Question

A wheel is rotating at the rate of $20{\text{rev min}}^{-1}$ in clockwise direction. What should be the magnitude of the angular acceleration of the wheel so that it stops rotating in $10s$?

• A. $\frac{2\pi }{30}{\text{rad s}}^{-2}$
• B. $\frac{7\pi }{24}{\text{rad s}}^{-2}$
• C. $\frac{5\pi }{24}{\text{rad s}}^{-2}$
• D. $\frac{7\pi }{30}{\text{rad s}}^{-2}$

Answer 1

The correct option is A $\frac{2\pi }{30}{\text{rad s}}^{-2}$

Let us assume that the clockwise direction as positive,
it is given that the wheel is rotating with constant angular
velocity
${\omega }_0=20{\text{rev min}}^{-1}$

We know that,
$=1\text{rev/ min}=\frac{1}{60}\text{rev/s}$
and we know that $1$ complete revolution is $2\pi \text{rad}$ , so $1\text{rev /min}=\frac{2\pi }{60}\text{rad/s}$
Hence $20{\text{rev min}}^{-1}$$=\frac{2\pi }{3}\text{rad/s}$.
It is given that the wheel comes to rest after $10\text{s}$

$\omega =0$ is the final angular velocity, by the equation of motion
$\omega ={\omega }_0+\alpha t$
$\alpha =-\frac{2\pi }{30}$
$\alpha =\frac{2\pi }{30}{\text{rad/s}}^2$