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A uniform vertical turntable (mass M and radius R center O) is at rest on the xy plane and is mounted on a frictionless axle, which lies along the vertical z axis. I throw a lump of putty mass m with speed v towards the edge of the turntable, so it approaches along a line that passes within a distance b of O.

So imagine shooting this piece of putty horizontally with speed v at the edge of this turntable. (The putty sticks to the edge of the turntable)

The question is, what is the angular velocity of the turntable? I understand before impact the angular momentum of the system is r(mv)sin(theta) = mvb. (theta is the angle from O to the point the putty was projected from measured along the horizontal) Once the putty hits the turntable however I don't understand how angular momentum is conserved. I understand afterward the turntable, with the putty on, has angular momentum of (m+M/2)R^2*w, but I don't understand why these two quantities are equal. I understand that angular momentum remains constant if the torque due to external forces at all times is zero...

Why during impact and afterwards is the external torque zero?

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Notice that you have implicitly chosen to measure angular momentum about the axle of the platform.

That means that all the forces exerted by the axle on the platform are applied through the axis for rotation, meaning the torque they exert is $$\text{force} \times \text{lever arm} = F \times 0 = 0\,.$$

And there are no other forces present expect those between the putty and the disk. From Newton's third law (in the strong form that can be applied in mechanics), the torque resulting from the forces the disk applies to the putty is equal and opposite to the torques resulting from the forces the putty applies on the disk. $$ \tau_\text{putty on disk} = - \tau_\text{disk on putty} \,.$$ The net from those interactions is also zero.

And that's it.

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