Conservation of Angular Momentum - Two ants on a massless rod

[HSSN19]

Homework Statement

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.

Homework Equations

L = mr²ω

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 - \alphat²)² + m2x)

Therefore, x= [(m1 + m2)(B/4)² - m1(B/4 - \alphat²)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?

Thanks in advance!

 

[Doc Al]

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

 

[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 - \alphat², what must the second ant, mass m2, do to keep the rod's angular velocity constant?

 

[Doc Al]

Initial L = (m1 + m2)(B/4)²ω
Final L= ω(m1(B/4 - \alphat²)² + m2x)

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?)

 

[HSSN19]

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