Question

A 1.8 kg , 20-cm-diameter turntable rotates at 130 rpm on frictionless bearings. Two 510 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

Mass of the turntable = M = 1.8 kg

Diameter of the turntable = D = 20 cm = 0.2 m

Radius of the turntable = R = D/2 = 0.2/2 = 0.1 m

Moment of inertia of the turntable = I

I = MR²/2

I = (1.8)(0.1)²/2

I = 9 x 10^-3 kg.m²

Initial angular speed of the turntable = ₁ = 130 rpm

Converting the angular speed to rad/s,

₁ = 13.614 rad/s

Mass of each block = m = 510 g = 0.51 kg

Number of blocks = n = 2

The blocks hit the turntable at opposite ends of a diameter therefore each block is at a distance equal to the radius from the rotation axis.

Final angular speed of the turntable = ₂

By conservation of angular momentum,

I₁ = (I + mR² + mR²)₂

(9x10^-3)(13.614) = [9x10^-3 + (0.51)(0.1)² + (0.51)(0.1)²]₂

₂ = 6.381 rad/s

Converting the angular speed to rpm,

₂ = 60.93 rpm

Angular velocity of the turntable after the event = 60.93 rpm