Question

A particle is rotating with constant angular acceleration on a circular track. If its angular velocity changes from $20\pi rad/s$ to $40\pi rad/s$ in $10s$, what are the number of revolutions that the particle has completed during this time?

• A. 100
• B. 150
• C. 250
• D. 1000

Answer 1

Answer: (B)

From equation of rotational motion
$\omega ={\omega }_0+\alpha t$
Here, $\omega =40\pi rad/s,$
${\omega }_0=20\pi rad/s,t=10s$
$\therefore 40\pi =20\pi +\alpha +10$
$\Rightarrow \alpha =\frac{20\pi }{10}=2\pi rad/s^2$
Again, $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$
$=20\pi \times 10+\frac{1}{2}\times 2\pi \times (10{)}^2$
$=200\pi \times 10+\frac{1}{2}\times 2\pi \times (10{)}^2$
$=200\pi +100\pi =300\pi$
$\therefore$ Number of revolutions
$n=\frac{\theta }{2\pi }=\frac{300\pi }{2\pi }=150$