A 1.9 kg, 20-cm-diameter turntable rotates at 150 rpm on frictionless bearings. Two 520 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable's angular velocity, inrpm, just after this event?

Answer: 73.08 rpm

Explanation:

Given

Mass of the table, m (t) = 1.9 kg

Radius of table, r = 20/2 cm = 10 cm = 0.1 m

Mass of the block, m (b) = 500 g = 0.5 kg

Speed of rotation, w (i) = 150 rpm

Using conservation of angular momentum, where, initial angular momentum is equal to final angular momentum

L (i) = L (f)

I (i) w (i) = I (f) w (f)

w (f) = I (i) w (i) / I (f)

also, moment of inertia a disk or cylinder is I = 1/2mr²

Thus, I (i) = 1/2 mr²

I (i) = 1/2 * 1.9 * 0.1²

I (i) = 0.0095 kgm²

I (f) = Σm (i) r (i) ² = I (i) + I (block1) + I (block2)

I (f) = I (i) + I (a) + I (b)

I (f) = 0.0095 + m (a) r (a) ² + m (b) r (b) ²

Remember, r (a) = r (b) = R

m (a) = m (b) = M, so that

I (f) = 0.0095 + MR² + MR²

I (f) = 0.0095 + 2MR²

I (f) = 0.0095 + 2 * 0.5 * 0.1²

I (f) = 0.0095 + 0.01

I (f) = 0.0195 kgm²

Substituting for values in the first equation

w (f) = I (i) w (i) / I (f)

w (f) = (0.0095 * 150) / 0.0195

w (f) = 1.425 / 0.0195

w (f) = 73.08 rpm

Thus, the turntable's angular velocity is 73.08 rpm