Answer to Question #150772 in Physics for Anne

Question

A motor exerts torque on a carousel for it to attain a speed of 2.5 rev/s starting from rest in 3.5s. Find the work done by the motor if the carousel has a radius of 27m and mass of 1.75 x 10^5 kg. Consider the carousel to be a thin-walled cylinder rotating about its center.

Answer 1

Since the carousel is a thin-walled cylinder rotating about its center, its moment of inertia is:

$$I = mr^2$$

where $r = 27m$ is the radius of the carousel, and $m = 1.75\times 10^5kg$ is its mass.

According to work-kinetic theorem for rotation, the amount of work done by the torquу acting on a rigid body under a fixed axis rotation (pure rotation) equals the change in its rotational kinetic energy.

The initial rotational kinetic energy of the carousel was zero started from rest, the final rotational kinetic energy is:

$$T = \dfrac{I\omega^2}{2}$$

where $\omega =2\pi \cdot 2.5\space rev/s = 5\pi \space rad/s$ is the final angular speed of the carousel. Thus, the change in otational kinetic energy and hence the work is:

$$A = T = \dfrac{mr^2\omega^2}{2}\\ A = \dfrac{1.75\times 10^5\cdot 27^2\cdot (5\pi)^2}{2} \approx 1.57\times 10^{10}\space J$$

Answer. $1.57\times 10^{10}\space J$.