Question

A record on a record player spins at 78.3 rpm. Two small 25.0 g balls of clay are dropped on the record and hit it simultaneously, at the edge of the record and at opposite ends from one another. Treat the record as a disk with radius 5.50 cm and mass 530. g, and the clay balls as particles.

What is the record's angular velocity just after the clay balls hit it? Give your answer in rad/s.

Answer 1

This is a conservation of angular momentum

Angular momentum = I * ω

For a solid disk, I = ½ * m * r^2
I = ½ * 0.530* 0.055^2 = 8.016*10^-4

As the record rotates one time, it rotates an angle of 2 π radians. One minute is 60 seconds.
One rpm = 2 π/60 = π/30 rad/s
ω = 78.3 * π/30 = 2.61 * π
This is approximately 8.2 rad/s.

Initial angular momentum = 8.016*10^-4 * 2.61 * π
This is approximately 6.572*10^-3 kg * m/s.

For each clay ball, I = m * r^2
I = 0.025 * 0.055^2 = 7.562*0^-5
Total I = 8.016*10^-4 + 2 * 7.562*10^-5 = 9.528*10^-4

Final angular momentum = 9.528*10^-4 * ωf

9.528*10^-4 * ωf = 6.572*10^-3
ωf = 6.572*10^-3 ÷ 9.528*10^-4
This is approximately 6.897 rad/s