Question

A 2.5 kg , 20-cm-diameter turntable rotates at 150 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick.
What is the turntable's angular velocity, in rpm, just after this event?

Answer 1

The concepts required to solve this problem are those related to the conservation of the angular momentum and the moment of inertia of the disk. We will begin by calculating the moment of inertia of the disc, then the moment of inertia of the disc after the two two blocks hits and sticks to the edges of the turn table. In the end we will apply the conservation theorem.

The radius is given as,

$R = \frac{20cm}{2} = 10cm = 0.1m$

When a block falls from above and sticks to the turn table, the moment inertia of the turntable increases.

Since two blocks are stick to the turn table, the total final moment of inertia of the turntable is the sum moment of inertias of individual turntable, and two blocks.

$I_1 = \frac{1}{2} MR^2$

$I_1 = \frac{1}{2} (2.5)(0.1)^2$

$I_1 = 0.0125kg \cdot m^2$

The moment of inertia of each block is

$I_0 = mR^2$

Total moment of inertia of two block is

$I_0' = 2mR^2$

The final moment of inertia of the turn table is

$I_2 = I_1 +I'0$

$I_2 = I_1 +2mR^2$

$I_2 = 0.01kg\cdot m^2 + 2(500*10^{-3}kg)(0.1m)^2$

$I_2 = 0.0225kg\cdot m^2$

From the conservation of the angular momentum, the initial angular momentum of the system is equal to final angular momentum of the system,

Rearrange the equation we have that

$I_1\omega_2 = I_2\omega_2$

$\omega_2 = \frac{I_1\omega_2}{I_2}$

$\omega_2 = \frac{0.01*150rpm}{0.0225}$

$\omega_2 = 66.67rpm$

The magnitude of the turntable's angular velocity is 66.67rpm