Question

A wheel rotating at a speed of $600\text{rpm}$ (revolutions per minute) about its axis is brought to rest by applying a constant torque for $10$ seconds. The angular velocity $5$ seconds after the application of the torque is $n\pi \text{rad/s}$ where $n$ is

Answer 1

Given,
Angular speed of wheel at $t=0$
${\omega }_0=600\text{rpm}$
$=600\times \frac{2\pi }{60}\text{rad/s}=20\pi \text{rad/s}$
Angular speed of wheel at $t=10\text{s}$
$\omega =0$ (brought to rest)

From eqn. $\omega ={\omega }_0+\alpha t$
$0=20\pi +\alpha \times 10$
$\alpha =-2\pi {\text{rad/s}}^2$
Angular speed of the wheel after $5$ seconds will be
$\omega ={\omega }_0+\alpha t$
$=20\pi +(-2\pi )\times 5$
$=10\pi \text{rad/s}$
So, $n=10$