Solution to Second Midterm Exam Solution to Second Midterm Exam
Part I (in class)
I. In the last homework problem you showed H/t = - L/t, where H and L are the Hamiltonian and the Lagrangian, respectively. Also, our text proves in general that dH/dt = H/t (p. 530). One could then try to argue thus:
"Suppose L is time-independent (L/t = 0, the usual case). Then
(1) H is constant. But also
(2) q·i = H/pi therefore
(3) q·i = 0, all coordinates qi are constant in time."
But that can't be true in general. In fact,
1. Give a simple example of a Lagrangian L for which q(t) = const is not a general solution (almost any L you ever met has this property).
2. A good example that many of you chose is 1D motion under gravity, L = 1/2mq·2 - mgq, which yields motion with constant acceleration -g, so q can never be constant.
3. So where is the error? Are any of the statements (1)-(3) wrong? Or does the error in the reasoning occur when going from one of those statements to the next?
4. The main error occurred in going from (1) to (2). For a function, the meaning of "constant" depends on the argument of the function. In (1) H is considered as a function of the time t. That is, a solution of the equations of motion, p(t), q(t) is subsituted in H(p, q), and the result is t-independent. But in (2), H is what it properly should be, a function of p and q. As such it is not constant, and its partial derivatives do not vanish. (A certain sum - what you get from the chain rule when forming dH/dt, see (13.26) - does vanish, but not the separate terms.)
5. What is the Hamiltonian H for the case of the Lagrangian you gave above? So is this H constant or not? Or does it depend what we mean by "constant"?
H(p,q) = p2/2m + mgq
This is constant in time, but not constant in p an q. Yes, it depends on what we mean by "constant"
II. A body (treated as a point particle) is moving in a circular orbit in a 1/r2 force field - for example, gravity due to a large, central mass. In the equivalent one-dimensional problem its effective potential energy is the sum of the actual gravitational potential U(r) = -Gm1m2/r and the centrifugal term Ucf = l2/2mr2, and the body is at the minimum of this potential (Text p. 302).
In a frame that is rotating about the central mass at the same angular velocity W as the orbiting body, the body is at rest. It is of course also in free fall, and its (vanishing) acceleration is correctly given by the effective force (Text p. 345/6), Feff = Fgrav + Fcf = -GM1m2/r2 +mW2 r = 0.
Now we make a small radial perturbation. According to the effective potential energy the motion will be a stable oscillation about the minimum. But the effective force indicates that the initial acceleration is away from the circular orbit, like an instability.
1. So, which is it: is the circular orbit in the 1/r2 force field stable or not?
2. According to the typical textbook (Marion & Thornton, for example) a circular orbit is stable if it is at a minimum of the effective potential. That is the case here. But all one can really say in this case is that the radial motion is stable, and that the curve of the orbit always stays in the same (toroidal) neighborhood of the circular orbit. Even that statement needs qualification, because the effective potential compares only orbits with the same angular momentum, so one needs to know in addition that small changes in l do not change the location of the effective potential's minimum by much. Our text is careful and does not say in so many words that the orbit is stable. And it is not stable in the usual sense that the distance between the perturbed and the unperturbed particle never gets large: a perturbation can change the orbital period, so after sufficient time the orbiting particles will be at very different angular positions.
But all you needed to say is that it's stable.
3. What is wrong or misleading with the argument above that says the opposite? Can it be "fixed up" so there is no longer a disagreement?
4. The opposite argument uses Feff. The subscript is very accomodating, lets itself be used in many contexts, and so it is not true that Feff and Ueff have much to do with each other (the first is certainly not the gradient of the second). This may be a contradiction in notation, but it is no physical contradiction. Instead, the argument neglects that the quoted equation gives only the initial acceleration. When the particle starts moving in the rotating frame, a Coriolis force will also act on it, bringing it back near its starting position - at least in radius.
So, to fix up the argument you could include the Coriolis force. Or you could do the same thing as the "correct" argument, namely compare only orbits with the same angular momentum. That is, when the initial conditions are perturbed, you would also change W appropriately. The initial acceleration should then point toward the old orbit. But you would not get it from a single force field Feff.
5. How can one change the physical set-up so that the "wrong" prediction actually becomes the correct one?
6. Now you have to change the system to accomodate the given Feff (with constant W), for example by forcing the particle always to rotate at that rate, enclosing it in a rotating tube or stringing it on a rotating wire.
III. A solid disk is suspended from a point on its circumference by a universal joint, so that it can swing in any direction.
1. Describe the two normal modes of vibration (for small angular displacements of the disk), and say which normal mode has the longer period, giving reasons.
2. There a really three normal modes, because a rigid body with one point fixed has three degrees of freedom. So the disk can move in its own plane, or perpendicular to its plane, or it can rotate about a vertical axis. The rotation has w = 0, hence infinite period (though the disk comes back to its initial position, the rotation never reverses, which would be necessary for a finite period). If you said that this rotation has the longest period you would be counted correct, but legalistically the rotation is excluded because it is not for all times confined to small angular displacements :)
For the other two ways of oscillating we recall the compoud pendulum (treated in the assigned problem 10.13; also in your elementary physics course) with larger moment of inertia has the lower period. Parallel axes contributes the same amount to I for both ways of oscillating, but problem 10.23 tells you which ICM is larger: when the disk swings in its own plane.
3. Show that the ratio of the periods of the two normal modes is Ö{6/5}.
4. The key thing to remember or derive is that w2 µ 1/I. So now we have to find the I's for a disk. By problem 10.14 and various lecture examples you know for the axis perpendicular to the disk through its CM, Izz = 1/2 Ma2, where a is the radius of the disk. Problem 10.23 and symmetry, or common sense, or plain memory tells you for an axis in the plane of the disk, Ixx = 1/4 Ma2. We have to shift the axis by a for each of these, yielding
 I¢zz = Ma2 + 1/2Ma2 = 3/2Ma2        I¢xx = Ma2 + 1/4Ma2 = 5/4Ma2
Their ratio is I¢zz/I¢xx = 6/5 and by the above remark this is also the ratio of the w2's.
5. Show that the same mode remains longer-period even for large angular displacements, provided that motions with the same amplitude are compared (for example, the CM is initially raised to the same height). Or show that the result b still applies for such large amplitudes.
6. At each height the torque is the same, but the I's are different as computed in b. So the accelerations are different by the same factor. Hence the longer period should at least stay longer as the amplitude is increased beyond small angles. To show that the factor is the same, note that the EOM is q·· = -ga/I sinq. Suppose a solution with some maximum amplitude is given by some function q(t). Consider q¢(t) = q(ct) for some constant c, which has the same maximum amplitude (but reaches it at a different time t). We have
 ×× q ¢(t) = c2 ×× q (ct) = - c2 ga I sin(q(ct)) = ga I/c2 sin(q¢(t))
so q¢(t) solves the equation for moment of inertia I/c2. In our case c2 = 5/6. The period of q¢ scales as 1/c, and that proves it.
Part II (take-home) Part II (take-home)
IV For simplicity we can assume that the car is "all wheels", and consider only one pair of wheels, each of mass M, as shown. The wheels have radius a and the axle has length d. The CM is being steered in a circle C of radius r. (To be totally explicit, this means that the axle is horizontal and always points toward a point at height a above the center of the circle C, so that one wheel is rolling along a circle of radius r+d/2, the other in a circle of radius r-d/2.) Let vc be the speed of the CM while it is moving in a circle, and let vs be its speed when the car later moves straight with the same total kinetic energy.
1. Maybe you can think of a limiting case where the answer can be guessed. (It's always a good idea to know the answer before starting a calculation.)
2. If d = 0 we have effectively just one wheel, but we have to assume it stays vertical, does not lean like a bicycle. When it goes around the curve it has KE due to its forward motion and due to rotation about a vertical axis. The latter vanishes when it goes straight, so it must be converted into KE of forward motion to conserve energy. Hence vs > vc.
Or we can take r = d/2, so that one wheel is fixed. Then the speed of the moving wheel is 2vc, so its energy is µ 4 vc2. When going straight two (2) wheels go at vs, with energy µ 2 vs2. If the constant of proportionality is the same in both cases (nearly true), v2 ~ Ö2 vc > vc.
3. As a warm-up and in order to check yourself later, find the total kinetic energy of a disk of mass 2M and radius a rolling in a straight line without slipping, with CM speed vs (this is your car when d = 0).
4. For a rigid body T = TCM + Trotation = 1/2(2M)vs2 + 1/2Iw2. We know (see in-class exam) I = 1/2(2M)a2 and w = vs/a, so T = 3/2 Mvs2.
5. Now for the calculation itself: Find the kinetic energy of each wheel, add it up for the two wheels to get the total kinetic energy of the system while going in the circle. You will probably use various components of the angular velocity. Distinguish them by subscripts or otherwise, and define them so I can tell what you are doing. The final answer should not contain any symbols you defined, it should be expressed entirely in terms of the given quantities M, vc, d, a, and r.
6. Note that the problem asks you to work out the KE for each wheel separately and add the results. You could also decompose the KE into motion of the CM of the pair of wheels and motion relative to that CM. But that relative motion is not given by the rotational KE expression because the body with that CM is not rigid (the wheels rotate at different angular speeds when going around the curve). So for the relative motion you would have to treat the two wheels separately anyway.
The wheel spins about the car's axle with wx = vw/a where vw is the speed of the wheel's CM, and about its vertical axis at the same rate as the car's CM is rounding the curve, wz = vc/r. Clearly vw = (r+d/2)vc/r and the relevant moments of inertia are Ixx=1/2Ma2 and Izz=1/4Ma2. Putting it all together,
 Tone wheel = 1/2Mvw2 + 1/2 Ixxwx2 + 1/2 Izzwz2 = 3 4 M æè 1+ d 2r öø 2 vc2 + 1 8 M æè a r öø 2 vc2
For the other wheel, d/2 is replaced by -d/2. When you square out the sums and add, the double products cancel so the total energy becomes
 Tcar = éë 3 2 M æè 1+ d2 4r2 öø + 1 4 M æè a r öø 2 ùû vc2
(*)
7. Check that the dimension of your answer to (c) is correct. Also check that when d=0, your answer is not the same as b with vc=vs.
8. Each term in the round parentheses is dimensionless, so the square bracket has dimension of mass, and the whole expression mass ×velocity2 which is energy.
When d = 0 we get the expression in b plus and additional term, due to rotation about the vertical axis. Note that latter contribution is the same for both wheels, since they are rotating at the same rate about that axis.
9. Same thing for the car going in a straight path, with speed vs. But that's easy: a straight path is like a circle of infinite radius. The answer should be the same as for (b) above.
10. When r®¥, (*) becomes 3/2 Mvs2, the same as b.
11. Finally, the solution to the cliff-hanger: Put the two kinetic energies equal and see which speed v (if any) is larger.
12. The additional terms in (*) when r ¹ ¥ are all positive, so vc < Vs.
13. By the way: As the car goes around the curve, the direction of the wheels' angular momentum is changing, so a torque must be applied. But the total energy does not change, so the torque cannot do work. Explain in detail how this comes about. One way to start is with the equation for the rate at which work is being done by a torque. I have not found that formula in our text, so you should derive it.
14. By analogy to dW/dt = F ·v we have for rotation dW/dt = G·w. It's not hard to derive the latter from the former if the force is applied at just one point r, but it's valid more generally. So it may be easiest to use conservation of energy and argue
 dW/dt = dT/dt = d(1/2 w·Iw) = w· × L = w·G.
While the car is in the curve, L is constant in size and changes direction due to rotation about the vertical axis, so L· is horizontal and in the direction of vc (the y-direction). However w has only x- and z-components, so the scalar product vanishes.
V. This problem is rather open-ended (unlike the previous one).
You have four particles of equal mass m confined to a thin, horizontal, circular tube of radius R. Four springs (drawn as zig-zags below) of equal spring constant k connect adjacent masses. The springs are also confined to the tube, so that the length of each spring, and hence its force, is proportional to the difference in angular position of its ends. Find out as much as you can about the normal modes and the normal frequencies of vibration of this system.
First identify the number of degrees of freedom: four. So you need four coordinates. The angular differences between the masses may be tempting, but they are not all independent, because they add up tp 2p. (You could still use them if you write the kinetic energy appropriately - NOT by fixing one of them, but by implementing the constancy of angular momentum. You would then miss the w = 0 mode.) The angular position of the masses away from the equilibrium position (shown in the problem's figure) is certainly the most straightforward, and without loss of generality we may assume that the springs are unstreched at equilibrium, so
 L = 1/2m R2 4å i=1 × q 2i - 1/2kR2 4å i=1 (qi+1-qi)2
where q5 = q1. This gives us the (M) and (K) matrices,
(M) = mR2 diag(1,1,1,1)        (K) = kR2 æ
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 2
 -1
 0
 -1
 -1
 2
 -1
 0
 0
 -1
 2
 -1
 -1
 0
 -1
 2
ö
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ø
and the secular equation,
0 = R2 ê
ê
ê
ê
ê
ê
ê
 2k - mw2
 -k
 0
 -k
 -k
 2k - mw2
 -k
 0
 0
 -k
 2k - mw2
 -k
 -k
 0
 -k
 2k - mw2
ê
ê
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ê
= R2m w(w- 2w0)(w2 - 2w02)2
where w02 = k/m.
So there is one mode with w = 0 (uniform rotation of the 4 masses with unstretched springs, (q) = (1, 1, 1, 1); two degenerate modes with w = Ö2w0, (q) = (1, 0, -1, 0) and (0, 1, 0, -1) or any linear combination of these; and one mode with w = 2w0, (q) = (1, -1, 1, -1). (to be continued)

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On 02 May 2005, 23:33.