Q137 / 180 · COMEDK 2020 · Physics
A particle of mass $m$ is moving in a horizontal circle of radius $R$ under a centripetal force equal to $-\frac{A}{R^{2}}(A=$ constant $)$. The total energy of the particle is
- A$\frac{A}{R}$
- B$-\frac{A}{R}$
- C$\frac{A}{2 R}$
- Correct answerD$-\frac{A}{2 R}$
Solution
Given, centripetal force, $F=-\frac{A}{R^{2}}$
$\therefore$ Potential energy, $U=-\frac{d F}{d R}$
$$
\begin{aligned}
&=-\frac{d}{d R}\left(-\frac{A}{R^{2}}\right) \\
&=-\frac{A}{R}
\end{aligned}
$$
In circular motion, centripetal force is responsible for the motion of object.
$$
\begin{aligned}
&\therefore \quad \frac{m v^{2}}{R}=\frac{A}{R^{2}} \\
&\Rightarrow \quad m v^{2}=\frac{A}{R} \\
&\Rightarrow \quad \frac{1}{2} m v^{2}=\frac{A}{2 R} \\
&\therefore \text { Kinetic energy, } K=\frac{1}{2} m v^{2}=\frac{A}{2 R} \\
&\therefore \text { Total energy }=U+K \quad=-\frac{A}{R}+\frac{A}{2 R}=-\frac{A}{2 R}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.