A particle of mass $\mathrm{m}$ starts moving in a circular path of constant radius $\mathrm{r}$, such that its centripetal acceleration $a_c$ is varying with time t as $a_c=k^{2}r{t}^2$, where $\mathrm{k}$ is a constant. What is the power delivered to the particle by he forces acting on it?

As $a_c=\left(v^2/r\right)$ so $\left(v^{2}r\right)=k^{2}r{t}^2$

$\therefore \therefore$ Kinetic energy $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{k}^2{r}^2{t}^2$

Now, from work-energy theorem

$W=\Delta K=\frac{1}{2}m{k}^2{r}^2{t}^2-0$ 

$P=\frac{dW}{dt}=\frac{d}{dt}\left(\frac{1}{2}m{k}^2{r}^2{t}^2\right)=m{k}^2{r}^{2}t$

[as at t=0, K=0 ]

Alternate solution : Given that $a_c=k^{2}r{t}^2$, so that

$F_c=m{a}_c=m{k}^{2}r{t}^2$

$a_c=\left(t^2/r\right)$, so $\left(v^2/r\right)=k^{2}r{t}^2$

Now, as

v=k r t

or

$a_i=\left(dvdt\right)=kr$

So, that

$F_1=m{a}_t=mkr$

i.e.

$\mathrm{F}={\mathrm{F}}_{\mathrm{c}}+{\mathrm{F}}_{\mathrm{t}}$

Now, as

$\mathrm{P}=\frac{\mathrm{dW}}{\mathrm{dt}}=\mathrm{F}·\mathrm{v}=\left({\mathrm{F}}_{\mathrm{e}}+{\mathrm{F}}_t\right)·\mathrm{v}$

In circular motion, ${\mathrm{F}}_{\mathrm{e}}$ is perpendicular to $\mathrm{v}$ while ${\mathrm{F}}_{,}$parallel to it, so

$\mathrm{P}={\mathrm{F}}_t\mathrm{v}$

$\mathrm{P}={\mathrm{mk}}^2{\mathrm{r}}^2\mathrm{t}$