A 2.5 kg, 20-cm-diameter turntable rotates at 80 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 speed, in rpm, just after this event?

the turntable's angular speed after the event is 50 rpm

Explanation:

Since the turntable rotates on frictionless bearings, this means that there are no external torques that will act on the table. Therefore, the table and the two blocks can be considered as an isolated system.

Therefore we will use the conservation of angular momentum which states that the initial angular momentum of the system is equal to the final angular momentum of the system.

L_i = L_f

I_i ω_i = I_f ω_f

ω_f = (I_i ω_i) / I_f (1)

Therefore, we must determine the initial moment of inertia, before dropping two blocks I_i, and the final moment of inertia, after dropping the two blocks I_f.

The moment of inertia for a disk is

I_table = 1/2 M R²

I_i = 1/2 M R²

= 1/2 (2.0) (0.1) ²

= 0.01 kg·m² (2)

After dropping the blocks, we will add their moment of inertia about the axle of rotation to the moment of inertia of the table, in order to get to the final moment of inertia

I_f = ∑ m_i r_i² = I_table + I_block, 1 + I_block, 2

I_f = I_t + I₁ + I₂

I_f = 0.01 + (m₁ r₁²) + (m₂ r₂²)

r₁ = r₂ and m₁ = m₂

Thus,

I_f = 0.01 + 2 m R²

I_f = 0.01 + 2 (0.5) (0.1) ²

I_f = 0.02 kg·m² (3)

Therefore, substitute (2) and (3) into (1):

ω_f = (I_i ω_i) / I_f

ω_f = (0.01) (100) / 0.02

ω_f = 50 rpm

Therefore, the turntable's angular speed after the event is 50 rpm