Question

The motor of an engine is rotating its axis with an angular velocity of $100\text{rev per min}$. It comes to rest in $15\text{s}$ after being switched off, assuming constant angular deceleration. What is the number of revolutions made by it before coming to rest ?

• A. $12.5$
• B. $40$
• C. $32.6$
• D. $15.6$

Answer 1

The correct option is A $12.5$

Given:

The angular velocity of motor,

${\omega }_i=100\text{rev/min}=100\text{RPM}$

$\Rightarrow {\omega }_i=\frac{100\times 2\pi }{60}=\frac{10\pi }{3}\text{rad/s}$

Motor comes to rest in $15\text{sec}$, applying kinetic equation:

${\omega }_f={\omega }_i+\alpha \Delta t$

putting the value

$\Rightarrow 0=\frac{10\pi }{3}+\alpha \left(15\right)$

$\Rightarrow \alpha =-\frac{10\pi }{45}{\text{rad/s}}^2$

Again applying kinetic equation,

${\omega }_f^2={\omega }_i^2+2\alpha \Delta \theta$

$\Rightarrow 0={\left(\frac{10\pi }{3}\right)}^2+2\left(\frac{-10\pi }{45}\right)\Delta \theta$

$\Rightarrow \frac{20\pi }{45}\Delta \theta =\frac{100{\pi }^2}{9}$

$\Delta \theta =25\pi \text{rad}$

Total number of revolutions made before coming to rest is,

$n=\frac{\Delta \theta }{2\pi }$

$\Rightarrow n=\frac{25\pi }{2\pi }=12.5$

Hence, option $\left(a\right)$ is correct.