# Homework Help: Some Physics problems HELP ME!

1. Apr 5, 2004

### PrudensOptimus

A cylinder of mass 12.0 kg rolls without slipping on a horizontal surface. At the instant its center of mass has a speed of 12.0 m/s, determine the following energies.

ROTATIONAL KE, and toltal energy.

#2: The position vector of a particle of mass 3.50 kg is given as a function of time by r = (7.00 i + 8.00t j ) m. Determine the angular momentum of the particle as a function of time.

#3: A student sits on a rotating stool holding two weights, each of mass 3.00 kg. When his arms are extended horizontally, the weights are 0.900 m from the axis of rotation and he rotates with an angular speed of 0.750 rad/s. The moment of inertia of the student plus stool is 3.00 kg · m2 and is assumed to be constant. The student pulls the weights horizontally to 0.300 m from the rotation axis.

Find angular velicty of student,

Find KE of student before weights pulled in,

and Find KE of student after weights pulled in.

#4 A space station shaped like a giant wheel has a radius of 100 m and a moment of inertia of 6.00 108 kg · m2. A crew of 150 are living on the rim, and the station's rotation causes the crew to experience an acceleration of 1g (Fig. P10.47). When 100 people move to the center of the station for a union meeting, the angular speed changes. What acceleration is experienced by the managers remaining at the rim? Assume that the average mass of each inhabitant is 75.0 kg.

2. Apr 5, 2004

### harsh

Hi, before I start to help you, what do you know so far? What formulas do you think you should use to solve these problems?

3. Apr 5, 2004

### ShawnD

This one has no answer. You need to know the radius of the cylinder.

I don't understand this question too well.

Remember that the energy in the system does not change. The energy of the weights is

$$\frac{1}{2}mv^2$$

v is wr so is looks like this

$$\frac{1}{2}m \omega ^2 r^2$$

r decreases and w increases.

The energy before is (1/2)Iw^2 + (1/2)mw^2r^2. The first term is the kid and the stool, the second term is the weight the kid is holding.
The energy after is the same.

Have you ever stood on one of those rotating circular platforms with a pole through the centre? If you hold your arms out and spin, you'l spin slow. If you try to pull yourself closer to the pole, your moment of inertia decreases so your rotation speed has to increase.

Last edited: Apr 5, 2004
4. Apr 5, 2004

### Janitor

Shawn,

It seems to me that the potential energy of the weights in the rotating stool problem gets increased by the muscular work done by the person sitting on the stool. This means that KE_f + PE_f = KE_i + PE_i, so that KE_f is going to be less than KE_i, right?

Certainly we can say that the angular momentum of the system remains constant, ignoring friction-caused torque in the bushings of the stool. That is:

m r_f^2 w_f = m r_i^2 w_i.

Padon my LaTex laziness. I know this looks ugly. The point here is that this equation involves omega ("w") only to the first power, unlike your equation.

5. Apr 5, 2004

### harsh

For problem # 3, imagine the following situation: Say you are ice skating, and you are spinning really fast with your arms extended. As you start to bring your arms closer in, you start to rotate faster. This is similar. Initially, when you are rotating with the weights some distance apart, you along with the weights have some moment of inertia.

If L = I * (omega), and if it is held constant (which it should be), then as your I increases or decreases, your omega should change accordinly. In this case, since you are bringing the weights closer towards you, you are decreasing the moment of inertia of this system. therefore, you would start spinning faster. Kinetic energy then would obviously increase, since you are rotating faster than before.
KE= .5 (I) (omega)^2.

6. Apr 5, 2004

### Janitor

If my analysis above is correct, then the final KE of just the weights alone is

KE_f = (1/2) m w_i^2 (r_i^4/r_f^2).

Because r_f < r_i, this is greater than the inital kinetic energy of just the weights alone. The KE of the rest of the system doesn't change, ignoring the fact that the person's arms change position.

EDIT: fixed 'f' vs. 'i' goofup and "less" --> "greater" completely changing my direction on this thing!

Last edited: Apr 5, 2004
7. Apr 5, 2004

### harsh

But the weights have some initial angular momentum, which must be kept constant. In that case, the final moment of inertia is less than the initial one, and therefore the kinetic energy must increase. No?

8. Apr 5, 2004

### Janitor

Harsh,

I see your point! If r_f = (1/3) r_i, then my equation says KE_f = 9 KE_i.

Hmmm.

EDIT: Add the comment that the factor of 9 applies to the KE of just a system of weights, pretending the rest of the system does not exist.

Last edited: Apr 5, 2004
9. Apr 5, 2004

### Janitor

I'm thinking Shawn was right, so I withdraw my complaint.

10. Apr 5, 2004

### ShawnD

Actually I was wrong. Once again, I have confused momentum and energy.
Momentum is the same, energy is not.

11. Apr 5, 2004

### Janitor

Now I'm really confused!

It seems like a simple problem. But my original reasoning said that while angular momentum remains the same (which I still believe), the final KE is less because the final PE is more. But then I convinced myself, with help from Harsh, that final KE is more, not less.

Last edited: Apr 5, 2004
12. Apr 5, 2004

### Janitor

Upon further review...

I think my introduction of potential energy was a red herring. I am imagining that instead of a human, a system of springs is on the stool, with a latch that keeps the springs extended in tension. After the system is set to spinning, the latch is tripped. The springs pull the weights inward, and in so doing, the springs lose potential at the same rate as the weights gain potential, so "it's a wash," as the saying goes. In the case of a human, the muscle fibers are doing something similar to what the springs are doing. Does that make sense?

13. Apr 5, 2004

### Janitor

Calling the moment of inertia of the person plus the stool "I," conservation of angular momentum for the process is given by:

I w_f + m r_f^2 w_f = I w_i + m r_i^2 w_i.

Here m is the combined mass of the two weights, while r is the length of just one arm (or better yet, the distance of either weight from the axis). Everything in that equation is given as data except the final angular speed w_f, so you can solve for w_f.

Once you have w_f, you can get the KE of the student plus stool by using KE=(1/2) I w_f^2, which will be pretty close to the KE of just the student. Not enough information is given to separate out just the KE of the student.

Last edited: Apr 5, 2004
14. Apr 6, 2004

### Janitor

After sleeping on it

There is no axial field in the problem, so I never should have brought up potential energy. The change in the total system angular momentum is zero, because there is no outside torque acting on the system, but I believe the change in total system kinetic energy must be equal to the work done by the arm muscles. At any rate, the constraint of constant angular momentum determines what the final kinetic energy has to be, per my previous post.