Question

The motor of an engine is rotating about its axis at an angular speed of $120\text{rpm}$. It comes to rest in $10\text{s}$, after being swiched off. Assuming constant acceleration, the number of revolutions made by it before coming to rest are

• A. $5$
• B. $10$
• C. $15$
• D. $20$

Answer 1

The correct option is B $10$

Given:
$n_1=120\text{rpm}=\frac{120}{60}\text{rps}=2\text{rps}$;
$n_2=0;t=10\text{s}$

From equation of rotational motion

${\omega }_2={\omega }_1+\alpha t$

$\Rightarrow \alpha =\frac{{\omega }_2-{\omega }_1}{t}=\frac{2\pi (n_2-n_1)}{t}$

$\Rightarrow \alpha =\frac{2\pi (0-2)}{10}=\frac{-2\pi }{5}{\text{rad/s}}^2$

We also have,

${\omega }_2^2-{\omega }_1^2=2\alpha \theta$

$\Rightarrow \theta =\frac{{\omega }_2^2-{\omega }_1^2}{2\alpha }=\frac{4{\pi }^2(n_2^2-n_1^2)}{2\alpha }$

$\Rightarrow \theta =\frac{4{\pi }^2(0-2^2)}{2(-2\pi /5)}=20\pi \text{rad}$

So the number of revolutions made by motor are

$m=\frac{\theta }{2\pi }=\frac{20\pi }{2\pi }=10$

So, $\left(b\right)$ is the correct option.