Question

A particle moves in a circle of radius $R$ with a constant speed under a centripetal force $F$. The magnitude of work done in completing a full circle will be:
$\begin{aligned} & A)2RF \\\ & B)\pi {{R}^{2}}F \\\ & C)2\pi RF \\\ & D)Zero \\\ \end{aligned}$

Answer 1

When a particle is moving with uniform velocity in a circle, the direction of force acting on the particle is towards the centre of the circle. Work done on a particle is defined as the product of force acting on the particle and the displacement caused on the particle, due to the applied force. Displacement refers to the distance between the initial position of a particle and the final position of the particle.

Formula used:
$1){{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
$2)W=Fd\cos \theta$

Complete step by step answer:
When a particle is moving with uniform velocity in a circular direction, the force acting on the particle is termed as centripetal force. The direction of centripetal force on the moving particle is towards the centre of the circular path, in which the particle is moving. Magnitude of centripetal force is given by
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
where
${{F}_{c}}$ is the centripetal force acting on a particle in circular motion
$v$ is the velocity of particle
$m$ is the mass of particle
$r$ is the radius of the circular path, in which the particle is moving
Let this be equation 1.
Now, we know that work done on a particle is equal to the product of force acting on the particle and the displacement caused, due to the applied force. Mathematically, work done is equal to the dot product of force acting on a particle and the displacement caused on the particle, as given below:
$W=Fd\cos \theta$
Let this be equation 2.
Coming to our question, we are provided that a particle is moving at constant speed or uniform velocity, in a circular path of radius $R$. We are also given that the magnitude of centripetal force is equal to $F$. We are required to determine the work done on the particle when it completes one cycle of rotation.
Using equation 1, centripetal force can be expressed as
$F=\dfrac{m{{v}^{2}}}{R}$
where
$F$ is the centripetal force acting on the given particle
$v$ is the velocity of the particle
$m$ is the mass of the particle
$R$ is the radius of the circular path, in which the particle is moving
Let this be equation 3.
Now, when the particle completes one cycle of rotation, the displacement caused on the particle is equal to zero because the particle is back to its initial position after the completion of single rotation. Therefore, if $d$ represents the displacement of the given particle, it is given by
$d=0$
Let this be equation 4.
If the work done on the particle by the centripetal force to complete one cycle of rotation is denoted as $W$, from equation 2, we have
$W=Fd\cos \theta =\dfrac{m{{v}^{2}}}{R}\times 0\times \cos \theta =0$
where
$F=\dfrac{m{{v}^{2}}}{R}$, from equation 3
$d=0$, from equation 4
Therefore, the magnitude of work done in completing a full circle is equal to zero. The correct option to be marked is $D$.

Note:
As already mentioned, work done on a particle is the dot product of force acting on the particle and the displacement caused due to the applied force. Work done is given by
$W=Fd\cos \theta$
Here,
$\theta$ is the angle between the direction of the force acting on the particle and the direction of motion of the particle. In our case, the direction of centripetal force acting on the particle is towards the centre of the circle and the direction of motion of the particle at any point on the circle is along the tangent drawn at that point, acting outwards the circle. Clearly, the angle between centripetal force and the tangent at any point on the circle is $90{}^\circ$, and hence, work done on the particle turns out to be zero. Students can approach the question in this way too.