Angular Momentum/Impulse Problem

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A 1.8 kg turntable with a diameter of 20 cm rotates at 160 rpm when two 500 g blocks fall and stick to it. The discussion revolves around whether to apply conservation of angular momentum or another method, with a focus on the blocks' velocities before impact. It is suggested that the blocks do not carry angular momentum, as they fall vertically. The key principle highlighted is that no vertical torque affects the turntable, indicating that angular momentum remains constant. The final angular velocity of the turntable can be determined using the conservation of angular momentum.
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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.
 
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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).
 
Thanks! that's all I needed!
 
hi cdbowman42! :smile:
cdbowman42 said:
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)
 
I would think about it in the following way

\int \tau dt = \Delta ( I \omega)

where \int \tau dt is the angular impulse and \Delta (I \omega) 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

(I \omega)_{initial} = (I \omega)_{final}
 
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