Question

A semicircular wire has a length L and mass M. A particle of mass m is placed at the centre of the circle. Find the gravitational attraction on the particle due to the wire.

Answer 1

Consider a small mass element of length dl subtending dθ angle at the centre.

In the semicircle, we can consider a small element dθ.
Then length of the element, dl = R dθ
Mass of the element, dm \[= \left( \frac{M}{L} \right)R d\theta\]

Force on the mass element is given by

\[dF = \frac{Gm}{R^2}dm = \frac{GMRm}{L R^2}{d}{\theta}\]

The symmetric components along AB cancel each other. 
Now, net gravitational force on the particle at O is given by

\[F = \int2dF\sin \theta\]

\[ = \int\frac{2GMm}{LR}\sin \theta d\theta\]

\[ \therefore F = \int_0^\pi/2 \frac{- 2GMm}{LR}\sin \theta d\theta\]

\[ = \frac{2GMm}{LR} \left[ - \cos \theta \right]_0^\pi/2 \]

\[ = - 2\frac{GMm}{LR}\left( - 1 \right)\]

\[ = \frac{2GMm}{LR} = \frac{2GMm}{LL/\pi}\]

\[ = \frac{2\pi G Mm}{L^2}\]