Phys 7221 Homework # 8
|
|
- Nathaniel Brown
- 6 years ago
- Views:
Transcription
1 Phys 71 Homework # 8 Gabriela González November 15, 6 Derivation 5-6: Torque free symmetric top In a torque free, symmetric top, with I x = I y = I, the angular velocity vector ω in body coordinates with axes along the principal axes, is given by ω x = ω cos Ωt ω y = ω sin Ωt with Ω = ω 3 (I 3?I)/I. components ω z = ω 3 Also in body coordinates, the angular momentum vector L has L x = Iω x = Iω cos Ωt L y = Iω y = Iω sin Ωt L z = I 3 ω z = I 3 ω 3 We see that the component of the angular momentum along the top symmetry axis, L z, is constant, and the the components perpendicular to the symmetry axis rotate about the z axis with angular velocity Ω. Since the top is free of external torques, the angular momentum vector L is constant in an inertial system. In an inertial system, it is the tops symmetry axis the one that rotates about the L direction. We can use Euler angles to describe the rotation between body axes and inertial axes. Comparing the expressions for the components of ω in body axes we obtained and the expressions for the same components in terms of Eulers angles in (4.87): ω x = ω cos Ωt = φ sin θ sin ψ + θ cos ψ ω y = ω sin Ωt = φ sin θ cos ψ + θ sin ψ ω z = ω 3 = φ cos θ + ψ we recognize that the Euler angle θ is constant, the bodys rotation about the symmetry axis is ψ = π/ Ω, and φ cos θ = ω 3 ψ = ω 3 + Ω = I 3 ω 3 /I φ sin θ = ω 1
2 The angle θ is the angle between the symmetry axis and the angular momentum vector, and is determined by initial conditions. The bodys symmetry axis rotates about the angular momentum with constant angular velocity φ = I 3 ω 3 /I cos θ. We now use the Euler angles we obtained to calculate the components of ω in the inertial system, using the expression from Derivation 15 in Chapter 4: ω x = θ cos φ + ψ sin θ sin φ = Ω sin θ sin( φt + φ ) ω y = θ sin φ ψ sin θ cos φ = Ωsinθcos( φt + φ ) ω z = ψ cos θ + φ = Ω cos θ + I 3 ω 3 /I We see that the component of the angular momentum along the z-axis, or the direction of the angular momentum vector L, is constant, and the component of the angular velocity perpendicular to L is rotating with angular velocity φ. The angle between ω and L is given by sin θ = ωx + omega y/ω = Ω sin θ/ω. The angle between ω and the symmetry axis is given by sin θ = ωx + ω y /ω = ω /ω. Then we have sin θ = Ω sin θ/ω = Ω sin θ sin θ /ω = Ω sin θ / φ where we have used ω = φ sin θ, which we had obtained when solving for the Euler angles. We can also use moments of inertia for an expression of sin θ : sin θ = Ω sin θ / φ = ΩI cos θ sin θ /I 3 ω 3 = ((I 3 I)/I 3 ) cos θ sin θ For the Earth considered as a free symmetric top, we have (I 3 I)/I 3 1 3, so the angle θ is very small, independent of values of θ, θ : the angular velocity vector ω is very close to the angular momentum vector L. The measured distance R sin θ is about 1m, so the distance R sin θ = (I 3 I)/IR sin θ cos θ 15mm cos θ < 1.5cm. As seen in the body axes, the angular velocity vector describes a cone of aperture angle θ about the symmetry axis: this is called the body cone. As seen in the inertial frame, the angular velocity vector describes a cone of aperture angle θ, about the angular momentum vector: this is called the space cone. Both cones share the angular momentum vector along their sides at any given instant. The angular velocity vector is the instantaneous axis of rotation, so the cones are rolling without slipping on each other. A very nice page with animations showing this example, by Prof. Eugene Butikov, can be found in from which Fig.1 is a snapshot. Exercise 5-15 Consider a flat rigid body in the shape of a right triangle with uniform mass density σ = M/A, and area A = a /, where a is the length of the equal sides of the triangle.
3 Figure 1: A rotating torque free symmetric top (left), and the associated space and body cones. The conserved angular momentum vector (blue) is along the z axis; the (red) instantaneous angular velocity vector is at the intersection of the cones, precessing about the z axis. From Figure : Exercise
4 Let us choose the right angle corner of the triangle as the origin of a coordinate system with the x, y axis along the sides of the triangle. The boundary of the triangle is given by x + y = a; the mass elements on the surface will have coordinates (x,y,) with (x,y) within the triangle. The elements of the inertia tensor in such a system are I xx = A σ(y + z )da = σ I yy = A I xx = I zz = I yx = σxyda = σ A a dx a x σ(x + z )da = σ A a y dy = σ a a (a x) 3 3 a y dy x dx = Ma 6 σ(x + y )da = I xx + I yy = Ma 3 xdx a x I xz = I = Ma 1 A ydy = σ a σxzda = = I yz (a x) xdx dx = σ a4 1 = Ma 6 = σ a4 4 = Ma 1 We look for eigenvalues of the inertia tensor, which will be solutions to the equation det(i I1) =, or k I k det(i I1) = k k I 4k I = (4k I)((k I) k ) with k = Ma /1. The three real, positive solutions are the principal moments of inertia: (I 1, I, I 3 ) = (k, 3k, 4k) = (Ma /1, Ma /4, Ma /3). The principal axes are the eigenvectors corresponding to each eigenvalue. If the eigenvectors have components n i = (n ix, n iy, n iz ), the equations are I n i = α i k n i with α i = 1, 3, 4. The equations for n 1 are: I n 1 = kn 1 k n 1x n 1y n 1x + n 1y 4n 1z = n 1x n 1y n 1z 4 n 1x n 1y n 1z = k n 1x n 1y n 1z n 1x = n 1y, n 1z =.
5 .. The equations for n are: k n x n y n x + n y 4n z The equations for n 3 are: k n x n y n x + n y 4n z = I n = 3kn n x n y n z n x n y n z I n 3 = 4kn 3 n 3x n 3y n 3z = 4 n 3x n 3y n 3z The eigenvectors with unit magnitude are then = 3k n x n y n z n 1x = n 1y, n 1z = = 4k n 3x n 3y n 3z n 3x = n 3y =, n 3z Exercise 5-17: A rolling cone n 1 = (1/ )(1, 1, ) n = (1/ )(1, 1, ) n 3 = (,, 1) The instantaneous axis of rotation, and thus the direction of the angular velocity ω is along the line of contact of the cone with the surface. The center of mass is at a distance 3h/4 along the axis of the cone, which then will be at a vertical height a = 3h sin α/4. The total mass of the cone is πρh 3 tan α/3. Each cross section of the cone is like a disk perpendicular to the cone s axis (and thus tilted with respect to the vertical), moving in a circle. The velocity of the center of mass has magnitude v, and is related to the angular velocity is ω = v/a sin α. We also know that, if there is no slipping, the velocity f the center of mass is related to the angular velocity as v = a θ cos α, where θ is the angle along the circle, and θ = π/τ. Then ω = θ cot α. One principal axis of the cone is along its axis, say x 3, with moment of inertia I 3 ; the other two axes are in the plane perpendicular to its axis, with moments of inertia 5
6 I 1 = I = I. The angular velocity will have a component ω cos α along the cone axis, and a component ω sin α on the plane perpendicular to the axis. The moments of inertia with respect to the cone s axis are I 3 = 3MR /1, I = 3M(h +R /4)/5, where R = h tan α is the radius of the base. The moments of inertia with respect to the center of mass at a distance a = 3h/4 along axis, are I 3 = I 3 = 3MR /1, I = I Ma = 3M(R + h /4)/. If we choose the center of mass as the origin of the body axes, the kinetic energy is T = (1/)Mv + (1/)I 3 ω cos α + (1/)I 1 ω sin α = 3Mh θ (1 + 5 cos α)/4. Exercise 5-18 A weightless bar of length l has two masses of mass m at the two ends, and is rotating uniformly about an axis passing through the bar s center, making an angle θ with the bar. Figure 3: Exercise 5-18 The principal body axes for a bar are one along the bar itself, say the z axis, and the other two axes, x and y, in the plane perpendicular to the bar. The principal moments of inertia are then I 3 =, and I 1 = I = ml. Since the axis of rotation is not perpendicular to the bar, the angular velocity (along the axis of rotation by definition) will have a component along the bar, ω // = ω 3 = ω cos θ and a component in the plane perpendicular to the the bar, ω = ω sin θ. We can choose the x axis along the component of the angular velocity in the plane perpendicular to the bar, so ω 1 = ω sin θ and ω =. For uniform 6
7 velocity, Euler equations are then N 1 = I 1 ω 1 ω ω 3 (I I 3 ) = N = I ω ω 3 ω 1 (I 3 I 1 ) = ml ω sin θ cos θ N 3 = I 3 ω 3 ω 1 ω (I 1 I ) = The torque is then along the y axis: a direction perpendicular to the bar, and perpendicular to the angular velocity (which is in the x z plane). We can also calculate the torque from the time derivative of the angular momentum vector in an inertial frame. If we set up an inertial frame with the origin in the center of the bar, and the z axis along the rotation axis, then the masses have position vectors The velocities are r 1 = l(sin θ cos(ωt), sin θ sin(ωt), cos θ) r = l( sin θ cos(ωt), sin θ sin(ωt), cos θ) = r 1. v 1 = lω( sin θ sin(ωt), sin θ cos(ωt), ) v = v 1. The angular momentum of each mass is L 1 = r 1 mv 1 = ml ω(sin θ cos(ωt), sin θ sin(ωt), cos θ) ( sin θ sin(ωt), sin θ cos(ωt), ) = ml ω sin θ( cos θ cos(ωt), cos θ sin(ωt), sin θ) L = r mv = L 1 The total angular momentum is L = L 1 + L = ml ω sin θ(cos θ cos(ωt), cos θ sin(ωt), sin θ) and the total external torque N = dl dt = ml ω sin θ cos θ(sin(ωt), cos(ωt), ) The torque is a rotating vector, perpendicular to the angular velocity (which is along the z axis), and perpendicular to the bar (which is along r 1 ), just as we had obtained from Euler s equations. 7
8 5-: A plane physical pendulum Consider a uniform rod of length l and mass m, suspended in a vertical plane by one end. At the other end, there is a uniform disk of radius a and mass M attached, which can rotate freely in the vertical plane. The systems configuration can be described at any time with two angles: the angle θ the rod makes with the vertical direction, and the angle φ a reference direction on the the disk makes with the vertical. If the disk is rigidly attached to the rod, then θ = φ. Let us take an inertial system with the origin at the suspension point, a horizontal x axis, and the y axis pointing down along the vertical direction. The velocity of a mass element dm a distance s along the rod will be s θ. The kinetic energy of the rod is then T rod = (1/)(m/l) l s θ ds = (1/)(m/l)(l 3 /3) θ = ml θ /6. The velocity of points in the disk are equal equal to ṙ a + ω a r, where r is the position vector of the mass element in the reference frame fixed to the rotating disk, r a is the position of the center of mass of of the disk, and the the angular velocity vector ω a describes the rotation of the disk in an inertial system. The magnitude of the angular velocity vector ω a is = φ, and its direction is perpendicular to the motion plane. The squared speed of mass elements will then be v = va + ṙ a (ω a r ) + ωar, and the kinetic energy of the disk will be T disk = 1 v dm = 1 (va + ṙ a (ω a r ) + ω ar )dm. We recognize that the third term will lead to a term (1/)I ω a in the kinetic energy, with I = Ma / the moment of inertia of the disk with respect to the center of mass. The integral of the second term will vanish, since r dm is the position of the center of mass in a coordinate system where the center of mass is at the origin. The velocity of the center of mass of the disk ṙ a is equal to ṙ + ω a a, where r is the position vector of the attachment point, and a is the position of the center of mass of the disk with respect to the attachment point. Then va = (ṙ + ω a a) (ṙ + ω a a) = v + v (ω a a) + ωaa = l θ + ω a (a v ) + a φ The velocity of the attachment point v is tangent to the disk, so the direction of the cross product a v is perpendicular to the plane of motion, and ω a (a v ) = ω a a v. The position vector of the attachment point is r = l(cos θ, sin θ, ), and its velocity vector is v = l θ( sin θ, cos θ, ). The position of the center of mass of the disk with respect to the attachment point is a = a(cos φ, sin φ, ). Thus, a v = a(cos φ, sin φ, ) l θ( sin θ, cos θ, ) 8
9 The kinetic energy of the disk is then = al θ(,, cos φ cos θ + sin φ sin θ) = al θ cos(φ θ)ˆk T disk = 1 Mv a + I ω a and the total kinetic energy is = 1 M(l θ + al θ φ cos(φ θ) + a φ ) + 1 Ma φ = 1 Ml θ + Mal θ φ cos(φ θ) Ma φ T = T rod + T disk = 1 6 ml θ + 1 Ml θ + Mal θdotφcos(φ θ) Ma φ = 1 6 (3M + m)l θ + Mal θ φcos(φ θ) Ma φ The potential energy of the rod is V rod = mg(l/) cos θ, and the potential energy of the disk is V disk = Mg(l cos θ + a cos φ), so the total potential energy is V = 1 mgl cos θ Mg(l cos θ + a cos φ) = 1 (m + M)gl cos θ Mga cos φ The Lagrangian is L = T V = 1 Ml θ + Mal θ φ cos(φ θ) Ma φ + 1 (m + M)gl cos θ + Mga cos φ Lagrange s equation for θ is: d ) (Ml θ + Mal φ cos(φ θ) dt (Mal θ φ sin(φ θ) 1 ) (m + M)glsinθ = Ml θ + Mal φ cos(φ θ) Mal φ sin(φ θ) + 1 (m + M)gl sin θ = Lagrange s equation for φ is d dt ( 3 Ma φ + Mal θ cos(φ θ) ) ( Mal θ φ ) sin(φ θ) Mgasinφ = 3 Ma φ + Mal θ cos(φ θ) + Mal θ cos(φ θ) + mga sin φ = 9
10 Exercise 5: A rolling sphere The sphere is a rigid body described by the position of its center of mass r = (x, y, z), and its orientation, defined by Euler angles θ, φ, ψ. The constraint of the sphere remaining on the surface (the effect of the normal force responding to gravity) is z = R (if the origin of the coordinate system is on the surface). At any instant, the sphere is rotating about an axis on its point of contact with the surface, which is the instantaneous axis of rotation and the direction of the angular velocity vector ω, and thus ω = (ω x, ω y, ω z ). If the sphere is not slipping, the angular velocity vector is perpendicular to the velocity of the center of mass, ω v =, which we can express as a condition for an angle Θ to exist (defining the direction of the horizontal component of the angular velocity) such that ω = (Ω cos Θ, Ω sin Θ, ω z ) and v = v(sin Θ, cos Θ). Moreover, not slipping also means that the instantaneous velocity of the contact point is zero: = v c = v + ω r = v(sin Θ, cos Θ, ) + (Ω cos Θ, Ω sin Θ, ω z ) (,, R) = (v RΩ)(sin Θ, cos Θ, ) which then means v = RΩ. The angular velocity vector is related to Euler angles, and thus our constraints are Rω x = v cos Θ = ẏ = R( θ cos φ + ψ sin θ sin φ) Rω y = v sin Θ = ẋ = R( θ sin φ ψ sin θ cos φ) It may be useful to compare these conditions with the problem of a rolling disk considered in Chapter 1: our equations are the same as (1.39), since the disk would have ψ =. We also know that ω z = ψ cos θ + φ, but that is not a constraint, of course. The angular velocity ω z is the instantaneous spinning velocity of the sphere about a vertical axis. The sphere then has 6 coordinates, and three constraints, so the system only has three degrees of freedom. We d like to choose these as the three Euler angles, for example, using the two costraints to solve for the x, y. However, the constraints are non-holonomic and do not allow us to integrate them for x, y. To prove the constraints are non-holonomic, we consider the differentials df x (x, θ, ψ, φ) = R(dθ sin φ dψ sin θ cos φ) dx df y (y, θ, ψ, φ) = R(dθ cos φ + dψ sin θ sin φ) + dy 1
11 and we prove they are not exact differentials (they are not derivatives of a function): d df x dφ dθ = d dφ R sin φ = R cos φ d df x dθ dφ = d df y dφ dθ = d dφ R cos φ = R sin φ d df y dθ dφ = We now want to write Lagrange s equations of motion, using Lagrange multipliers. The potential energy is constant, so we only have kinetic energy. The kinetic energy has a translational part, and a rotational part. The rotational energy is especially simple since the sphere has identical moments of inertia about the principal axes: L = T = 1 mv v + 1 Iω ω = 1 m(ẋ + ẏ ) + 1 I(( θ cos φ + ψ sin θ sin φ) + ( θ sin φ ψ sin θ cos φ) + ( ψ cos θ + φ) ) = 1 m(ẋ + ẏ ) + 1 I( θ + ψ + φ + φ ψ cos θ) The non-holonomic constraints are F x = R( θ sin φ ψ sin θ cos φ) ẋ F y = R( θ cos φ + ψ sin θ sin φ) + ẏ Lagrange s equations are mẍ = µ x F x ẋ = µ x F y mÿ = µ y ẏ = µ y I θ + I φ ψ F x sin θ = µ x θ + µ F y y θ = R(µ x sin φ + µ y cos φ) I ψ + I( φ cos θ φ θ F x sin θ) = µ x ψ + µ F y y ψ = R sin θ( µ x cos φ + µ y sin φ) I φ + I( ψ cos θ ψ θ sin θ) = = I d dt ( φ + ψ cos θ) = I dω z dt We have then 7(!) differential equations (five Lagrange equations and two constraints) for the eight unknowns x, y, φ, θ, ψ, µ x, µ y. 11
12 The last equation (Lagrange s equation for ψ) says that the angular velocity ω z is constant. This is because φ is a cyclical variable in the Lagrangian and because the constraints do not depend on φ. Notice that ψ is also a cyclical variable in the Lagrangian, but it is not associated with a conserved quantity because the constraints depend on ψ. Since there is no dissipation or forces doing any work, the energy (equal to the kinetic energy) is conserved: E = T + V = T tr + T rot + = 1 mv v + 1 Iω ω = 1 mv + 1 I(Ω + ω z) Due to the constraint v = RΩ, the energy is equal to E = 1 (mr +I)Ω + 1 Iω z. Since ω z is constant, then we know that Ω (and thus also v) are constant. Therefore, the translational kinetic energy 1 mv and the rotational energy 1 I(Ω + ω z) are separately conserved. Exercise 3: Closing tilted door The door s axis of rotation is along the hinges, which make an angle θ with the vertical (normally, on firm ground and for a well aligned door, of course, θ = ). The position of the door as a rigid body is determined by the angle about its axis of rotation. The principal axes of the door are an axis perpendicular to the door, and two axes in the door plane, parallel to the door s sides (assumed to be straight). We choose a body coordinate system with an origin in the bottom door s corner along the hinged side. We choose an inertial coordinate system with the same origin, a vertical axis z pointing up, and two perpendicular axis x, y in the horizontal plane. We choose body axes x along the short side of the door, z along its hinged side (the axis of rotation), and y perpendicular to the door. Following the convention for Euler angles, and placing the door in the x z plane in Figure 4.7, we see that the angle θ is the angle between the door and the vertical; the angle φ =, and the angle ψ is the one that defines the position (angle) of the door. The initial position of the door ( lifted by 9 degrees) has an angle ψ =. The final equilibrium position of the door (when the door is shut ) has an angle ψ = π/. The gravitational potential energys is V = mg r = mgz, where r is the position of the center of mass in the inertial system, and z its vertical component. The position of the center of mass in the body axes system is r = (x, y, z ) = (w,, h)/, where w is the door s width and h is the doors height. In the inertial system, the position of the center of mass is given by the transformation (4.47), r = A 1 r. The vertical coordinate will be z = x sin θ sin ψ + y sin θ cos ψ + z cos θ = (w/) sin θ sin ψ + (h/) cos θ The gravitational potential energy is then V = mgz = mg(w sin θsinψ + h cos θ)/ 1
13 Since the motion is a pure rotation, the kientic energy has the form T = (1/)I ω, where I is the moment of inertia about the axis of rotation (the door s hinged side, the z axis), and ω = ψ is the angular velocity: T rot = 1 I z ψ. The moment of inertia of the door with respect to the z axis, is calcualted integrating over the points on the door (all with y = coordinates: I z = (x + y )σdx dz = m wh The kinetic energy is then The total energy is w x dx h T = 1 I z ψ = 1 6 mw ψ. dz = m w 3 wh 3 h = 1 3 mw. E = T + V = 1 6 mw ψ + mg (w sin θsinψ + h cos θ). In the initial position, ψ = and ψ =. Since the energy is conserved, we obtain E = 1 6 mw ψ + mg mgh (w sin θsinψ + h cos θ) = cos θ, an expression we can use as a differential equation for ψ: ψ 3g sin θ = sin ψ w Notice that since π/ < ψ <, the expression on the left hand side is a positive expression. The angular velocity when the door reaches the equilibrium position at ψ = π/ (where it will not stop, but oscillate about, if it can go through the shut position) is then 3g sin θ ψ f = w. The time it will take to reach that position can be obtained the equation for ψ: dψ dt = dt = t = = 3g sin θ sin ψ w w dψ 3g sin θ sin ψ w π/ dψ 3g sinψ w dψ 3g sin θ sin ψ π/ 13
14 w = F (π/4, ) 3g sin θ w = 1.51 g sin θ sin θ = w ( ) 1.51 =.9m ( ) 1.51 g t 9.8m/s = 1.3 3s Three seconds is a looong time for a door to close, so we obtain a small hinge angle. The smallest time for the door to close is when the hinges are horizontal, θ = π/ and t = 1.51 w/g =.45sec. The wider the door, the longer it takes to close. 14
CP1 REVISION LECTURE 3 INTRODUCTION TO CLASSICAL MECHANICS. Prof. N. Harnew University of Oxford TT 2017
CP1 REVISION LECTURE 3 INTRODUCTION TO CLASSICAL MECHANICS Prof. N. Harnew University of Oxford TT 2017 1 OUTLINE : CP1 REVISION LECTURE 3 : INTRODUCTION TO CLASSICAL MECHANICS 1. Angular velocity and
More informationSOLUTIONS, PROBLEM SET 11
SOLUTIONS, PROBLEM SET 11 1 In this problem we investigate the Lagrangian formulation of dynamics in a rotating frame. Consider a frame of reference which we will consider to be inertial. Suppose that
More informationPhysics 106a, Caltech 4 December, Lecture 18: Examples on Rigid Body Dynamics. Rotating rectangle. Heavy symmetric top
Physics 106a, Caltech 4 December, 2018 Lecture 18: Examples on Rigid Body Dynamics I go through a number of examples illustrating the methods of solving rigid body dynamics. In most cases, the problem
More informationPart 8: Rigid Body Dynamics
Document that contains homework problems. Comment out the solutions when printing off for students. Part 8: Rigid Body Dynamics Problem 1. Inertia review Find the moment of inertia for a thin uniform rod
More information1/30. Rigid Body Rotations. Dave Frank
. 1/3 Rigid Body Rotations Dave Frank A Point Particle and Fundamental Quantities z 2/3 m v ω r y x Angular Velocity v = dr dt = ω r Kinetic Energy K = 1 2 mv2 Momentum p = mv Rigid Bodies We treat a rigid
More informationProblem 1. Mathematics of rotations
Problem 1. Mathematics of rotations (a) Show by algebraic means (i.e. no pictures) that the relationship between ω and is: φ, ψ, θ Feel free to use computer algebra. ω X = φ sin θ sin ψ + θ cos ψ (1) ω
More informationMarion and Thornton. Tyler Shendruk October 1, Hamilton s Principle - Lagrangian and Hamiltonian dynamics.
Marion and Thornton Tyler Shendruk October 1, 2010 1 Marion and Thornton Chapter 7 Hamilton s Principle - Lagrangian and Hamiltonian dynamics. 1.1 Problem 6.4 s r z θ Figure 1: Geodesic on circular cylinder
More informationQuestion 1: A particle starts at rest and moves along a cycloid whose equation is. 2ay y a
Stephen Martin PHYS 10 Homework #1 Question 1: A particle starts at rest and moves along a cycloid whose equation is [ ( ) a y x = ± a cos 1 + ] ay y a There is a gravitational field of strength g in the
More information12. Rigid Body Dynamics I
University of Rhode Island DigitalCommons@URI Classical Dynamics Physics Course Materials 015 1. Rigid Body Dynamics I Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License
More informationClassical Mechanics. Luis Anchordoqui
1 Rigid Body Motion Inertia Tensor Rotational Kinetic Energy Principal Axes of Rotation Steiner s Theorem Euler s Equations for a Rigid Body Eulerian Angles Review of Fundamental Equations 2 Rigid body
More informationVariation Principle in Mechanics
Section 2 Variation Principle in Mechanics Hamilton s Principle: Every mechanical system is characterized by a Lagrangian, L(q i, q i, t) or L(q, q, t) in brief, and the motion of he system is such that
More informationChapter 9 Notes. x cm =
Chapter 9 Notes Chapter 8 begins the discussion of rigid bodies, a system of particles with fixed relative positions. Previously we have dealt with translation of a particle: if a rigid body does not rotate
More informationP321(b), Assignement 1
P31(b), Assignement 1 1 Exercise 3.1 (Fetter and Walecka) a) The problem is that of a point mass rotating along a circle of radius a, rotating with a constant angular velocity Ω. Generally, 3 coordinates
More informationHomework 7: # 4.22, 5.15, 5.21, 5.23, Foucault pendulum
Homework 7: # 4., 5.15, 5.1, 5.3, Foucault pendulum Michael Good Oct 9, 4 4. A projectile is fired horizontally along Earth s surface. Show that to a first approximation the angular deviation from the
More informationPhysics 351, Spring 2018, Homework #9. Due at start of class, Friday, March 30, 2018
Physics 351, Spring 218, Homework #9. Due at start of class, Friday, March 3, 218 Please write your name on the LAST PAGE of your homework submission, so that we don t notice whose paper we re grading
More information13. Rigid Body Dynamics II
University of Rhode Island DigitalCommons@URI Classical Dynamics Physics Course Materials 2015 13. Rigid Body Dynamics II Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License
More informationPHY 5246: Theoretical Dynamics, Fall Assignment # 9, Solutions. y CM (θ = 0) = 2 ρ m
PHY 546: Theoretical Dnamics, Fall 5 Assignment # 9, Solutions Graded Problems Problem (.a) l l/ l/ CM θ x In order to find the equation of motion of the triangle, we need to write the Lagrangian, with
More informationClassical Mechanics III (8.09) Fall 2014 Assignment 3
Classical Mechanics III (8.09) Fall 2014 Assignment 3 Massachusetts Institute of Technology Physics Department Due September 29, 2014 September 22, 2014 6:00pm Announcements This week we continue our discussion
More information9 Kinetics of 3D rigid bodies - rotating frames
9 Kinetics of 3D rigid bodies - rotating frames 9. Consider the two gears depicted in the figure. The gear B of radius R B is fixed to the ground, while the gear A of mass m A and radius R A turns freely
More information7 Kinematics and kinetics of planar rigid bodies II
7 Kinematics and kinetics of planar rigid bodies II 7.1 In-class A rigid circular cylinder of radius a and length h has a hole of radius 0.5a cut out. The density of the cylinder is ρ. Assume that the
More informationM2A2 Problem Sheet 3 - Hamiltonian Mechanics
MA Problem Sheet 3 - Hamiltonian Mechanics. The particle in a cone. A particle slides under gravity, inside a smooth circular cone with a vertical axis, z = k x + y. Write down its Lagrangian in a) Cartesian,
More informationRigid bodies - general theory
Rigid bodies - general theory Kinetic Energy: based on FW-26 Consider a system on N particles with all their relative separations fixed: it has 3 translational and 3 rotational degrees of freedom. Motion
More informationGeneral Physics I. Lecture 10: Rolling Motion and Angular Momentum.
General Physics I Lecture 10: Rolling Motion and Angular Momentum Prof. WAN, Xin (万歆) 万歆 ) xinwan@zju.edu.cn http://zimp.zju.edu.cn/~xinwan/ Outline Rolling motion of a rigid object: center-of-mass motion
More informationTwo-Dimensional Rotational Kinematics
Two-Dimensional Rotational Kinematics Rigid Bodies A rigid body is an extended object in which the distance between any two points in the object is constant in time. Springs or human bodies are non-rigid
More informationRotational & Rigid-Body Mechanics. Lectures 3+4
Rotational & Rigid-Body Mechanics Lectures 3+4 Rotational Motion So far: point objects moving through a trajectory. Next: moving actual dimensional objects and rotating them. 2 Circular Motion - Definitions
More informationClassical Mechanics Comprehensive Exam Solution
Classical Mechanics Comprehensive Exam Solution January 31, 011, 1:00 pm 5:pm Solve the following six problems. In the following problems, e x, e y, and e z are unit vectors in the x, y, and z directions,
More informationRotational Kinematics and Dynamics. UCVTS AIT Physics
Rotational Kinematics and Dynamics UCVTS AIT Physics Angular Position Axis of rotation is the center of the disc Choose a fixed reference line Point P is at a fixed distance r from the origin Angular Position,
More informationPLANAR KINETIC EQUATIONS OF MOTION (Section 17.2)
PLANAR KINETIC EQUATIONS OF MOTION (Section 17.2) We will limit our study of planar kinetics to rigid bodies that are symmetric with respect to a fixed reference plane. As discussed in Chapter 16, when
More informationPHY6426/Fall 07: CLASSICAL MECHANICS HOMEWORK ASSIGNMENT #1 due by 9:35 a.m. Wed 09/05 Instructor: D. L. Maslov Rm.
PHY646/Fall 07: CLASSICAL MECHANICS HOMEWORK ASSIGNMENT # due by 9:35 a.m. Wed 09/05 Instructor: D. L. Maslov maslov@phys.ufl.edu 39-053 Rm. 4 Please help your instructor by doing your work neatly.. Goldstein,
More informationRigid body simulation. Once we consider an object with spatial extent, particle system simulation is no longer sufficient
Rigid body dynamics Rigid body simulation Once we consider an object with spatial extent, particle system simulation is no longer sufficient Rigid body simulation Unconstrained system no contact Constrained
More informationRigid Body Dynamics, SG2150 Solutions to Exam,
KTH Mechanics 011 10 Calculational problems Rigid Body Dynamics, SG150 Solutions to Eam, 011 10 Problem 1: A slender homogeneous rod of mass m and length a can rotate in a vertical plane about a fied smooth
More informationAssignments VIII and IX, PHYS 301 (Classical Mechanics) Spring 2014 Due 3/21/14 at start of class
Assignments VIII and IX, PHYS 301 (Classical Mechanics) Spring 2014 Due 3/21/14 at start of class Homeworks VIII and IX both center on Lagrangian mechanics and involve many of the same skills. Therefore,
More informationRotational motion problems
Rotational motion problems. (Massive pulley) Masses m and m 2 are connected by a string that runs over a pulley of radius R and moment of inertia I. Find the acceleration of the two masses, as well as
More informationA set of N particles forms a rigid body if the distance between any 2 particles is fixed:
Chapter Rigid Body Dynamics.1 Coordinates of a Rigid Body A set of N particles forms a rigid body if the distance between any particles is fixed: r ij r i r j = c ij = constant. (.1) Given these constraints,
More informationSymmetries 2 - Rotations in Space
Symmetries 2 - Rotations in Space This symmetry is about the isotropy of space, i.e. space is the same in all orientations. Thus, if we continuously rotated an entire system in space, we expect the system
More informationOscillatory Motion. Solutions of Selected Problems
Chapter 15 Oscillatory Motion. Solutions of Selected Problems 15.1 Problem 15.18 (In the text book) A block-spring system oscillates with an amplitude of 3.50 cm. If the spring constant is 250 N/m and
More information28. Pendulum phase portrait Draw the phase portrait for the pendulum (supported by an inextensible rod)
28. Pendulum phase portrait Draw the phase portrait for the pendulum (supported by an inextensible rod) θ + ω 2 sin θ = 0. Indicate the stable equilibrium points as well as the unstable equilibrium points.
More informationConstraints. Noninertial coordinate systems
Chapter 8 Constraints. Noninertial codinate systems 8.1 Constraints Oftentimes we encounter problems with constraints. F example, f a ball rolling on a flo without slipping, there is a constraint linking
More informationTorque and Rotation Lecture 7
Torque and Rotation Lecture 7 ˆ In this lecture we finally move beyond a simple particle in our mechanical analysis of motion. ˆ Now we consider the so-called rigid body. Essentially, a particle with extension
More informationRotational Motion and Torque
Rotational Motion and Torque Introduction to Angular Quantities Sections 8- to 8-2 Introduction Rotational motion deals with spinning objects, or objects rotating around some point. Rotational motion is
More information7 Pendulum. Part II: More complicated situations
MATH 35, by T. Lakoba, University of Vermont 60 7 Pendulum. Part II: More complicated situations In this Lecture, we will pursue two main goals. First, we will take a glimpse at a method of Classical Mechanics
More informationLecture 38: Equations of Rigid-Body Motion
Lecture 38: Equations of Rigid-Body Motion It s going to be easiest to find the equations of motion for the object in the body frame i.e., the frame where the axes are principal axes In general, we can
More informationPhysical Dynamics (SPA5304) Lecture Plan 2018
Physical Dynamics (SPA5304) Lecture Plan 2018 The numbers on the left margin are approximate lecture numbers. Items in gray are not covered this year 1 Advanced Review of Newtonian Mechanics 1.1 One Particle
More informationPhysics 351, Spring 2015, Final Exam.
Physics 351, Spring 2015, Final Exam. This closed-book exam has (only) 25% weight in your course grade. You can use one sheet of your own hand-written notes. Please show your work on these pages. The back
More informationPhysical Dynamics (PHY-304)
Physical Dynamics (PHY-304) Gabriele Travaglini March 31, 2012 1 Review of Newtonian Mechanics 1.1 One particle Lectures 1-2. Frame, velocity, acceleration, number of degrees of freedom, generalised coordinates.
More informationDynamics of Rigid Bodies
CHAPTER Dynamics of Rigid Bodies -. The calculation will be simplified if we use spherical coordinates: x rsin cos φ y rsin sin φ z rcos ( z y Using the definition of the moment of inertia, x we have or,
More information16.07 Dynamics Final Exam
Name:... Massachusetts Institute of Technology 16.07 Dynamics Final Exam Tuesday, December 20, 2005 Problem 1 (8) Problem 2 (8) Problem 3 (10) Problem 4 (10) Problem 5 (10) Problem 6 (10) Problem 7 (10)
More informationLecture 38: Equations of Rigid-Body Motion
Lecture 38: Equations of Rigid-Body Motion It s going to be easiest to find the equations of motion for the object in the body frame i.e., the frame where the axes are principal axes In general, we can
More information2.003 Engineering Dynamics Problem Set 4 (Solutions)
.003 Engineering Dynamics Problem Set 4 (Solutions) Problem 1: 1. Determine the velocity of point A on the outer rim of the spool at the instant shown when the cable is pulled to the right with a velocity
More informationRotation. Rotational Variables
Rotation Rigid Bodies Rotation variables Constant angular acceleration Rotational KE Rotational Inertia Rotational Variables Rotation of a rigid body About a fixed rotation axis. Rigid Body an object that
More informationLecture AC-1. Aircraft Dynamics. Copy right 2003 by Jon at h an H ow
Lecture AC-1 Aircraft Dynamics Copy right 23 by Jon at h an H ow 1 Spring 23 16.61 AC 1 2 Aircraft Dynamics First note that it is possible to develop a very good approximation of a key motion of an aircraft
More informationChapter 10. Rotation of a Rigid Object about a Fixed Axis
Chapter 10 Rotation of a Rigid Object about a Fixed Axis Angular Position Axis of rotation is the center of the disc Choose a fixed reference line. Point P is at a fixed distance r from the origin. A small
More informationPHYS2330 Intermediate Mechanics Fall Final Exam Tuesday, 21 Dec 2010
Name: PHYS2330 Intermediate Mechanics Fall 2010 Final Exam Tuesday, 21 Dec 2010 This exam has two parts. Part I has 20 multiple choice questions, worth two points each. Part II consists of six relatively
More informationRotational Motion. Chapter 4. P. J. Grandinetti. Sep. 1, Chem P. J. Grandinetti (Chem. 4300) Rotational Motion Sep.
Rotational Motion Chapter 4 P. J. Grandinetti Chem. 4300 Sep. 1, 2017 P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 1 / 76 Angular Momentum The angular momentum of a particle with respect
More informationAnalytical Mechanics ( AM )
Analytical Mechanics ( AM ) Olaf Scholten KVI, kamer v8; tel nr 6-55; email: scholten@kvinl Web page: http://wwwkvinl/ scholten Book: Classical Dynamics of Particles and Systems, Stephen T Thornton & Jerry
More informationRotational Kinetic Energy
Lecture 17, Chapter 10: Rotational Energy and Angular Momentum 1 Rotational Kinetic Energy Consider a rigid body rotating with an angular velocity ω about an axis. Clearly every point in the rigid body
More informationPhysics 5A Final Review Solutions
Physics A Final Review Solutions Eric Reichwein Department of Physics University of California, Santa Cruz November 6, 0. A stone is dropped into the water from a tower 44.m above the ground. Another stone
More informationHomework 1. Due whatever day you decide to have the homework session.
Homework 1. Due whatever day you decide to have the homework session. Problem 1. Rising Snake A snake of length L and linear mass density ρ rises from the table. It s head is moving straight up with the
More informationAssignment 2. Goldstein 2.3 Prove that the shortest distance between two points in space is a straight line.
Assignment Goldstein.3 Prove that the shortest distance between two points in space is a straight line. The distance between two points is given by the integral of the infinitesimal arclength: s = = =
More informationHomework 1. Due Tuesday, January 29.
Homework 1. Due Tuesday, January 29. Problem 1. An ideal rope (no friction) lying on a table slides from its edge down to a scales lying on the floor. The table s height is h. Find a stationary velocity
More informationPhysics 201, Lecture 18
q q Physics 01, Lecture 18 Rotational Dynamics Torque Exercises and Applications Rolling Motion Today s Topics Review Angular Velocity And Angular Acceleration q Angular Velocity (ω) describes how fast
More informationHandout 7: Torque, angular momentum, rotational kinetic energy and rolling motion. Torque and angular momentum
Handout 7: Torque, angular momentum, rotational kinetic energy and rolling motion Torque and angular momentum In Figure, in order to turn a rod about a fixed hinge at one end, a force F is applied at a
More informationidentify appropriate degrees of freedom and coordinates for a rigid body;
Chapter 5 Rotational motion A rigid body is defined as a body (or collection of particles) where all mass points stay at the same relative distances at all times. This can be a continuous body, or a collection
More informationLecture D16-2D Rigid Body Kinematics
J. Peraire 16.07 Dynamics Fall 2004 Version 1.2 Lecture D16-2D Rigid Body Kinematics In this lecture, we will start from the general relative motion concepts introduced in lectures D11 and D12, and then
More informationChapter 10. Rotation
Chapter 10 Rotation Rotation Rotational Kinematics: Angular velocity and Angular Acceleration Rotational Kinetic Energy Moment of Inertia Newton s nd Law for Rotation Applications MFMcGraw-PHY 45 Chap_10Ha-Rotation-Revised
More informationAngular velocity and angular acceleration CHAPTER 9 ROTATION. Angular velocity and angular acceleration. ! equations of rotational motion
Angular velocity and angular acceleration CHAPTER 9 ROTATION! r i ds i dθ θ i Angular velocity and angular acceleration! equations of rotational motion Torque and Moment of Inertia! Newton s nd Law for
More informationHomework 1. Due Thursday, January 21
Homework 1. Due Thursday, January 21 Problem 1. Rising Snake A snake of length L and linear mass density ρ rises from the table. It s head is moving straight up with the constant velocity v. What force
More informationPHY 5246: Theoretical Dynamics, Fall Assignment # 10, Solutions. (1.a) N = a. we see that a m ar a = 0 and so N = 0. ω 3 ω 2 = 0 ω 2 + I 1 I 3
PHY 54: Theoretical Dynamics, Fall 015 Assignment # 10, Solutions 1 Graded Problems Problem 1 x 3 a ω First we calculate the moments of inertia: ( ) a I 1 = I = m 4 + b, 1 (1.a) I 3 = ma. b/ α The torque
More informationPhys 7221, Fall 2006: Midterm exam
Phys 7221, Fall 2006: Midterm exam October 20, 2006 Problem 1 (40 pts) Consider a spherical pendulum, a mass m attached to a rod of length l, as a constrained system with r = l, as shown in the figure.
More informationArtificial Intelligence & Neuro Cognitive Systems Fakultät für Informatik. Robot Dynamics. Dr.-Ing. John Nassour J.
Artificial Intelligence & Neuro Cognitive Systems Fakultät für Informatik Robot Dynamics Dr.-Ing. John Nassour 25.1.218 J.Nassour 1 Introduction Dynamics concerns the motion of bodies Includes Kinematics
More information2.003 Engineering Dynamics Problem Set 6 with solution
.00 Engineering Dynamics Problem Set 6 with solution Problem : A slender uniform rod of mass m is attached to a cart of mass m at a frictionless pivot located at point A. The cart is connected to a fixed
More information14. Rotational Kinematics and Moment of Inertia
14. Rotational Kinematics and Moment of nertia A) Overview n this unit we will introduce rotational motion. n particular, we will introduce the angular kinematic variables that are used to describe the
More informationHandout 6: Rotational motion and moment of inertia. Angular velocity and angular acceleration
1 Handout 6: Rotational motion and moment of inertia Angular velocity and angular acceleration In Figure 1, a particle b is rotating about an axis along a circular path with radius r. The radius sweeps
More informationE = K + U. p mv. p i = p f. F dt = p. J t 1. a r = v2. F c = m v2. s = rθ. a t = rα. r 2 dm i. m i r 2 i. I ring = MR 2.
v = v i + at x = x i + v i t + 1 2 at2 E = K + U p mv p i = p f L r p = Iω τ r F = rf sin θ v 2 = v 2 i + 2a x F = ma = dp dt = U v dx dt a dv dt = d2 x dt 2 A circle = πr 2 A sphere = 4πr 2 V sphere =
More informationPhysics 106b/196b Problem Set 9 Due Jan 19, 2007
Physics 06b/96b Problem Set 9 Due Jan 9, 2007 Version 3: January 8, 2007 This problem set focuses on dynamics in rotating coordinate systems (Section 5.2), with some additional early material on dynamics
More informationCEE 271: Applied Mechanics II, Dynamics Lecture 27: Ch.18, Sec.1 5
1 / 42 CEE 271: Applied Mechanics II, Dynamics Lecture 27: Ch.18, Sec.1 5 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Tuesday, November 27, 2012 2 / 42 KINETIC
More informationIf the symmetry axes of a uniform symmetric body coincide with the coordinate axes, the products of inertia (Ixy etc.
Prof. O. B. Wright, Autumn 007 Mechanics Lecture 9 More on rigid bodies, coupled vibrations Principal axes of the inertia tensor If the symmetry axes of a uniform symmetric body coincide with the coordinate
More informationPart II. Classical Dynamics. Year
Part II Year 28 27 26 25 24 23 22 21 20 2009 2008 2007 2006 2005 28 Paper 1, Section I 8B Derive Hamilton s equations from an action principle. 22 Consider a two-dimensional phase space with the Hamiltonian
More informationChapter 11. Angular Momentum
Chapter 11 Angular Momentum Angular Momentum Angular momentum plays a key role in rotational dynamics. There is a principle of conservation of angular momentum. In analogy to the principle of conservation
More informationPhysics 121, March 27, Angular Momentum, Torque, and Precession. Department of Physics and Astronomy, University of Rochester
Physics 121, March 27, 2008. Angular Momentum, Torque, and Precession. Physics 121. March 27, 2008. Course Information Quiz Topics to be discussed today: Review of Angular Momentum Conservation of Angular
More informationDistance travelled time taken and if the particle is a distance s(t) along the x-axis, then its instantaneous speed is:
Chapter 1 Kinematics 1.1 Basic ideas r(t) is the position of a particle; r = r is the distance to the origin. If r = x i + y j + z k = (x, y, z), then r = r = x 2 + y 2 + z 2. v(t) is the velocity; v =
More informationRigid Body Motion and Rotational Dynamics
Chapter 13 Rigid Body Motion and Rotational Dynamics 13.1 Rigid Bodies A rigid body consists of a group of particles whose separations are all fixed in magnitude. Six independent coordinates are required
More informationLab #4 - Gyroscopic Motion of a Rigid Body
Lab #4 - Gyroscopic Motion of a Rigid Body Last Updated: April 6, 2007 INTRODUCTION Gyroscope is a word used to describe a rigid body, usually with symmetry about an axis, that has a comparatively large
More informationClassical Mechanics [Taylor, J.R.] Solution Manual. Written by JGSK
Written by JGSK Last Updated December 1, 216 Contents 1 Newton s Laws of Motion 2 2 Projectiles and Charged Particles 3 3 Momentum and Angular Momentum 4 4 Energy 5 5 Oscillations 6 6 Calculus of Variations
More informationLecture 19: Calculus of Variations II - Lagrangian
Lecture 19: Calculus of Variations II - Lagrangian 1. Key points Lagrangian Euler-Lagrange equation Canonical momentum Variable transformation Maple VariationalCalculus package EulerLagrange 2. Newton's
More informationVIBRATING BASE PENDULUM. Math 485 Project team Thomas Bello Emily Huang Fabian Lopez Kellin Rumsey Tao Tao
VIBRATING BASE PENDULUM Math 485 Project team Thomas Bello Emily Huang Fabian Lopez Kellin Rumsey Tao Tao Agenda Midterm Recap Equation of Motion & Energy Modeling Effective Potential Stability Analysis
More informationPHYSICS 221, FALL 2011 EXAM #2 SOLUTIONS WEDNESDAY, NOVEMBER 2, 2011
PHYSICS 1, FALL 011 EXAM SOLUTIONS WEDNESDAY, NOVEMBER, 011 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively. In this
More informationTorque. Introduction. Torque. PHY torque - J. Hedberg
Torque PHY 207 - torque - J. Hedberg - 2017 1. Introduction 2. Torque 1. Lever arm changes 3. Net Torques 4. Moment of Rotational Inertia 1. Moment of Inertia for Arbitrary Shapes 2. Parallel Axis Theorem
More informationPhysics 121, March 25, Rotational Motion and Angular Momentum. Department of Physics and Astronomy, University of Rochester
Physics 121, March 25, 2008. Rotational Motion and Angular Momentum. Physics 121. March 25, 2008. Course Information Topics to be discussed today: Review of Rotational Motion Rolling Motion Angular Momentum
More informationProblem 1 Problem 2 Problem 3 Problem 4 Total
Name Section THE PENNSYLVANIA STATE UNIVERSITY Department of Engineering Science and Mechanics Engineering Mechanics 12 Final Exam May 5, 2003 8:00 9:50 am (110 minutes) Problem 1 Problem 2 Problem 3 Problem
More information= 2 5 MR2. I sphere = MR 2. I hoop = 1 2 MR2. I disk
A sphere (green), a disk (blue), and a hoop (red0, each with mass M and radius R, all start from rest at the top of an inclined plane and roll to the bottom. Which object reaches the bottom first? (Use
More informationDYNAMICS OF RIGID BODIES
DYNAMICS OF RIGID BODIES Measuring angles in radian Define the value of an angle θ in radian as θ = s r, or arc length s = rθ a pure number, without dimension independent of radius r of the circle one
More informationHW3 Physics 311 Mechanics
HW3 Physics 311 Mechanics FA L L 2 0 1 5 P H Y S I C S D E PA R T M E N T U N I V E R S I T Y O F W I S C O N S I N, M A D I S O N I N S T R U C T O R : P R O F E S S O R S T E F A N W E S T E R H O F
More informationName Student ID Score Last First. I = 2mR 2 /5 around the sphere s center of mass?
NOTE: ignore air resistance in all Questions. In all Questions choose the answer that is the closest!! Question I. (15 pts) Rotation 1. (5 pts) A bowling ball that has an 11 cm radius and a 7.2 kg mass
More informationAB-267 DYNAMICS & CONTROL OF FLEXIBLE AIRCRAFT
FLÁIO SILESTRE DYNAMICS & CONTROL OF FLEXIBLE AIRCRAFT LECTURE NOTES LAGRANGIAN MECHANICS APPLIED TO RIGID-BODY DYNAMICS IMAGE CREDITS: BOEING FLÁIO SILESTRE Introduction Lagrangian Mechanics shall be
More informationGeneral Definition of Torque, final. Lever Arm. General Definition of Torque 7/29/2010. Units of Chapter 10
Units of Chapter 10 Determining Moments of Inertia Rotational Kinetic Energy Rotational Plus Translational Motion; Rolling Why Does a Rolling Sphere Slow Down? General Definition of Torque, final Taking
More information27. Euler s Equations
27 Euler s Equations Michael Fowler Introduction We ve just seen that by specifying the rotational direction and the angular phase of a rotating body using Euler s angles, we can write the Lagrangian in
More informationKinematics (special case) Dynamics gravity, tension, elastic, normal, friction. Energy: kinetic, potential gravity, spring + work (friction)
Kinematics (special case) a = constant 1D motion 2D projectile Uniform circular Dynamics gravity, tension, elastic, normal, friction Motion with a = constant Newton s Laws F = m a F 12 = F 21 Time & Position
More informationLecture 37: Principal Axes, Translations, and Eulerian Angles
Lecture 37: Principal Axes, Translations, and Eulerian Angles When Can We Find Principal Axes? We can always write down the cubic equation that one must solve to determine the principal moments But if
More informationLecture D20-2D Rigid Body Dynamics: Impulse and Momentum
J Peraire 1607 Dynamics Fall 004 Version 11 Lecture D0 - D Rigid Body Dynamics: Impulse and Momentum In lecture D9, we saw the principle of impulse and momentum applied to particle motion This principle
More information