Question

A particle is moving in a circular path of radius $a$ under the action of an attractive potential $U=-\frac{k}{2r^2}$. Its total energy is

• A. Zero
• B. $\frac{-3k}{2a^2}$
• C. $\frac{-k}{4a^2}$
• D. $\frac{k}{2a^2}$

Answer 1

The correct option is A

Zero

Step 1: Given data

Potential, $U=-\frac{k}{2r^2}$

Step 2: Find total energy

By the concept of Force and potential energy gradient,

$$\begin{gathered} F=-\frac{dU}{dr} \\ F=-\frac{k}{2}\frac{(-2)}{r^3} \\ F=\frac{k}{r^3} \end{gathered}$$

Centripetal force, $F_c$ is given by formulae,

$\begin{array}{rcl}{F}_c& =& \frac{m{v}^2}{r}\\ \frac{k}{r^3}& =& \frac{m{v}^2}{r}\\ \frac{k}{2r^2}& =& \frac{m{v}^2}{2}\\ m{v}^2& =& \frac{k}{r^2}\end{array}$ [ $m$ is mass, $v$ is velocity]

By using the formula of the kinetic energy, $K.E.$

$K.E.=\frac{1}{2}m{v}^2$

$K.E.=\frac{k}{2r^2}$ [ from the above value]

Total energy, $E$ is given by

$$\begin{gathered} E=U+K.E. \\ E=\frac{-k}{2r^2}+\frac{k}{2r^2} \\ E=0 \end{gathered}$$

Hence, option $\left(A\right)$ is correct.