Solutions for Problem Set 14

Last revised 1999/12/08

Chapter 14

#5.

First we should think about what we expect the answer to be. We split mass M into two pieces, and nothing distinguishes the pieces. If we get a single solution for the size of the pieces, the pieces should be the same size. The only way to get pieces of unequal size is if there are two solutions, one with the first piece bigger and the other with the second piece bigger by the same amount. The two-solution, different-sizes possibility is called "spontaneous symmetry breaking" and is not found very often.

Let's determine the force between the two pieces after they are broken up, and manipulate the size of the pieces until the force is as large as possible. If the size of one is m, the size of the other is M - m, and the force is

We find the maximum by using dF/dm = 0 , which gives

#21.

To determine the weight on the moon of an object that weighs 100 N on the Earth, we need the ratio of the gravitational force on the oject near the moon to the gravitational force near the earth. We have

which gives a numerical result of W_M = 17 N. If we move the object to a distance from the center of the Earth of x R_E and demand that the gravitational force exerted by the earth be equal to the object's weight on the moon, the previous equation changes to

A little algebra gives

#23.
(a)

Following instructions, we set the gravitational force equal to the required centripetal force:

where T is the period of rotation. Solving for T² gives

which is just Kepler's Third Law. Next we use the definition of density, rho = M / V , to get M = (4/3) pi R³ rho. Using that relation to eliminate M

For rho = 3.0 gm/cm³, T =6.9 x 10³ sec = 1.9 hr

#53.

Although the problem is marked in the text as a difficult one, the hints take most of the sting out of it. Let the velocity at any given time of mass M be V and that of m be v, and let d be the distance between the masses. Energy and momentum conservation are then

#74.

The diagram shows the situation as if looking down at the north pole. We will first calculate the radius r₀ of the satellite's orbit that will produce an orbital period of 1 day and the speed of the point directly below the satellite for an arbitrary orbital radius. We will then be able to get the speed of the point on the ground if the radius of the orbit is slightly incorrect by use of calculus.

We can relate the speed of the satellite to the speed of the point on the ground below the satellite by geometry and the speed of a satellite at a given radius by using Newton's Second Law:

The second of these gives v(r) and the first then gives v_g(r):

At the radius r₀ that we want, v_g = 2 pi R / T  where T = 1 day . Substituting that value for v_g into the relations for v and v_g as functions of r gives a desired radius of

so

Now that we know the radius that we want, we can return to the equation for v_g as a function of r. The change in v_g if there is an error in r is dv_g = (dv_g/dr) dr , where all quantities in the derivative should be evaluated at the intended value of r. Thus

The negative sign means that v_g is less than intended so that the drift of the point below the satellite with respect to the ground is westward. Numerically,

v_g = 4.63 x 10² m/sec

r₀ = 4.23 x 10⁷ m

dv_g = - 1.64 x 10^-2 m/sec

#80.[The previous diagram shows all the quantities that occur in this problem.]

Let R be the radius of the Earth, and r be the distance of the satellite from the center of the Earth. The altitude of the satellite is then r - R . We need expressions for the kinetic and potential energies in order to determine how they change, and we need to equate the gravitational force to the required centripetal force in circular motion in order to find the velocity of the satellite as a function or r :

which, substituted into the first gives

Call DV the change in Potential Energy and DK the change in Kinetic Energy between the ground and the orbit. The kinetic energy of the satellite on the ground due to the speed of rotation of the Earth is fairly small, so neglect it. Then

Setting the change in potential energy equal to the change in potential energy sets a critical orbital radius of

Below that height, more kinetic than potential energy is required to orbit a satellite; above that height more potential energy is required than kinetic. There is, of course, no abrupt change in the total energy required.

Chapter 16

19.

Clearly we are working with the behavior of a spring, and we want to determine the mass attached to the spring given the period of oscillation. So the general problem is solved using

where T is the period of oscillation. Solving the first for the mass and eliminating omega using the second:

(a) Hence if M is the mass of the astronaut and m is the effective mass of the remainder of the mechanism,

as required. [The quantity m can be measured by determining the period of the system with no astronaut in the chair.]

(b) Using M=0 and substituting the other numbers given, m = 12.47 kg

(c) Substituting numbers, M = 54.43 kg. This number is unreasonably accurate, but corresponds to the number of significant figures claimed in the problem.

26.

There are two pieces of physics here: the movement under harmonic motion and the maximum acceleration permitted by friction given the static coefficient of friction. The latter tells you to calculate the acceleration under harmonic motion.

In this problem the normal force is m g, so

Then from x = A cos (omega t) we get

and comparing the two values for a_max we have

71.

Use the diagram in the book for the definition of variables. There should be a force mg pointing down acting through the center of the stick, and an unknown force of arbitrary direction acting through the point O. Since the point O is fixed, treating the problem as involving pure rotation around an axis perpendicular to the page and passing through point O will be the fastest way of getting results.

Torque: tau = I d²(theta)/dt²

SHO: d²(theta) + omega' ² theta = 0, where omega' is the frequency of oscillation of the pendulum [not d(theta)/dt]

Period: T = 2 pi / omega'

Parallel axis theorem: I = I_CM + M d², where d is the distance from the new axis to the axis through the center of mass.

Taking the axis at O, the only force exerting a torque is mg, and that torque is tau = m g x theta if the angle theta is small. Using the parallel axis theorem and looking up the moment of inertia of a stick about its center of mass, I = m (L² + 12 x²) / 12.

The torque equation then gives

which is the equation for simple harmonic motion with an oscillation frequency of

giving the answer to part (a):

Now finding the minimum value for T² immediately finds the minimum for T and involves much less labor. So take

The result for part (c) is gotten by substituting the values for L and x = L / sqrt(12) into the formula above for T.