2 I x (y z )dm B = + 2 2 I y (x z )dm B = + 2 2 I z (x y )dm B R B Hx Ix Ixy Ixz ωx R B H y = Iyx Iy Iyz ω y R B Hz Izx Izy I z ωz Mass Moments of Inertia Inertia Matrix Mass Products of Inertia I = (xy)dm= I Rigid Body Dynamics K. Craig 52 xy xz yz B I = (xz)dm= I B I = (yz)dm= I Note: the elements of the inertia matrix are for a particular point and a particular orientation of the xyz body-fixed axes. B yx zx zy

Parallel Axis Theorem of Inertia Matrix There is an inertia matrix associated with every point of a rigid body. Consider two parallel coordinate systems fixed to a rigid body: x 1 y 1 z 1 and x 2 y 2 z 2. Let point 1 coincide with the mass center C. 2 2 b c ab ac 2 2 I = I + m ab c a bc 2 C + 2 2 [ ] [ ] x = x + a 2 1 y = y + b 2 1 z = z + c 2 1 + ac bc a + b Rigid Body Dynamics K. Craig 53

Principal Axes It is often convenient to deal with rigid-body dynamics problems using the coordinate system fixed in the body for which all products of inertia are zero simultaneously, i.e., the inertia matrix is diagonal. The 3 mutually perpendicular axes are called principal axes. The 3 mass moments of inertia are called principal moments of inertia. The 3 planes formed by the principal axes are called principal planes. Rigid Body Dynamics K. Craig 54

Plane of Symmetry Many rigid bodies have a plane of symmetry. For example, if the xy plane is a plane of symmetry, then for every mass element with coordinates (x, y, z) there exists a mass element with coordinates (x, y, -z). Hence I = I = 0 yz Translation Theorem for Angular Momentum The angular momentum of a body B about any point P(on or off the body, fixed or moving) can be expressed as: Rigid Body Dynamics K. Craig 55 xz PC H P = (r L) + H

H = (r v)dm P B Derivation R C R B = r v + ( ω ρ) dm B PC R C R B = (r +ρ ) v + ( ω ρ) dm B PC RC RC = (r v )dm + ( ρ v )dm+ B B Rigid Body B PC R B R B r ( ω ρ ) dm + ρ ( ω ρ) dm B B Y PC R C 0 RC = r m v v ρ dm + B PC R B 0 R B r ω ρ dm + ρ ( ω ρ) dm Z X B B Ground R PC H P = (r L) + H Rigid Body Dynamics K. Craig 56 r P dm r PC ρ C

Equations of Motion The six scalar equations of motion for a rigid body are given by the two vector equations: F R R R C d d v F= L= m dt dt R R d d M= H or MO = H dt dt is the resultant of all external forces acting on the body. M(M ) is the resultant moment of external forces O and couples about the mass center C (fixed point O). O Rigid Body Dynamics K. Craig 57

Let s express these equations in terms of the bodyfixed xyz coordinate system. R B R B R B R B ω = ω ˆ H= H ˆ ˆ ˆ xi+ Hyj+ Hzk xi+ ω ˆ yj+ ωzkˆ R C R C R C R C v = v ˆi+ v ˆj+ v kˆ H = H ˆi+ H ˆj+ H kˆ x y z R C R C R B R C R B x z y y z R C R C R B R C R B y x z z x R C R C R B R C R B z y x x y O O O O R R C B R C d v d v R B R C = + ( ω v ) dt dt = ˆ ˆ ˆ ˆ ˆ ˆ = ( v + v ω v ω )i ˆ x y z R C R C R C R B R B R B R C ( vi x + vyj+ vzk) + ( ω xi+ ω yj+ ω zk) v + ( v + v ω v ω )j ˆ + ( v + v ω v ω )kˆ Rigid Body Dynamics K. Craig 58

R B dh dh R ( B = + ω H) dt = dt ˆ ˆ ˆ ω R B ( Hi x + Hj y + Hk) z + ( H) R B R B = (H x + H ˆ z ωy H y ωz)i + (H + H ω H ω )j ˆ R B R B y x z z x + (H + H ω H ω )kˆ R B R B z y x x y The inertia matrix is constant with respect to time since it is expressed in the bodyfixed coordinate system. So we can write: R B Hx Ix Ixy Ixz ωx R B H y = Iyx Iy Iyz ω y R B Hz Izx Izy Iz ω z R B H x Ix Ixy Ixz ω x R B H y = Iyx Iy Iyz ω y R B H z Izx Izy I z ω z Rigid Body Dynamics K. Craig 59

The velocity terms refer to: R The angular velocity terms refer to: The xyz axes are body-fixed axes. C v R Six Scalar Equations of Motion M = I ω + I ( ω ω ω ) + I ( ω +ω ω ) + x x x xy y x z xz z x y (I I ) ω ω + I ( ω ω ) 2 2 z y y z yz y z M = I ω + I ( ω +ω ω ) + I ( ω ω ω ) + y y y xy x y z yz z x y 2 2 (Ix I z) ωxω z + I xz( ωz ωx) M = I ω + I ( ω ω ω ) + I ( ω +ω ω ) + z z z xz x y z yz y x z (I I ) ω ω + I ( ω ω ) 2 2 y x x y xy x y Rigid Body Dynamics K. Craig 60 ω B F = m v + v ω v ω F = m v + v ω v ω F = m v + v ω v ω x x z y y z y y x z z x z z y x x y The moments and inertia terms are with respect to axes fixed in the body with origin at C, the mass center.

If we assume that the xyz body-fixed axes are principal axes (1, 2, and 3), then all the products of inertia are zero, and the mass moments of inertia are identified as: Ix = I1 Iy = I2 Iz = I3 The three rotational equations then are: I 1ω+ 1 (I3 I 2) ωω 2 3 = M1 I 2ω 2 + (I1 I 3) ωω 1 3 = M2 I ω + (I I ) ωω = M 3 3 2 1 1 2 3 ω ω Euler s Equations Note: If only z and z are nonzero in the general equations, then: 2 Mx = Ixzω z Iyzω For these to be zero, z the xy plane must be a 2 My = Iyzω z + Ixzωz plane of symmetry: M = I ω I = I = 0 z z z xz yz Rigid Body Dynamics K. Craig 61

Kinetic Energy T = Kinetic Energy v is the velocity of particle dm with respect to R 1 1 = = R B 2 2 ω =ω 1 y = (v ) (v ) dm 2 +ω ρ i +ω ρ 1 T = (v +ω ρ ) (v +ω ρ) dm 2 i B 1 1 = m(viv) + [ ( ω ρ) i( ω ρ) ] dm 2 2 2 dt v dm (v i v)dm = T + T translation B rotation Note: ρ dm = 0 dm Rigid Body Dynamics K. Craig 62 B x C Y z Rigid Body B Z Ground R ρ X

1 Ttranslation = m(v i v) 2 1 T [ ] rotation = ( ω ρ) i( ω ρ) dm 2 B 1 = [ ωi( ρ ( ω ρ)) ] dm 2 B 1 = ω i H 2 vector identity 1 [ ] T T= m v [ v] 2 1 T + ω ω 2 [ ] I [ ] 1 2 2 2 = (I xω x + I yω y + I zωz ) 2 + I ω ω + I ω ω + I ω ω xy x y yz y z zx z x Rigid Body Dynamics K. Craig 63

T 1 T = ω ω O 2 [ ] [ I] [ ] If the body has a fixed point O in inertial space and the origin of the xyz coordinate system is at this point, then the total kinetic energy T is entirely due to rotational motion about the fixed point. Rigid Body Dynamics K. Craig 64

Work-Energy Equation F M U1 2 The resultant of all external forces acting on the rigid body The resultant moment of external forces and couples acting on the rigid body about the center of mass C The work done by all external forces and couples in time interval from t 1 to t 2 t 2 2 U 1 2 = (Fiv)dt + (M iω)dt t1 t1 U = T T 1 2 2 1 t Rigid Body Dynamics K. Craig 65

Impulse-Momentum Principle Integration of the force equation with respect to time yields the theorem that the linear impulse of a rigid body is equal to the change in linear momentum. d v = = dt R R C t 2 F m Fdt m v(t t 2) v(t 1) 1 [ ] Similarly, integration of the moment equation with respect to time yields the theorem that the angular impulse of a rigid body is equal to the change in angular momentum. R t d = = dt 2 M H Mdt t H(t 2) H(t 1) 1 Rigid Body Dynamics K. Craig 66

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Rigid Body 3D Kinetics Example Rigid Body Dynamics K. Craig 78

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