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Conservation of Angular Momentum - Two ants on a massless rod

  1. Apr 13, 2012 #1
    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 - [itex]\alpha[/itex]t2)2 + m2x)

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

    Thanks in advance!
     

    Attached Files:

    Last edited: Apr 13, 2012
  2. jcsd
  3. Apr 13, 2012 #2

    Doc Al

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    Staff: Mentor

    Can you at least state the problem. (Where's the screenshot?)
     
  4. Apr 13, 2012 #3
    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 - [itex]\alpha[/itex]t2, what must the second ant, mass m2, do to keep the rod's angular velocity constant?
     
  5. Apr 13, 2012 #4

    Doc Al

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    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?)
     
  6. Apr 13, 2012 #5
    Oh right, forgot the square. Thank you very much!
     
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