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 $m{k}^2{r}^{2}t$

$\text{Step 1: Calculate velocity V}$

Given, Centripetal acceleration $a_c=K^{2}r{t}^2$ $....\left(1\right)$

Also, $a_c=\frac{V^2}{r}$ $....\left(2\right)$

From eqn $\left(1\right)$ and $\left(2\right)$

$\frac{V^2}{r}=K^{2}r{t}^2$

$\Rightarrow$ $V=Krt$ $....\left(3\right)$

$\text{Step 2: Calculate tangential acceleration V}$

$a_t=\frac{dv}{dt}=\frac{d}{dt}\left(Krt\right)$

$=Kr$

Now, Tangential force acting on the particle $F=m{a}_t$ $=mKr$ $....\left(4\right)$

$\text{Step 3: Calculate power delivered}$

Power of a force is given by, $P=\overrightarrow{F}.\overrightarrow{v}=Fvcos\theta$

Since centripetal force is perpendicular to the velocity($\theta ={90}^o$), therefore power due to centripetal force will be zero.

So calculating power due to tangential force

$P=FV\cos \theta$ $=mKr\left(Krt\right)cos{0}^o$ (From equation $\left(3\right)$ and $\left(4\right)$)

$=m{K}^2{r}^{2}t$

Hence option $B$ is correct.