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Angular momentum ball problem

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A 5.00-kg ball is dropped from a height of 12.0 m above
    one end of a uniform bar that pivots at its center. The bar has mass
    8.00 kg and is 4.00 m in length. At the other end of the bar sits
    another 5.00-kg ball, unattached to the bar. The dropped ball sticks
    to the bar after the collision. How high will the other ball go after
    the collision?

    2. Relevant equations
    L= r X p
    L = I [tex]\omega[/tex]
    mgh =1/2 mv^2

    3. The attempt at a solution
    The solution seems to require the conservation of angular momentum. But i am not sure how angular momentum could be conserved? Doesnt the weight of the falling ball provide a net external torque to the system?
  2. jcsd
  3. Nov 15, 2008 #2


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    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi boredaxel! Welcome to PF! :smile:

    Yes, it does provide a torque, but that doesn't matter, because it isn't an external torque …

    at least, if you consider the whole system together, it isn't external! :wink:

    Angular momentum, like momentum, is always conserved.

    Hint: treat this exactly the same way as you would if the ball was hitting a block with another ball directly the other side …

    you'd use conservation of (linear) momentum, and v1f = v2f, wouldn't you?

    Well, do the same, except … "angularly"! :biggrin:
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