Question

A grinding wheel attained a velocity of $40\text{rad/s}$ in $4s$ starting from rest. Find the number of revolutions made by the wheel.

• A. $\frac{40}{\pi }$ revolutions
• B. $\frac{50}{\pi }$ revolutions
• C. $\frac{20}{\pi }$ revolutions
• D. $\frac{25}{\pi }$ revolutions

Answer 1

The correct option is A $\frac{40}{\pi }$ revolutions

Method $\text{I}$

It is given that angular velocity of $40\text{rad/s}$ in $4s$ starting from rest ${\omega }_0=0$,
by the equation of motion by constant angular acceleration,

$\omega ={\omega }_0+\alpha t$
$40=\alpha t$
$\alpha =10\text{rad/s}$
Now we know that, $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2=\frac{1}{2}\times 10\times 16$
$\theta =80\text{rad}$, we know that $1\text{rad}=\frac{1}{2\pi }$ revolution
$\Rightarrow \theta =\frac{80}{2\pi }=\frac{40}{\pi }$ revolutions.


Method $\text{II}$
We have for the angular displacement $\theta =\left(\text{initial velocity+final velocity}\right)\frac{t}{2}$
Hence we get, $\theta =(0+40)\frac{4}{2}=80\text{rad}$
we know that $1\text{rad}=\frac{1}{2\pi }$ revolution
$\Rightarrow \theta =\frac{80}{2\pi }=\frac{40}{\pi }$ revolutions.

​​​​​​​Option A is the correct answer.