Question

1. A centrifuge in a medical laboratory rotates at an angular speed of 3650 rev/min. When switched off, it rotates through 50.0 revolutions before coming to rest. Find the constant angular acceleration of the centrifuge.

2. A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?

3. A potter's wheel moves from rest to an angular speed of 0.19 rev/s in 35 s. Find its angular acceleration in radians per second per second.

Answer 1

1.Divide the initial ANGULAR velocity in radians/sec by the time required to stop.
The initial angular velocity is 3650 rev/min*(2 pi rad/rev)/60 (s/min) = 382.2 rad/s. The time required to stop is
(Number of revolutions)/(average speed) = 50 rev/1825 rev/min = .0274 min = 1.644 s.
angular acceleration = -382.2 rad/s/1.644 s = ? rad/s^2
The minus sign is there becasue it is slowing down.

2. Impulse = change in momentum
Don't forget that the sign of the momentum changes. You will need to calculate the velocity sqrt(2gH) when the ball hits the floor. It will rebound with a lower velocity that you also need to calculate. You can use the same sqrt(2gH) formula with a lower value for H.

3. Use method similar to (1), but the initial velocity is in rev/s this time, not rev/min