A 2.5 kg, 20-cm-diameter turntable rotates at 70 rpm on frictionless bearings. Two 480 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick.

Question: What is the turntable's angular velocity, in rpm, just after this event?

A: 39.6 rpm

Explanation:

Assuming that no external torques act on the system (turntable + blocks) during the collision, total angular momentum must be conserved.

Just before that the two blocks fell down on the turntable, the angular momentum is as follows:

L = I * ω

For a solid disk, the moment of inertia is mr²/2, so we can write I as follows:

I = (2.5 * ((0.2) ²/4)) kg. m² = 0.0125 kg. m²

Taking ω as 70 rpm, we get the value of the initial L as follows:

L = 0.0125 kg. m² * 70 rpm = 0.875 kg. m²*rpm

After the collision, the system has a new moment of inertia, that can be expressed as follows:

If = 0.0125 kg. m² + (2*0.48 kg * ((0.2) ² / 4)) = 0.0221 kg. m²

As the total angular momentum must be conserved, we can write the following equation:

L = If * ωf

Solving for ωf:

⇒ ωf = L/If = 0.875 kg. m². rpm / 0.0221 kg. m² = 39.6 rpm