Question

A thin rod of length $L$ is bent to form a semicircle. The mass of the rod is $M$. What will be the gravitational potential at the centre of the semicircle?

• A. $-\frac{GM}{L}$
• B. $-\frac{GM}{2\pi L}$
• C. $-\frac{\pi GM}{2L}$
• D. $-\frac{\pi GM}{L}$

Answer 1

The correct option is D $-\frac{\pi GM}{L}$

Given that,
Length of the rod $=L$
Mass of the rod $=M$
Let the radius of the rod be $R$.
Then, $L=\pi R$
$\Rightarrow R=\frac{L}{\pi }$


All the mass is at the periphery of the semicircle, and equidistant from the centre.
Let us take a small mass $dm$ and calculate the gravitational potential at the centre of the ring.
$dV=-\frac{Gdm}{R}$
$dV=-\frac{Gdm}{L/\pi }\dots \left(i\right)$
Now to get the potential due to entire mass $M$, integrate the equation $\left(i\right)$
$\Rightarrow V=-\int \frac{Gdm}{L/\pi }$
$\Rightarrow V=-\frac{G\pi }{L}\int dm$
The integration of $dm$ will be $M$.
$\Rightarrow V=-\frac{G\pi M}{L}$

Alternate:
When the rod is bent into a semicircle, the entire mass is at equidistant from its centre.
The gravitational potential is a scalar quantity and it depends on distance only.

The gravitational potential at the centre of semicircle is obtained by using the formula
$V=-\frac{GM}{R}$
$\Rightarrow V=-\frac{GM}{L/\pi }$
$\Rightarrow V=-\frac{GM\pi }{L}$

Why this question? To familiarize students with the calculation of potential energy for continuous mass distribution.