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

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

Step.1 Given Data,

Centripetal acceleration $\left(a_c\right)=k^{2}r{t}^2$

Step 2. Formula used,

$a_c=\frac{v^2}{r}$, $a_c$ is the centripetal acceleration. --(A)

$v$ is the velocity

$r$ is the radius

Tangential acceleration,$a_t=\frac{dv}{dt}$.

Power, $P=F\times v$

The tangential force acting on the particle, $F=m{a}_t$

$m$ is the mass of the particle.

Step 3. Calculating the power,

Centripetal acceleration$\left(a_c\right)=k^{2}r{t}^2$

Putting the values of $a_c$ from (A) we get,

$k^{2}r{t}^2=\frac{v^2}{r}$

$v=krt........1$

Tangential acceleration is defined as the rate of change of tangential velocity of the matter in the circular path.

$a_t=\frac{dv}{dt}$

$a_t=kr..........2$

$a_t$ is the tangential acceleration
The tangential force acting on the particle
$F=m{a}_t=mkr$
Power delivered is
$P=F.v=Fv\cos \theta$
$$\begin{gathered} P=mkr\times krt(\because \theta =0^o) \\ P=m{k}^2{r}^{2}t \end{gathered}$$)

Hence, the power delivered to the particle by the force acting on it is $m{k}^2{r}^{2}t$.

Hence, option B is the correct option.