Question

A particle is moving in a circle of radius r under the action of a force \[F=\alpha {{r}^{2}}\]which is directed towards the center of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy=0 for r=0).
A. $\dfrac{5}{6}\alpha {{r}^{3}}$
B. $\alpha {{r}^{3}}$
C. $\dfrac{1}{2}\alpha {{r}^{3}}$
D. $\dfrac{4}{3}\alpha {{r}^{3}}$

Answer 1

Hint: Apply the knowledge of circular motion. Write down formula for kinetic energy and potential energy for a particle performing circular motion. Centripetal force is the force acting on particle performing circular motion, which is along radius of a circle and direction towards the centre of circle. Since particle is moving in a circle and directed towards centre, use formula of centripetal force which is given as $-\dfrac{m{{v}^{2}}}{r}=F$. Negative sign in formula shows that particle is moving towards centre. Then use total energy formula which is sum of potential energy and kinetic energy.

Complete step-by-step answer:
Given that, a particle is moving in a circle of radius r under the action of a force \[F=\alpha {{r}^{2}}\]which is directed towards the centre of the circle. Therefore this force \[F=\alpha {{r}^{2}}\]is providing centripetal force.
Thus,
\[\begin{align}
  & \text{centripetal force = }F=\alpha {{r}^{2}} \\
 & \dfrac{m{{v}^{2}}}{r}=\alpha {{r}^{2}} \\
 & m{{v}^{2}}=\alpha {{r}^{3}} \\
\end{align}\]
Kinetic energy of the motion is
$K.E.=\dfrac{1}{2}m{{v}^{2}}$
Substituting the value of $m{{v}^{2}}$in the above equation, we get
$K.E.=\dfrac{\alpha {{r}^{3}}}{2}$
The potential energy of a particle is nothing but the work done by force F in moving particles from r=0 to r=r. Thus,
$\begin{align}
  & P.E.=\int\limits_{0}^{r}{\vec{F}.d\vec{r}} \\
 & P.E.=\int\limits_{0}^{r}{(\alpha {{r}^{2}})dr} \\
 & \therefore P.E.=\dfrac{\alpha {{r}^{3}}}{3} \\
\end{align}$
We know,
\[\begin{align}
  & \text{Total energy = Kinetic energy + Potential energy} \\
 & \text{Total energy = }\dfrac{\alpha {{r}^{3}}}{2}+\dfrac{\alpha {{r}^{3}}}{3} \\
 & \text{Total energy = }\dfrac{5}{6}\alpha {{r}^{3}} \\
\end{align}\]
Total mechanical energy (kinetic energy + potential energy) of the particle is $\dfrac{5}{6}\alpha {{r}^{3}}$
Answer is ( A )

Note: In circular motion the body moves around a fixed point. The motion of a body along the circular path is known as circular motion. There should be net force acting on a body in order to perform a circular motion. Two main forces are acting the body along the radius; one centripetal force which acts inward while the other is centrifugal force which acts in outward direction. The centripetal force attracts the body towards the centre whereas centrifugal force throws our body away from the centre. The direction of velocity and hence acceleration changes continuously during the motion