Question

A motor exerts torque in a carousel for it to attain a speed of 2.5 rev/s starting

from rest in 3.5 s. Find the work done by the motor if the carousel has a radius

of 27 m and a mass of 1.75 X 105 kg. Consider the carousel to be a thin-walled

cylinder rotating about its center

Answer 1

The work done is equal to the change in kinetic energy

$$W=\frac{I\omega^2}{2}=\frac{mr^2(2\pi n)^2}{4}$$

$$W=\frac{1.75*10^5*27^2*(2\pi *2.5)^2}{4}=6.2*10^9\:\rm N\cdot m$$