Ph 425 (Classical Mechanics II) : Final Exam (open-book)

12:30 – 2:20 pm, 6 June 2000 Dr. R. Egerton

--------------------------------------------- g » 9.8 m/s² ------------------------------------------

1. Starting from rest (and zero potential energy), a particle of mass m slides down a frictionless plane inclined at an angle q to the horizontal.

(a) Provide an expression for the Lagrangian L of the system when the particle has traveled a distance x along the plane.

(b) Deduce the generalized momentum associated with coordinate x and incorporate this into an expression for the Hamiltonian H of the system.

(c) Give the Hamilton canonical equations for this system.

2. An elastic pendulum consists of a particle of mass m attached to an elastic string of unstretched length a and stiffness k .

1. Derive an expression for the Lagrangian L in terms of the instantaneous length r angle q of the string (from the vertical) and any other necessary parameters, taking the potential energy to be zero for q = p /2 radians and r = a.
2. Derive the two Lagrangian equations of motion for the elastic pendulum.

3. A top consists of a thin disk of radius a and mass m attached to a massless rod, perpendicular to the plane of the disk and passing through its center. It spins with the rod vertical and the disk supported at a distance h above a horizontal surface.

(a) Deduce its moments of inertia (I_v and I_h respectively) about the vertical axis of rotation and about an axis in the plane of the supporting surface.

(b) For m = 3 g, h = 1cm and a = 2 cm, compute the minimum rotational speed (revolutions per second) for stable rotation about the vertical axis.

4. (a) Name two circumstances in which an object can undergo free rotation.

(b) Under what conditions does the angular momentum of an object coincide with its axis of rotation ?

(c) Write down the moment of inertia tensor for a cube of side a and mass m which is rotating about an axis passing through its center of mass.

5. (a) A uniform rod, of length 1m, is initially pivoted about one end. Find its period of oscillation T₁ , assuming that the amplitude of swing is small.

(b) Calculate the position (relative to the center of mass) of a second pivot point which gives the same period of swing T₁.

(c) Calculate the smallest possible period of swing T_min for a third pivot point located somewhere along the rod.

(d) What information could be obtained by precisely measuring the distance between the first and second pivot points, even for a non-uniform rod? a:P425fi00.doc

Ph 425 Final (June 2000) Answers (R. Egerton)

1. (a) T = ½ m (x’)² , V = - mgx sinq , so: L = T – V = ½ m (x’)² - mgx sinq

(b) p = dL/dx’ = mx’, giving x’ = p/m and H = T + V = p²/(2m) + mgx sinq

(c) The Hamilton equations are: x’ = dH/dp , giving x’ = p/m

and -p’ = dH/dx = -mg sinq , giving p’ = mg sinq

2. (a) T = (m/2) [(r’)² + (rq ’)²] , V = (k/2)(r – a)² – mgr cosq

L = (m/2) [(r’)² + (rq ’)²] - (k/2)[r - a]² + mgr cosq

(b) The first equation is: (d/dt)(dL/dr’) = dL/dr

dL/dr = (m/2)(2r q ’²) – k (r - a) + mg cosq and (d/dt)(dL/dr’) =(d/dt)(mr’) = mr’’

so the equation becomes: mr’’ = mr(q ’)² – k (r - a) + mg cosq

The second equation is: (d/dt)(dL/dq ’) = dL/dq

dL/dq = -mgr sinq , (d/dt)(dL/dq ’) = (d/dt)(mr^2q ’)

so the equation becomes (d/dt)(mr^2q ’) = - mgr sinq

3. (a) I_v = M of I of a disk = (1/2)ma²

From perpendicular-axis theorem, I_d + I_d = I_v where I_d is the M of I about a horizontal axis in the plane of the disk, giving I_d = (1/4) ma²

From the parallel-axis theorem, I_h = I_d + mh² = = (¼)ma² + mh²

(b) From Eq. (9.7.12) on p.373, the minimum angular velocity (radians per sec) S is given by: S² = 4mglI/I_s² where l = h = 0.01 m (using SI units),

I = I_h = (0.25)(0.003)(0.02)² + (0.003)(0.01)² = 3E-7 + 3E-7 = 6E-7 kg m²

and I_s = I_v = (0.5)(0.003)(0.02)² = 6E-7 kg m²

Therefore S² = (4)(0.003)(9.8)(0.01)[(6E-7)/(6E-7)²] = 1.18E-3/6E-7 = 1960

S = 44.27 rad/s, minimum revolutions per second = 44.27/(2p ) = 7.05

4. (a) Free rotation occurs if the object is suspended by a frictionless pivot at its center of mass, or is in a field-free region, or is in free fall under gravity.

(b) The axis of rotation must be one of the three principal axes.

(c) The 3´ 3 tensor contains equal diagonal components: (1/6)ma² ; the off-diagonal components are zero (see F & C p.346).

5. (a) T = 2p [I/mgl]^1/2 = 2p [k²/gl]^1/2

For pivot at one end, l = 0.5 and k² = a²/3 = 1/3, giving T₁ = 2p /[(1.5)(9.8)]^1/2 = 1.64 s

(b) For same period, ll’ = k_cm² = a²/12 = 1/12, giving l’ = (1/12)(1/0.5) = 1/6 = 0.167 m

(c) For minimum period, l = l’ = k_cm = a/(12)^1/2

T_min= 2p [k²/gl]^1/2 = 2p [(k_cm²+l²)/gl]^1/2 = 2p [2k_cm²/gk_cm]^1/2 = 2p [2(1/12)^1/2/g]^1/2 = 2p [0.577/g]^1/2 = 1.525 s

1. By measuring the distance l – l’, the value of g can be calculated (Kater’s pendulum).