Question

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

• A. $-\frac{GM}{\pi L}$
• 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,
Length of the rod $=L$
Mass of the rod $=M$


Let, $R$ is the radius of the semicircle, so $L=\pi R$
$\Rightarrow R=\frac{L}{\pi }$

Now, all the mass of the semicircle is concentrated at its perimeter equidistant from the centre of the circle.

So, the potential at the centre due to the mass $M$ is $$V=-\frac{GM}{R}$$$\Rightarrow V=-\frac{GM}{\left(L/\pi \right)}$

$\Rightarrow V=-\frac{G\pi M}{L}$

Hence, option (d) is the correct answer.

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