Question

A particle of mass $m$ moves in a circular path of radius $r$ , under the action of force which delivers it constant power $p$ and increases its speed. The angular acceleration of particle at time $\left(t\right)$ is proportional

• A. $\frac{1}{\sqrt{t}}$
• B. $\sqrt{t}$
• C. $t^0$
• D. $t^{3/2}$

Answer 1

Answer: (A)

$\frac{1}{\sqrt{t}}$

The power is given as,

$p=m\times \frac{dv}{dt}\times v$

$\frac{dv}{dt}\times v=\frac{p}{m}$

By integrating both sides, we get

$\frac{1}{2}{v}^2=\left(\frac{p}{m}\right)\times t$

$\frac{1}{2}{\left(r\omega \right)}^2=\left(\frac{p}{m}\right)\times t$

$\omega =\sqrt{\left(\frac{2p}{m{r}^2}\right)\times t}$

$\omega =c{t}^{\frac{1}{2}}$

The angular acceleration is given as,

$\frac{d\omega }{dt}=\frac{1}{2}c{t}^{\frac{-1}{2}}$

Thus, the angular acceleration of particle is proportional to $\frac{1}{\sqrt{t}}$.