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 centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy $=0$ for $r=0)$

• A. $\frac{5}{6}\alpha r^3$
• B. $\alpha r^3$
• C. $\frac{1}{2}\alpha r^3$
• D. $\frac{4}{3}\alpha r^3$

Answer 1

Answer: (A)

$\frac{5}{6}\alpha r^3$

The force F is providing centripetal acceleration to the particle
The kinetic energy is given as $\frac{1}{2}m{v}^2$
$F=\frac{m{v}^2}{r}=\alpha r^2$ multiplying both sides $r$ and dividing by $2$
$\Rightarrow \frac{1}{2}m{v}^2=\frac{\alpha r^3}{2}$
The potential energy is work done by force $F$ from moving the mass from $r=0$ to $r=r$
$U={\int }_0^r\overrightarrow{F}d\overrightarrow{r}$
$U={\int }_0^r\alpha r^{2}dr$
$U=\frac{\alpha r^3}{3}$
Total energy$=K.E.+P.E.=\frac{\alpha r^3}{2}+\frac{\alpha r^3}{3}=\frac{5\alpha r^3}{6}$
Hence correct answer is option A.