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Rolling/Angular Momentum

  1. Nov 21, 2005 #1
    A solid horizontal cylinder of mass 13.1 kg and radius 1.22m rotates with an angular speed of 2.99 rad/s about a fixed vertical axis through its center. A 0.258kg piece of putty is dropped vertically onto the cylinder at a point 0.853m from the center of rotation, and sticks to the cylinder. What is the final angular speed of the system. Answer in rad/s



    Now, I know the solution to this question involves conservation of momentum.
    mgh=the sum of rotational and translational kinetic energy

    i'm guessing the mass to be used initially is that of the cylinder, while the mass to be used to calculate the P.E from the height, is the mass of the cylinder and putty. Can anyone confirm this?
     
  2. jcsd
  3. Nov 21, 2005 #2

    Doc Al

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    Staff: Mentor

    You'll need to consider the rotational inertia of the system before and after the putty is dropped. How does adding the putty change the rotational inertia?
     
  4. Nov 21, 2005 #3
    I need help with this one also... any more incite would be appreciated. Not really understanding this problem.:grumpy:
     
  5. Nov 21, 2005 #4
    Any hint as to what equation to use? pleeeaaassee :)
     
  6. Nov 21, 2005 #5
    Before the putty is dropped its not rotating so it has no Angular momentum... after its dropped it has a angular momentum....

    you have to use the law of conservation of angular momentum where

    initial = just cylinder
    final = cylinder + putty
    for final: w(Icylinder+Iputty)

    L=Iw

    i think :X
     
  7. Jul 13, 2007 #6
    but how do you find the moment of inertia for the putty? is it just MR^2?
     
  8. Jul 13, 2007 #7

    Doc Al

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    Staff: Mentor

    Yes. Just treat it as a point mass. (Note that this thread is quite old!)
     
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