Question

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

• A. $2 \pi \, mk^2 \, r^2$
• B. $mk^2 \, r^2 t$
• C. $\frac{(mk^4 \, r^2 t^5)}{3}$
• D. zero

Answer 1

The Correct Option is B

$a_{c}=\frac{v^{2}}{r}=k^{2}\, r\, t^{2} $
$\Rightarrow v^{2}=k^{2} \,r^{2}\, t^{2} $
$K E=\frac{1}{2} m v^{2}=\frac{1}{2} m k^{2} \,r^{2}\, t^{2}$
According to work-energy theorem,
change in kinetic energy is equal to work done.
$\therefore W=\frac{1}{2} m k^{2}\, r^{2}\, t^{2}$
Thus, power delivered to the particle,
$P=\frac{d W}{d t}=m k^{2} \,r^{2} \,t$