Question

A particle of mass m is moving in a horizontal circle of radius r under centripetal force equal to $-\dfrac{K}{{{r}^{2}}}$, where K is a constant. The total energy of the particle is:
A. $-\dfrac{K}{2{{r}^{2}}}$
B. $-\dfrac{K}{2{{r}^{2}}}$
C. $-\dfrac{K}{2{{r}^{2}}}$
D. $-\dfrac{K}{2{{r}^{2}}}$

Answer 1

Hint: When a particle is moving in a horizontal circle then the total energy of the particle will be the sum of kinetic energy and potential energy of the particle.

Complete answer:
Given,

When the particle is moving in a horizontal circle than the centripetal force act on the body

= $\dfrac{m{{v}^{2}}}{r}=\dfrac{-K}{{{r}^{2}}}$(negative sign indicates the direction only)

$\Rightarrow m{{v}^{2}}=\dfrac{K}{r}$
$\text{Kinetic Energy =}\dfrac{\text{1}}{\text{2}}\text{m}{{\text{v}}^{\text{2}}}\text{=}\dfrac{K}{\text{2r}}$
$\Rightarrow Potential\,Energy\text{=-}\int\limits_{\infty }^{\text{r}}{\dfrac{K}{{{\text{r}}^{\text{2}}}}\text{dr}}$
$\Rightarrow Potential\,Energy=-K\int\limits_{\infty }^{r}{{{r}^{-2}}dr}$
$\Rightarrow Potential\,Energy=-K\left[ \dfrac{{{r}^{-1}}}{-1} \right]_{\infty }^{r}$
$\Rightarrow Potential\,Energy=-K\left[ \dfrac{-1}{r}+\dfrac{1}{\infty } \right]$
$\Rightarrow Potential\,Energy=\dfrac{K}{r}$

Total energy = kinetic energy +potential energy

$\Rightarrow \dfrac{K}{2r}-\dfrac{K}{r}=\dfrac{-K}{2r}$

Therefore, the correct choice is : (C) $-\dfrac{K}{2r}$

Note:
In the given data a negative sign indicates only the direction. so we have to exclude it in the calculation . Then the total energy of the particle will be the sum of the kinetic energy and potential energy of the system.