Question

a disc of moment of inertia $\frac{9.8}{{\pi }^2}kg-m^2$ is rotating at 600 rmp. If the frequency of rotation changes from 600 rpm to 300 rpm, then what is the work done?

• A. 1467 J
• B. 1452 J
• C. 1567 J
• D. 1632 J

Answer 1

The correct option is A 1467 J

Work done by a force acting on a body is equal to the change produced in the kinetic energy of the body.
According to work energy theorem,
Work done = change in rotational kinetic energy
$W=(\Delta K{E}_r{)}_1-(\Delta K{E}_r{)}_2$ ...(i)
But rotational kinetic energy
$K=\frac{1}{2}I{\omega }^2$
From Eq. (i), we get
$W=\frac{1}{2}I{\omega }_1^2-\frac{1}{2}{\omega }_2^2$
$=\frac{1}{2}I({\omega }_1^2-{\omega }_2^2)$
As, $\omega =2\pi n$
Hence, we get
$W=\frac{1}{2}I\left[\right(2\pi n_1{)}^2-\left(2\pi n_2{)}^2\right]$
$=\frac{1}{2}I\times 4{\pi }^2(n_1^2-n_2^2)$ ...(ii)
Given, $I=\frac{9.8}{{\pi }^2}kg-m^2$
$n_1=600rpm=10rps$
$n_1=300rpm=5rps$
From Eq. (ii), we get
$w=\frac{1}{2}\times \frac{9.8}{{\pi }^2}\times 4{\pi }^2({10}^2-5^2)$
$=1467J$