Question

The angular velocity of a particle moving along positive direction varies as $\omega =\beta \sqrt{\theta }$ (in $rad/s$) where $\beta$ is a positive constant. Assuming that $\theta =0$ (at time $t=0$), the value of angular acceleration of the particle

• A. Is directly proportional to time
• B. Is constant with time
• C. Is inversely proportional with time
• D. Cannot be determined as data is insufficient

Answer 1

The correct option is B Is constant with time

Given, Angular velocity, $\omega =\beta \sqrt{\theta }$
$\Rightarrow \frac{d\theta }{dt}=\beta \sqrt{\theta }$
$\frac{d\theta }{\sqrt{\theta }}=\beta dt$
Integrating on both sides,
${\int }_0^{\theta }\frac{d\theta }{\sqrt{\theta }}={\int }_0^t\beta dt$
$2\sqrt{\theta }=\beta t$
$\therefore \theta =\frac{{\beta }^2{t}^2}{4}$
$\Rightarrow \frac{d\theta }{dt}=\frac{2{\beta }^{2}t}{4}=\frac{{\beta }^{2}t}{2}$
Then, angular acceleration
$\alpha =\frac{d^2\theta }{d{t}^2}=\frac{{\beta }^2}{2}$
$\therefore$ Angular acceleration is constant