Question

A uniform disk has a mass of

3.7

kg and a radius of

0.40

m. The disk is mounted on frictionless

bearings and is used as a turntable. The turntable is initially rotating at

30

rpm.

A thin

-

walled

hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the

turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for 0.20 s until it

acquires the same final angular velocity as the turntable. What is the final angular momentum of

the system?

Answer 1

For disk: Moment of Inertia I_d = mr² / 2 = 3.7kg * (0.40m)² / 2 = 0.296 kg·m²

Given, Initial angular velocity ω_i = 30rpm * 1min / 60s * 2π rad / rev = 3.14159 rad/s

For cylinder: Moment of Inertia Ic = mr² = 0.592 kg·m²

It doesn't matter how long it slips; momentum is conserved.

Initial momentum = final Momentum = 0.296kg·m² * 3.14159 rad/s = 0.9299 kg·m²/s

Therefore, Final angular momentum of the system = 0.9299 kg·m²/s