Rankers Physics
Topic: Gravitation
Subtopic: Gravitational Potential

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
\[ \frac{-GM}{L}\]
\[ \frac{-GM}{2\pi L} \]
\[ \frac{- \pi GM}{2L}\]
\[ \frac{- \pi GM}{L} \]

Solution:

The gravitational potential at the center due to a semicircular rod is given by:

\[
V = - \frac{GM}{R}
\]

Where \( R \) is the radius of the semicircle, which is related to the length of the rod:

\[
L = \pi R \quad \Rightarrow \quad R = \frac{L}{\pi}
\]

Substituting \( R \) into the potential formula:

\[
V = - \frac{GM}{\frac{L}{\pi}} = - \frac{\pi GM}{L}
\]

Thus, the gravitational potential at the center is:

\[
V = - \frac{\pi GM}{L}
\]

Leave a Reply

Your email address will not be published. Required fields are marked *