# Conservation of Angular Momentum - Two ants on a massless rod

1. Apr 13, 2012

### HSSN19

1. The problem statement, all variables and given/known data

This is a problem about angular momentum and torque from the physics textbook Don't Panic Volume I. Attached is a screenshot of the problem.

2. Relevant equations

L = mr2ω

3. The attempt at a solution

The angular momentum must be conserved to keep ω constant.

Initial L = (m1 + m2)(B/4)2ω
Final L= ω(m1(B/4 - $\alpha$t2)2 + m2x)

Therefore, x= [(m1 + m2)(B/4)2 - m1(B/4 - $\alpha$t2)2]/m2

The second ant must move toward the center so that its distance from it is x.

What do you guys think? This is the last problem so I thought it would not be this simple. Is there something missing or should I do something more?

#### Attached Files:

• ###### Don't Panic 14_6.JPG
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Last edited: Apr 13, 2012
2. Apr 13, 2012

### Staff: Mentor

Can you at least state the problem. (Where's the screenshot?)

3. Apr 13, 2012

### HSSN19

That's odd. I'm sure it was in the attachment... Oh well.

A massless rod of length B is pivoted at its center so that it can rotate in the horizontal plane. Two ants are riding this rod in the locations shown (both B/4 away from the center) and the rod is rotating with angular velocity ω in the counterclockwise direction, viewed from above. If ant 1, mass m1, starts (at t = 0) moving toward the center so that his distance from it is B/4 - $\alpha$t2, what must the second ant, mass m2, do to keep the rod's angular velocity constant?

4. Apr 13, 2012

### Staff: Mentor

Looks good except that x should be x2.

Looks like you have the right idea. (And that there's not much to it.) Perhaps the expression can be simplified a bit. (If ant 1 moves toward the center, which direction must ant 2 move?)

5. Apr 13, 2012

### HSSN19

Oh right, forgot the square. Thank you very much!