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1-19 Goldstein (3rd ed. 1.21) Start with a Cartesian coordinate system with the origin at the hole. these coordinates, the kinetic energy is 1 1 2 2 2 T = m1 (x + y ) + m2 z , 2 2 In
(1)
where m1 and m2 are the particle masses, x and y determine the position of particle 1 in the plane, and z determines the position of particle 2 below the hole. The system has only two degrees of freedom because the position of particle 2 on the z axis is constrained by the string length s. Use the polar coordinates r and as the two generalized coordinates for this system, x = r cos , (2)
y = r sin , and the constraint is z = r s. In terms of r and , we have the following expression for T
1 1 2 T = (m1 + m2 )r + m1 (r )2 , 2 2
(3)
(4)
(5)
and the potential energy is V = m2 gz = m2 g (r s). It follows that the Lagrangian, L = T V , is
1 2 1 L = mr + m1 (r )2 m2 g(r s) , 2 2
(6)
(7)
where m is the total mass m = m1 + m2 . (8)
1
The Equation of Motion The Lagrangian does not depend on . Hence, we can immediately identify the partial derivative L/ as a constant of the motion, the angular momentum about the z axis, L where l is a constant. The EOM itself would be the time derivative of Eq.(9), r + 2r = 0 ,
= m1 r2 = l ,
(9)
(10)
but EOM this is not of much interest here because we already have its rst integral in Eq.(9). The r Equation of Motion The required partial derivatives are
2 L = m1 r m2 g , r
(11)
and L r The resulting EOM is mr m1 r + m2 g = 0 , from which the dependence can be eliminated using Eq.(9) to give mr
2
= mr .
(12)
(13)
l2 + m2 g = 0 . m1 r3
(14)
This equation can be integrated readily after multiplying each term by rdt, realizing that rdt = dr, and rearranging it into the perfect dierential form, l2 1 m 2 dr + d + m2 gdr = 0 . 2 2m1 r2 (15)
Because the string length is constant, ds = 0, we may also write this equation as d[ l2 1 m 2 r+ + m2 g (r s)] = 0 . 2 2m1 r2 (16)
With the help of Eqs.(5), (6), (8), and (9), you should recognize Eq.(16) as a statement of the conservation of energy d(T + V ) = dE = 0, which integrates to l2 1 m 2 r+ + m2 g (r s) = E . 2 2m1 r2 2 (17)
In this case, the system is conservative, and you could have written down this rst integral without going through the trouble of nding and integrating the EOM. Incidentally, using the conservation law, Eq.(17), you can prove that, unless l = 0, r can never equal zero despite the implication of the problem statement in the book.
3

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Missouri S&T - PHYSICS - ph 409

GPS Problem 1.22 (2nd ed. 1.20) Planar Double Pendulum Because the motion of each mass point is conned to the same plane, the system has only two degrees of freedom. Use the two polar angles, 1 and 2 , dened in Fig. 1.4, as the two generalized coordinates

Missouri S&T - PHYSICS - ph 409

Particle Suspended on a Sti Fiber next to an Attractive Wall Because the motion of the mass point is conned to a plane and the ber has constant length L, the system has only one degree of freedom. In plane polar coordinates a suitable generalized coordina

Missouri S&T - PHYSICS - ph 409

Bead on the Wire ProblemFrom p. 29 in either edition of Goldstein, the kinetic energy T for the bead on the rotating straight wire is 1 2 m[r + (r )2 ], 2 where is the uniform (i.e., constant) rotation frequency, and the equation of motion is T= (1)r

Missouri S&T - PHYSICS - ph 409

Goldstein 1.10 (3rd ed. 1.14) This is a two particle system, and the kinetic energy is simply T= m2 2 (v + v2 ) . 21 (1)0 0 Now, we introduce the COM velocity v and the relative velocities v1 and v2 , 0 vi = v + vi ,(2)to obtain, T = mv2 + m 02 0 [(v )

Missouri S&T - PHYSICS - ph 409

Goldstein Problem 2.17 (3rd ed. # 2.18) The geometry of the problem: A particle of mass m is constrained to move on a circular hoop of radius a that is vertically oriented and forced to rotate about the vertical symmetry axis with angular frequency . The

Missouri S&T - PHYSICS - ph 409

Goldstein 2-16 (3rd ed. 2.17) The kinetic energy T and potential energy V for a system of n degrees of freedom are expressed as Xi 2T=fi (qi )qi , V =XiVi (qi ) ,(1)where qi and qi are the ith generalized coordinate and velocity, respectively, and

Missouri S&T - PHYSICS - ph 409

Constrained 2d Motion Problem (a) For 2-dimensional motion the kinetic energy is m 2 (1) [(x) + (y )2 ], 2 and with the positive y axis pointing down, the uniform gravitational potential is V = mgy. (2) T= The mass m is constrained to move on the curve y

Missouri S&T - PHYSICS - ph 409

Cone Geodesic Problem (a) Find the geodesic curve: the shortest path between two points. In cylindrical coordinates (, , z ), a dierential element of length in three dimensions is p ds = (d)2 + (d)2 + (dz )2 , (1)where is the radial distance measured fro

Missouri S&T - PHYSICS - ph 409

HW #4 Problem 3 The Rolling Constraint: A small circular hoop of radius R and mass m rolls without slipping up the inside of the parabola, y = ax2. Initially, point 1 is in contact with the origin (0) of the parabola. After the hoop has rolled up the para

Missouri S&T - PHYSICS - ph 409

Tunnel Problem The traversal time t12 can be expressed as Z2t12 =1ds , v(1)where ds is a dierential displacement along the tunnel and v is the particle velocity. By symmetry the tunnel always lies in a plane that includes the center of the earth. Th

Missouri S&T - PHYSICS - ph 409

Goldstein Problem 2.11 (3rd ed. # 13) While the motion of the mass point is constrained to follow the hoop, the system has only one degree of freedom. To calculate the reaction of the hoop on the particle, we treat the system as having two degrees of free

Missouri S&T - PHYSICS - ph 409

GPS 3.33 We can use a cylindrical coordinate system (, , z ) to describe the position of the particle. The polar (z )p axis points upwards, is the radial distance measured from the z axis ( = x2 + y 2 ), and is measured in the x-y plane from the x axis. B

Missouri S&T - PHYSICS - ph 409

Problem Goldstein 3-3 (3rd ed. 3.11) We can think of this two particle system as a single particle of reduced mass in a circular orbit of period , radius R, and angular momentum l. We can relate these quantities using Keplers Law of equal areas in equal t

Missouri S&T - PHYSICS - ph 409

Goldstein Problem 3-6 Problem GPS 3.13 If the center of force is located at the origin of an x-y coordinate system, then the orbit is a circle that passes through the origin as, for example, drawn below. This circle has radius a, and its origin is at (a,

Missouri S&T - PHYSICS - ph 409

Two-body central force problem for a logarithmic potential (a) This is a problem of motion in a central force eld determined by the potential V (r) = k ln(r/a) , where k and a are constants. The Lagrangian is m 2 2 r + (r) k ln(r/a) . L= 2(1)(2)Since i

Missouri S&T - PHYSICS - ph 409

Problem: Orbits with a 6-4 central potential (a) The equation of motion for the one-dimensional radial problem is r = f ef (r) = with the 6-4 potential is dened as V (r ) = k a6 r a 4 r , (2)dV l2 + 3, dr r(1)where k is a constant that determines the s

Missouri S&T - PHYSICS - ph 409

Problem Goldstein 3-12 (3rd ed. 3.18) From the result of Problem 2-9 (2.11 in 3rd ed.) we can relate the change in r incurred by the particle of reduced mass to the strength of the radial impulse S ,(r)2 (r)1 = S ,(1)where the subscripts 1 and 2 refer

Missouri S&T - PHYSICS - ph 409

Problem - Tangential Impulse Orbit Change (a) From the result of GPS Problem 2-11 (2.9 in 2nd ed.) we can relate the change in angular momentum incurred by the comet-star system to the strength of the tangential impulse S , (r2 )f (r2 )i = l l0 = S , (1

Missouri S&T - PHYSICS - ph 409

Problem: Orbits with a 1-3 central potential The equation of motion for the one-dimensional radial problem is r = f ef (r) = dV l2 + 3, dr r(1)where l is the constant angular momentum, r2 , and is the reduced mass of the two particle system. The 1-3 po

Missouri S&T - PHYSICS - ph 409

Problem GPS 3-31 (2nd ed. 3.26) Start with Eq.(3.97) (in both 2nd and 3rd editions), relating the scattering angle (s) to the interaction potential V (r) and the impact parameter s, Z um 1/2 (s) = 2s du. (1) 1 V (1/u)/E s2 u20For the repulsive central f

Missouri S&T - PHYSICS - ph 409

Problem Goldstein 4-1 (3rd ed. 4.1) Associativity Say we have three matrices, P, Q, R, whose dimensions are such that the matrix multiplications PQ = S , and QR = T , are dened. Then, to prove associativity, we need to show that SR = PT . Consider the ij

Missouri S&T - PHYSICS - ph 409

Problem Goldstein 4-2 (3rd ed. 4.2) Transposition For two matrices, A and B, we want to prove (AB)T = BT AT , (1)where superscript T indicates a matrix transpose. Consider the ji element of the left-hand-side T (AB)ji = (AB)ij . (2) By the rules of matri

Missouri S&T - PHYSICS - ph 409

Problem Goldstein 4-19 (3rd ed. 4.15) The most direct way to do this problem is to take the body frame components 0 derived in class or in the text and convert them into space frame components with the the help of the transposed Euler matrix = AT 0 . (1)

Missouri S&T - PHYSICS - ph 409

Problem GPS 4.20 (2nd ed. 4-9) In general, we would need six generalized coordinates to describe the motion of the sphere. Three of these would be for the center of mass (X , Y , Z ) and three for the orientation of a body xed frame rotating with the sphe

Missouri S&T - PHYSICS - ph 409

Problem GPS 4.21 (2nd ed. 4.22) If we neglect the centrifugal force term in GPS Eq.(4.91), the equation of motion for a particle of mass m in the rotating frame of the earth is m dv = mg + 2mv , dt (1)where v is the particle velocity, is the earths angul

Missouri S&T - PHYSICS - ph 409

Problem GPS 4.22 (2nd ed. 4.23)If we neglect the very small centrifugal force term in Eq.(4-129) [3rd ed. Eq.(4.91)], the equation of motion for a particle of mass m in the rotating frame of the earth is dv = mg + 2mv , (1) m dt where v is the particle v

Missouri S&T - PHYSICS - ph 409

Dropped Particle with Rebound Problem (a) If we neglect the very small centrifugal force term in GPS Eq.(4.91), the equation of motion for a particle of mass m in the rotating frame of the earth is dv m = mg + 2mv , (1) dt where v is the particle velocity

Missouri S&T - PHYSICS - ph 409

Problem GPS 5.3 (2nd ed. 5-7)We have a collection of mass points rotating about a xed point, and the positions ri and velocities vi of these points are measured in a space xed, inertial frame whose origin is at the xed point. Start with the denition of T

Missouri S&T - PHYSICS - ph 409

HW12 Problem 4(b) Start with the fundamental equation governing the rate of change of angular momentum from Chapter 1 or Chapter 5 dL = N. dt Next, take the dot product of Eq.(1) with L to get dL = L N. dt Now expand the left dot product of Eq.(2) as L 1

Missouri S&T - PHYSICS - ph 409

Problem - Inertia ellipsoid of a right circular cone The height of the cone is h and its base diameter is 2a. The cone is oriented so that its symmetry axis lies along the z axis of a Cartesian frame whose origin is at the vertex of the cone. In this orie

Missouri S&T - PHYSICS - ph 409

Problem Goldstein 5-4 (3rd ed. 5.15) You can save a lot of eort by orienting the triangle in a way that effectively exploits its symmetry. My preferred choice is shown in the gure.1Y-10X1 In this gure, the x and y variables are normalized by the fa

Missouri S&T - PHYSICS - ph 409

Problem Three mass points The three mass points are located at the vertices of the triangle shown in the gure. The x, y and z variables are normalized by the factor a, so that X = x/a, Y = y/a, and Z = z/a.ZXYWe are supposed to nd the principal moment

Missouri S&T - PHYSICS - ph 409

Solid Cube Problem (a) Inertia tensor in COM frame The cube is located with a vertex at the origin of a Cartesian coordinate system whose x, y and z axes lie along edges of the cube. The edges of the cube are of length a.zxywhere b = M a2 and M is the

Missouri S&T - PHYSICS - ph 409

Problem Goldstein 5-19 (3rd ed 5.6) (a) We will designate the body frame unit vectors as i, j, and k, and assume they lie along the principal axes of the body. Now we want to show that the angular momentum vector L rotates about the body symmetry axis k w

Missouri S&T - PHYSICS - ph 409

Problem: Space and Body Cones for the torque free top (a) One way to do this is as follows. Designate the body frame unit vectors as i, j, and k, and assume they lie along the principal axes of the body. In the body principal axis frame we can write the a

Missouri S&T - PHYSICS - ph 409

Problem Goldstein 5-24 (3rd ed 5.11) We will designate the body frame unit vectors as i, j, and k, and assume they lie along the principal axes of the body. In this body frame the angular momentum L is expressed as L = I1 1 i + I2 2 j + I3 3 k , and the m

Missouri S&T - PHYSICS - ph 409

Heavy Top Angular Momentum Problem (a)Long Way Designate the body frame unit vectors as i, j, and k, and assume they lie along the principal axes of the body. The space frame unit vectors will be designated as x, y, and z. Since it is easiest to write L i

Missouri S&T - PHYSICS - ph 409

Problem Goldstein 5-23 (3rd ed 5.10) We will designate the body frame unit vectors as i, j, and k, and assume they lie along the principal axes of the body. The space frame unit vectors will be designated as x, y, and z. In order to use Eulers equations o

Missouri S&T - PHYSICS - ph 409

GPS 5-19 or Goldstein 5-13 This problem is distantly related to the Spring-Pendulum Problem from Ch. 1, but here, the spring is constrained to oscillate vertically and the mass point is expanded into a rigid rod of mass M and length 2l that swings in a ve

Missouri S&T - PHYSICS - ph 409

Problem Kinetic Energy of a Suspended Sphere To move the origin of the coordinate system to the surface of the sphere, we construct the inertia tensor IR for the center of mass vector R in the new coordinate system with body frame unit basis vectors denot

Missouri S&T - PHYSICS - ph 409

Problem: Kinetic Energy of a Rod (a) First we need to nd the principal moments of the rod. This can be done with the origin either at the end of the rod (at the pivot) or at the center of the rod (the COM). There is a conceptual advantage with the latter

Missouri S&T - PHYSICS - ph 409

Problem Inertia Tensor and Kinetic Energy of Two Cones Joined at Their Tips The two cones are joined at their tips. Each cone is a prolate ellipsoid with I1 = I2 > I3 . One cone is oriented with its principal symmetry axis along the x axis of a body frame

Missouri S&T - PHYSICS - ph 409

Problem 8.1 (a) We can dene the Lagrangian L(qj , q j , t) as the Legendre transform of the Hamiltonian H (qj , pj , t), L= Xjpj q j H ,(1)where qj , qj , and pj are the j th generalized coordinate, velocity, and momentum, respectively. To nd the EO

Missouri S&T - PHYSICS - ph 409

Goldstein 8-21 (3rd ed. 8.3) We want to produce a o n new Hamiltonian, or Gamiltonian G(q, p, t), in which the quantities q, p are the natural independent variables. We can do this with the following Legendre transform of the Lagrangian L(q, q, t), G(q,

Missouri S&T - PHYSICS - ph 409

GPS 8-17 or Goldstein 8-7 Direct route: From Problem 5.13 (2nd ed.) or 5.19 (3rd ed.) the Lagrangian of this pendulum is L= M 2 I 2 1 y M ly sin + ky 2 + M g [y + l cos ] . 2 2 2 L y L (1)The generalized momenta, py and p , are py = and p = = M y M l s

Missouri S&T - PHYSICS - ph 409

GPS Problem 8.20 (2nd ed. 8.10) Modied Planar Double Pendulum This problem is a variation of the planar double pendulum problem, illustratrated in Fig. 1.4. The solution to that problem, GPS 1.22 or 2nd ed. 1.20, will be useful here. There are two major m

Missouri S&T - PHYSICS - ph 409

Goldstein 8-14 (3rd ed. 8.24) This view is looking down on the cylinder from above. The z axis is coming up out of the page through the center of the circular cylinder. The x axis is fixed in space. The x& axis is fixed in the cylinder. The angle measures

Missouri S&T - PHYSICS - ph 409

Goldstein 8-15 (3rd ed. 8.25) This problem diers from 8-14 because the cylinder is now driven to rotate at a constant angular speed , and its kinetic energy is no longer variable. Since the cylinders angular motion is prescribed, we only need to consider

Missouri S&T - PHYSICS - ph 409

GPS 8-27 (2nd ed. 8.17) (a) Using the identity, sin 2 = 2 sin cos , the given Lagrangian can be rewritten as m 2 2 22 L= q sin t + 2qq sin t cos t + q , (1) 2 from which we nd the generalized momentum p p= L q The energy function h is h=q = mq sin2 t +

Missouri S&T - PHYSICS - ph 409

Goldstein 9-28 (3rd ed. 9.30) (a) Let H be a Hamiltonian for some dynamical system. If A and B are two constants of the motion that explicitly depend on time, then it must be true that A = [A, H ] = [H, A] , t and B = [B, H ] = [H, B ] . t We want to prov

Missouri S&T - PHYSICS - ph 409

Goldstein 9-33 (3rd ed. 9.35) plus an additional part (b) (a) The Hamiltonian for this one dimensional system is H= p2 mk +2. 2m x (1)We want to nd y (t), where y is the quantity x2 , using the Poisson bracket operator formalism. The formal solution is y

Missouri S&T - PHYSICS - ph 409

Goldstein 9-37 (3rd ed. 9.37) Let r be the position vector of the bob of mass m. In spherical coordinates the components of r are x = r sin cos , (1) y = r sin sin , z = r cos , (2) (3)where r = |r| =constant, is the polar angle between r and the z axis,

Missouri S&T - PHYSICS - ph 409

Poisson Brackets for the Heavy Top The rst thing to do is dene the canonical variables, i.e., the generalized coordinates and momenta for the problem. The Euler angles , , and are the generalized coordinates. The corresponding canonical momenta are found

Missouri S&T - PHYSICS - ph 409

Goldstein 9-39a (3rd ed. 9.39a) (1) USING CARTESIAN COORDINATES Using Cartesian canonical variables, the Hamiltonian for the Kepler problem is H= where p2 = p2 + p2 + p2 , x y z and r= p x2 + y2 + z 2 . mk r. r (3) (2) k p2 , 2m r (1)The Laplace-Runge-Le

Missouri S&T - PHYSICS - ph 409

Physics 409: Classical MechanicsRoom 127TuTh 12:30-1:45Fall 10Instructor: G. Wilemski, 209 Physics, tel: 341-4409 wilemski@mst.edu Office Hours: M 3:30-6, W 3:30-6:00 or by appointment. Text: Goldstein, Classical Mechanics,3rd edition (2nd edition wil

Missouri S&T - PHYSICS - ph 409

Physics 409: Classical Mechanics 24 August Lecture schedule: 24 August - 2 September: Chapter 1 Homework Set #1: Due Tuesday, 31 August (1) Chap.1, GPS Problem 1 (#4 in 2nd ed) (2) Chap.1, GPS Problem 2 (#5 in 2nd ed) (3) Chap.1, GPS Problem 3 (#6 in 2nd

Missouri S&T - PHYSICS - ph 409

Physics 409: Classical Mechanics 31 August Lecture schedule: 31 August- 2 September: Chapter 1 7 September: Finish Ch. 1, start Ch.2, Sec. 2.2Fall 10Reading tips for 2 Sept.: Focus on: The Principle of Virtual Work and D'Alembert's Principle What is a g

Missouri S&T - PHYSICS - ph 409

Physics 409: Classical Mechanics 7 September Lecture schedule: L6 - 9 September: Ch.2, Sec. 2.2, 2.1, 2.3,2.5, 2.6, 2.7(3rd Ed.) L7 - 14 September 2.4 (but not the version in GPS) 16 September: Chapter Test #1 on Chapter 1Fall 10Reading tips for 9 Sept.

Missouri S&T - PHYSICS - ph 409

Physics 409: Classical Mechanics Fall 10 14 September Lecture schedule:L7 - 14 September 2.4 (but not the version in GPS) 16 September: Chapter Test #1 on Chapter 1 L8 - 21 September: Chap. 3 Review session: Tuesday, 14 September, 3:30 pm Rm 202 Reading t

Missouri S&T - PHYSICS - ph 409

Physics 409: Classical Mechanics Fall 10 21 September Lecture schedule: L8: 21 September Ch. 3, Sec. 1-3 (See typos and comments on other side.) L9: 23 September: Ch. 3, Sec. 3, 5, 6 L10: 28 September Ch.3 Sec. 7, 8 Review session for Ch. 2: 3:30-6, Rm202

Missouri S&T - PHYSICS - ph 409

Physics 409: Classical Mechanics 28 September 2010Fall 10Lecture schedule: 28 September: Chapter 3; 30 September: Chapter Test #2 on Ch. 2 (more problems setting up and solving Lagrangians, including Lagrange multipliers, plus calculus of variations pro