A 2.2 kg, 20 cm diameter turntable rotates at 120 rpm on frictionless bearings. Two 540 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is th

The initial angular momentum of the turntable is given by L = I?, where I is the moment of inertia and ? is the angular velocity. Since the turntable is rotating on frictionless bearings, we can assume that its initial angular momentum is conserved during the collision with the blocks. When the blocks stick to the turntable, the moment of inertia of the system increases due to the addition of the mass of the blocks. The moment of inertia of a thin disk is given by I = (1/2)mr?, where m is the mass and r is the radius. Adding the blocks increases the moment of inertia to I = (1/2)(2.2 + 0.54 + 0.54)(0.1)? = 0.018 kg·m?. During the collision, the blocks reduce the initial angular velocity of the turntable, but conserve the initial angular momentum. We can solve for the final angular velocity using the conservation of angular momentum equation: L = I? where L is the initial angular momentum of the turntable, and I is the final moment of inertia of the system (turntable + blocks). The initial angular momentum of the turntable is L = (1/2)I? = (1/2)(0.5)(2.2)(0.1)?(2?)(120/60) = 0.0133 kg·m?/s. The final moment of inertia of the system is I = (1/2)(2.2 + 0.54 + 0.54)(0.1)? = 0.018 kg·m?. Solving for the final angular velocity, we get: ? = L/I = 0.0133/0.018 = 0.74 rad/s Therefore, the turntable's angular velocity just after the collision with the blocks is 0.74 rad/s.