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Angular Momentum/Impulse Problem

  1. Apr 21, 2011 #1
    1. A 1.8 kg, 20 cm diameter turntable rotates at 160 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick. What is the turntable's angular velocity, in rpm, just after this event?


    2. Angular momentum(L)=angular velocity(w)*moment of inertia(I)


    3. I'm confused as to whether or not to appraoch this as a conservation of momentum problem or some other way. I think I would need to know the velocities of the blocks before they hit to to this.
     
  2. jcsd
  3. Apr 21, 2011 #2
    Do the blocks carry with them any angular momentum of their own? (It seems the problem is set up to hint that they do not).
     
  4. Apr 21, 2011 #3
    Thanks! thats all I needed!
     
  5. Apr 21, 2011 #4

    tiny-tim

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    hi cdbowman42! :smile:
    hint: what was their angular momentum about the axis, before they hit? :wink:

    (assuming they fell vertically at speed v)
     
  6. Apr 21, 2011 #5
    I would think about it in the following way

    [tex] \int \tau dt = \Delta ( I \omega) [/tex]

    where [tex] \int \tau dt [/tex] is the angular impulse and [tex] \Delta (I \omega) [/tex] is the change in angular momentum.

    We can see that no vertical torque is exerted on the turntable, so the angular momentum must remain constant.

    Then we must have

    [tex] (I \omega)_{initial} = (I \omega)_{final} [/tex]
     
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