Physics 410 Final Exam (solutions) May 19, 2003



Instructions: Do any four of the six problems. Open textbook. Use exam booklet.



(Picture Omitted)

The figure above shows two rails, crossing at an angle a. Two particles, each of mass m, are constrained to move each on its own rail. No gravity is acting in the plane of the figure. The figure applies to problems 1 and 2.


1. Let a = p/2  (90). The particles are initially at rest, and there is a central, attractive force between them, of otherwise unspecified nature (but action = reaction). Show that, no matter how this central attractive force depends on the distance r between the particles, they will always reach the intersection of the tracks together (that is, simultaneously).

For the special initial conditions x = 0, or y = 0, or x = y the result follows from symmetry. For a more general case, assume there is a potential and use Lagrangian approach with generalized coordinates r (the distance between the masses) and q, the angle between r and and an axis, which turns out to be cyclic; the concerved quantity is r2q·, vanishes initially, so theta is constant, which implies that the particles reach the origin together.
More generally, for any - even time-dependent - central force of magnitude F the Newtonian EOM is
m

x
 
=  F y

(x2+y2)
       m

y
 
=  F x

(x2+y2)
Multiply the first equation by x, the second by y and add,
m(

x
 
y +

y
 
x) =  d

dt
(

x
 
y -

y
 
x) = 0   hence   

x
 
y -

y
 
x = const
From the initial conditions the constant vanishes. Since the force is attractive the velocities will be non-vanishing after t = 0, so it x = 0, then also y = 0.


2. Let a be general, but let the force between the particles be a spring force, so the potential energy is 1/2kr2. Show that the system can then be described as two coupled oscillators, and find the eigenfrequencies w of the system, in terms of a and w0 = {k/m}.

An interesting way to do this is to guess the normal modes - they are the symmetrical motions, x = y, since by symmetry the particles must then reach the origin together (argument as in problem 1), so only one frequency can be involved in this motion, which therefore must be an eigenfrequency. The distance between the particles is then 2xsin(a/2) (resp. 2xcos(a/2)), so the force on the x-particle is -2kxsin2(a/2) (resp. -2kxcos2(a/2) and the frequency under this SH force is w0sin(a/2) resp. w0cos(a/2).
The more standard way is to write the distance between the particles for general x, y bia the law of cosines, r2 = x2 + y2 - 2xy cosa and e.g. via T = 1/2m(x·2 + y·2) and U = 1/2k r2 identify the m- and A-matrices,
(m) = m

1
0
0
1


       (A) = k

1
-cos a
-cos a
1


.
These lead to the secular equation (w02 - w2)2 - w04cos2a = 0, or w02 - w2 = w02cosa so w = w0(1cosa). By trigonometric identity this is the same as the answers above in terms of a/2.


3. When the first Earth satellite (``sputnik") was lauched by the then-USSR (``CCCP") in 1957, and orbited above the US (among other parts of the world), someone suggested ``Why don't we just shoot it down?" Maybe he thought a missile could disable the putative engine keeping it going. Rather than correct that misconception, it was explained that if we could launch something that could hit the sputnik, we could also put something in orbit - which at that time we were not quite ready yet to do.

Argue that, even without atmosphere, shooting a point-particle-like missile from the Earth cannot put it into an (approximately) circular orbit above the Earth. Here ``point-particle-like" means ``moving according to the point particle law of motion, not possessing rocket motors, warp drive, or other exotic propulsion." ``Shooting" means ``giving an initial velocity with upward component."

The orbit of the missile in the Earth's 1/r2 field will be an ellipse, with the Earth at one focus. Since the ellipse is closed, the particle will "try to" come back to its starting point, but it could do this only if the Earth were not in the way, so it will actually hit the Earth again - unless it is on a parabolic or hyperbolic escape orbit, which is far from circular.


4. Suppose we can shoot a missile of mass m straight up that just reaches a satellite's orbit, the satellite runs into it, and the two stick together. If the satellite mass is M, what is the minimum value of m such that the combination will fall to Earth? Where will it hit Earth, with respect to the missile lauch site? Neglect the Earth's rotation and atmospheric friction, and assume the satellite at a circular orbit of radius twice the Earth's radius.
By the way, for sputnik M was about 80 kg, and its orbital radius was only about 2% larger than Earth radius.

The collision between satellite and missile (which is momentarily at rest) will diminish the speed but not change the direction of the satellite. Therefore immediately after collision the satellite (and the missile) are at the apogee of the new orbit. The decrease in speed will be smallest if the satellite just reaches Earth at perigee (closest approach of orbit to Earth). Since apogee, focus and perigee lie on a straight line, the satellite will hit Earth at the antipode of the lauch site (180 away).
The semimajor axis a of the circular orbit was 2R (R = Earth radius, M = Earth mass), that of the elliptical crashing orbit, 3R/2. We have for the satellite's total energy GMM/2a = E = 1/2Mv2 - GMM/2R with different v and a before and after the collision, but the same potential energy. Substitute abefore = 2R and aafter = 3R/2 to find (vaft/vbef)2 = 1/3. The inelastic collision preserves momentum, Mvbef = (M+m)vafter. Eliminating the v's we find m = (3 - 1)M ~ 0.73 M.


5. A rigid body rotates about a fixed axis, not its principal axis, with angle of rotation q. Let e be a unit vector along the rotation axis, so that w = q·e, let T be the kinetic energy of rotation, and L the angular momentum.
Show that the projection of L on the rotation axis is given by

L
 
·

e
 
=  T

q·

Let I be the moment of inertia about the axis defined by e, that is, in a set of orthogonal axes that includes e we have one component of the inertia tensor Iee = I. Since w has only one component in those axes, we have

L
 
= (I)

w
 
= I

q
 
    and     T = 1/2

w
 
·(I)w = 1/2 I

q
 
2
 
and then it's just simple differentiation.
If you want to avoid the special frame, note that T = 1/2w·L = 1/2q· e·L so L·e = 2T/q·. Then argue that T is homogeneous quadratic in q·, so q· T/q· = 2T.


6. A winch stands on a horizontal plane surface on frictionless rollers. The winch and its rope are entirely massless except for the drum on which the rope is wound. That has mass M, radius R, and moment of inertia tensor, in principal axes, 1/4MR2 diag(2, 1, 1). It rotates on frictionless bearings. The rope runs over a massless pulley to a suspended load of mass m, as shown. All velocities remain in a vertical plane shown in the figure, with the rotation axis always perpendicular to that plane. Gravity g is acting.

Negect the big problem that the winch may eventually fall off the cliff. There is however the small problem that the drum is not properly balanced: although the axle passes through the drum's center of mass, the principal axis of the large moment of inertia makes an angle b with the axle.

What is the acceleration of the winch?

(Picture Omitted)

We evaluate the rotational part of the kinetic energy in principal axes, where the components of w are wcosb, wsinb. Let x be the position of the winch, y the position of the load. The contribution of the winch to the Lagrangian is

Twinch = Ttransl + Trot = 1/2M.
x
 
2
 
+ 1/2(1/2MR2w2cos2b+ 1/4MR2w2sin2b)
The load's Lagrangian is 1/2my·2 - mgy. The rope relates x, y, and w by w = (y· - x·)/R. Add the two Lagrangians and vary x and y:
M ..
x
 
- 1/2M( ..
y
 
- ..
x
 
)M(cos2b+ 1/2sin2b)
=
0
m ..
y
 
+ 1/2( ..
y
 
- ..
x
 
)M(cos2b+ 1/2sin2b) - mg
=
0
We solve for x¨ to obtain the ugly expression
..
x
 
= g(cos2b+ 1/2sin2b)
(3 + M/m)(cos2b+ 1/2sin2b)




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On 22 May 2003, 17:25.