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?