Question

A rod of mass $m$ and length $l$ is kept near a point mass $m$ which is at a distance of $a$ from the center of the rod as shown in the figure. Find the potential energy due to the rod at point mass.

• A. $U=\frac{-G{m}^2}{l}\ln \left(\frac{2a+l}{2a-l}\right)$
• B. $U=\frac{-G{m}^2}{l}\ln \left(\frac{a+l}{a}\right)$
• C. $U=\frac{-2G{m}^2}{l}\ln \left(\frac{2a+l}{2a-l}\right)$
• D. $U=\frac{-G{m}^2}{l}\ln \left(\frac{a}{a-l}\right)$

Answer 1

The correct option is A $U=\frac{-G{m}^2}{l}\ln \left(\frac{2a+l}{2a-l}\right)$

Consider a small element of the rod of length $dx$ at a distance $x$ from the from the centre of the rod.


Mass of the element, $dm=\frac{m}{l}dx$

Potential energy due to $dm$ and point mass $m$ will be

$dU=-\frac{Gm\left(dm\right)}{(a-x)}$

Substituting the value of $dm$,

$\Rightarrow dU=-\frac{G{m}^2}{l(a-x)}dx$

Potential energy due to the entire rod will be

$U={\int }_{-l/2}^{l/2}-\frac{G{m}^2}{l(a-x)}dx$

$\Rightarrow U=-\frac{G{m}^2}{l}{\int }_{-l/2}^{l/2}\frac{dx}{a-x}$

$\Rightarrow U=-\frac{G{m}^2}{l}(-\ln (a-x\left)\right){|}_{-l/2}^{l/2}$

$\Rightarrow U=\frac{-G{m}^2}{l}\ln \left(\frac{2a+l}{2a-l}\right)$

Hence, option (a) is the correct answer.

Why this question: To familiarize students with the calculation of potential energy for continuous mass distribution by using the integration.