Question

A particle of mass $m$ is moving in a circular path of constant radius $r$ such that centripetal acceleration is varying with time $t$ as $k^{2}r{t}^2,$ where $k$ is a constant. The power delivered to the particle by the force acting on it is

• A. $m^2{k}^2{r}^2{t}^2$
• B. $m{k}^2{r}^{2}t$
• C. $m{k}^{2}r{t}^2$
• D. $mk{r}^{2}t$

Answer 1

The correct option is B $m{k}^2{r}^{2}t$

Given that,
Centripetal acceleration, $a_c=k^{2}r{t}^2$
Where, $a_c=\frac{v^2}{r}$
$\Rightarrow \frac{v^2}{r}=k^{2}r{t}^2$
$\Rightarrow v=krt.......\left(1\right)$
The work done by the centripetal force is zero because the direction of centripetal force is perpendicular to the velocity.

Tangential acceleration of the particle, $a_t=\frac{dv}{dt}=kr..........\left(2\right)$

Tangential force acting on the particle, $F=m{a}_t=mkr$
The power delivered to the particle is
$P=\overrightarrow{F}.\overrightarrow{v}=Fv\cos \theta$
$\therefore P=Fv=\left(mkr\right)\times krt(\because \theta =0^{\circ })$
$\Rightarrow P=m{k}^2{r}^{2}t$

Hence option (b) is correct.