Physics 410 First Exam
March 7, 2003

Instructions: Do any two of the three problems, using Lagrangian methods for at least one of them.
                          Use exam booklet.

1. The figure shows an "Atwood's machine" with a frictionless, massless pulley and a massless rope. Gravity g is acting. The pulley is not fixed, instead it is being pulled upward with constant acceleration a (opposite in direction to g). For the masses, assume M > m. Initially M is at rest.

Find the value of a so that M remains forever at rest (or as long as the rope holds out)

The "quick and dirty way" (really consists of treating first M as a free body, then m): Since M is at rest, the net force on it must vanish. If T is the tension, T - Mg = 0 so T = Mg. Also since M is at rest, the acceleration of m is 2a, so Newton's law for m says m(2a) = T - mg = (M - m)g hence a = 1/2((M/m)-1)g.

By Lagrangians: there is only one degree of freedom, one coordinate, choose the position of M with respect to the pulley, call it x (positive down). Then the position of m with respect to pulley is l - pR - x (where l is the length of the rope, R the radius of the pulley). To get the inertial speeds we have to add the speed of the pulley, -at, to the relative speeds of the masses: vM = dx/dt - at, vm = -dx/dt -at. The inertial positions are, by integration, xM = x - 1/2at2,  xm = -x - 1/2at2. Therefore
L = 1/2(M+m)(dx/dt)2 + (m - M)(dx/dt)at + (M-m)gx
where terms that are functions of t only were omitted, and the EOM is

(M+m)(d2x/dt2) + (m-M)a - (M-m)g = 0
Since M is to be at rest, vM = 0 and d2x/dt2 = a. Substitute into the above to get the same result.

2. Two particles of equal mass m are constrained to move without friction along two straight, horizontal lines, at right angles to each other. (Even where the lines cross, each particle remains on its own line, and if they reach the crossing point together, they do not collide but somehow manage to pass through each other.) A massless spring of spring constant k and zero unstretched length connects the two particles.

(a) Choose coordinates to describe this system, in number equal to the number of degrees of freedom. Specify your choice explicitly and clearly.

The simplest coordinates are x and y, the distances of the masses from the crossing point.

(b) Write the (differential) equations of motion for this system, in terms of the given constants and the coordinates (and their time derivatives) chosen in (a). If your EOM contains any other symbol, define it in terms of those constants and coordinates.

Lagrangians are simpest here:
L = 1/2 m



+ 1/2k(x2+y2)
leading to

= -kx        m

= - ky ,
that is, the EsOM are the same as those for a 2D SHO.

(c) Solve for the motion with the initial condition that one particle is at rest a distance a from the crossing point, and the other particle is at the crossing point with speed v.

The SHO equations have sine and cosine solutions, we need one of each for the given initial conditions, for example

x = acoswt,        y = (v/w)sinwt        where        w = Ök/m.

(d) Energy is of course conserved for this system, but in addition there is another conserved quantity. What is that conserved quantity (it can, and should, be written in terms of the coordinates and their time derivatives only), and what is the cyclic coordinate conjugate to that conserved quantity?

Since equation (*) looks like a single 2D particle in a SHO central force, we expect the quantity analogous to angular momentum, mx(dy/dt) - my(dx/dt) to be conserved. Taking its time derivative and using (*) confirms this. Hence the conjugate variable should be an angle - the only angle available is that of the spring, e.g. tan-1(y/x).

3. A rod is rotated in a vertical plane with constant angular speed dq/dt = w about a horizontal axis. A bead of mass m slides without friction on the rod. Gravity g is acting. Use the distance r of the bead from the center of rotation as the generalized coordinate for the bead.

(a) Write the equation of motion for r.

The Lagrangian is that of a particle in polar coordinates with prescribed dq/dt = w:
L = 1/2m

+ r2w2
- mgrcos(wt)
leading to

- rw2 + gcos(wt)
= 0

(b) Solve, or describe the motion in words, for two special (easy) cases:

        (i) w = 0, g 0, rod vertical

So d2r/dt2 = -g, r = r0 - 1/2gt2 if started at rest: free fall under gravity.

        (ii) w 0, g = 0.

Here d2r/dt2 = w2 r which has sinh or cosh solutions, r = r0cosh(wt) if started at rest: the particle moves at increasing speed away from the rotation axis.

(c) Show that for w 0, g 0, the bead can move in a circle of suitable radius R, for example x = R(1 + cos(2wt), y = R(sin2wt), as shown in the figure. Do you expect this orbit to be stable?

Here one way is to boldly try a circle function, even though we just saw hyperbolic ones: r = A cos(wt) (because cos appears in (f)) works if

A(-w2 - w2) + g = 0        A = g/2w2        r = (g/2w2)coswt

If we take x = rcosq = rcoswt and y = rsinwt we get x = (g/2w2)cos2wt = g/4w2(1+cos2wt) and y = (g/4w2)sin2wt, which is the required circle (with R = g/4w2).