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 $10\, s$, 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=10 s$
$\therefore 40 \,\pi=20 \pi+\alpha+10$
$\Rightarrow \alpha=\frac{20 \pi}{10}=2 \,\pi\, rad / s ^{2}$
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
$n=\frac{\theta}{2 \pi}=\frac{300 \pi}{2 \pi}=150$