Angular Momentum/Impulse Problem

  • Thread starter cdbowman42
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  • #1
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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.
 

Answers and Replies

  • #2
olivermsun
Science Advisor
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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).
 
  • #3
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Thanks! thats all I needed!
 
  • #4
tiny-tim
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hi cdbowman42! :smile:
I think I would need to know the velocities of the blocks before they hit to to this.

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

(assuming they fell vertically at speed v)
 
  • #5
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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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