A centrifuge rotor rotating at 9600 rpm is shut off and is e

A centrifuge rotor rotating at 9600 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 2.30 mN .

Part A

If the mass of the rotor is 3.54 kg and it can be approximated as a solid cylinder of radius 0.0610 m , through how many revolutions will the rotor turn before coming to rest?

Express your answer to three significant figures.

Part B

How long will it take?

Express your answer to three significant figures and include the appropriate units.

Solution

v = 9600 rpm = 9600 * (2*)/60 = 1005 rad/s
Fr = 2.30 Nm
m = 3.54 kg
r = 0.0610 m

We know,
torque = I*
2.30 = 1/2 * m*r^2 *
2.30 = 1/2 * 3.54 * 0.061^2 *
= 349.2 rad/s^2

^2 = o^2 + 2**
0 = 1005^2 - 2*349.2*
= 1446.2 rad
= 1446.2 / (2*) rev
= 230 rev
revolutions will the rotor turn before coming to rest, = 230 rev

(b)
= o + t
0 = 1005 - 349.2*t
t = 2.89 s
time it will take, t = 2.89 s