Question

A particle of mass M is moving in a circle of fixed radius R in such a way that its centripetal acceleration at time t is given by ${n^2}R{t^2}$ where n is a constant. The power delivered to the particle by the force acting on it is:
A. $\dfrac{1}{2}M{n^2}{R^2}{t^2}$
B. $M{n^2}{R^2}t$
C. $Mn{R^2}{t^2}$
D. $Mn{R^2}t$

Answer 1

Hint: When an object is rotating in a circle its direction of velocity keeps on changing. When velocity changes there will be an acceleration and this contributes for the force. Now the force which is responsible for the change in direction in case of circular motion is called a centripetal force. Power is the dot product of force and the velocity. We will solve the question by using these two hints.
Formula used:
$\eqalign{
  & {a_c} = \dfrac{{{v^2}}}{R} \cr
  & {a_t} = \dfrac{{dv}}{{dt}} \cr} $

Complete step-by-step solution:
Centripetal force is given by the formula ${F_C} = \dfrac{{m{v^2}}}{R}$
Where ‘v’ is the magnitude of the velocity of rotation and ‘R’ is the radius of the circle of the circular motion and ‘m’ is the mass of the object which is rotating in a circle.
Force will be the product of mass and acceleration. So the centripetal acceleration will be
${a_c} = \dfrac{{{v^2}}}{R}$
The expression for the centripetal acceleration is given as ${n^2}R{t^2}$
So by equating the two expressions we will get
$\eqalign{
  & \dfrac{{{v^2}}}{R} = {n^2}R{t^2} \cr
  & \Rightarrow {v^2} = {n^2}{R^2}{t^2} \cr
  & \Rightarrow v = nRt \cr} $
The centripetal force will be acting along the center of the circle. That is in the radial direction.
But the velocity of the particle in the circular motion will be along the tangential direction. So the power delivered by the centripetal force will be zero as the dot product of velocity and the centripetal force will be zero as both are perpendicular.
So power will be delivered by the tangential force only and the tangential force is given by
${F_t} = m{a_t}$
$ \Rightarrow {a_t} = \dfrac{{dv}}{{dt}}$
$\eqalign{
  & \Rightarrow {F_t} = m\dfrac{{dv}}{{dt}} \cr
  & \Rightarrow {F_t} = mnR \cr} $
The power delivered will be a dot product of tangential force and the velocity.
$\eqalign{
  & P = {F_t}.v \cr
  & \Rightarrow P = \left( {mnR} \right)\left( {nRt} \right)\cos {0^0} \cr
  & \Rightarrow P = \left( {mnR} \right)\left( {nRt} \right) \cr
  & \Rightarrow P = M{n^2}{R^2}t \cr} $
The mass of the particle is given by M so we will replace ‘m’ with ‘M’
Hence option B will be the answer.

Note: Here the speed of the particle is varying with time. So there is tangential acceleration. But if the speed is constant then we call it a uniform circular motion. It doesn’t matter if the centripetal force is constant or not, it can never do the work on the particle which is in a circular motion as it will always be perpendicular to the displacement.