Question

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

• A. $m{k}^2{r}^4{t}^2$
• B. $m{k}^2{r}^2{t}^2$
• C. $m{k}^2{r}^{3}t$
• D. $mk{r}^4{t}^2$

Answer 1

The correct option is C $m{k}^2{r}^{3}t$

As the particle starts from rest so initial kinetic energy is zero.
Given,
Centripetal acceleration, $a_c=k^2{r}^2{t}^2$
We know that, centripetal acceleration is given by, $a_c=\frac{v^2}{r}$
So, $\frac{v^2}{r}=k^2{r}^2{t}^2$
$\Rightarrow v^2=k^2{r}^3{t}^2$
Now, Kinetic energy, $K=\frac{1}{2}m{v}^2$
$=\frac{1}{2}m{k}^2{r}^3{t}^2$
Power delivered,
$P=\frac{dK}{dt}=\frac{d}{dt}\left(\frac{1}{2}m{k}^2{r}^3{t}^2\right)$
$=m{k}^2{r}^{3}t$
Hence option $\left(c\right)$ is correct.