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प्रश्न
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.
उत्तर
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
\[F = \int2dF\sin \theta\]
\[ = \int\frac{2GMm}{LR}\sin \theta d\theta\]
\[ \therefore F = \int_0^\pi/2 \frac{- 2GMm}{LR}\sin \theta d\theta\]
\[ = \frac{2GMm}{LR} \left[ - \cos \theta \right]_0^\pi/2 \]
\[ = - 2\frac{GMm}{LR}\left( - 1 \right)\]
\[ = \frac{2GMm}{LR} = \frac{2GMm}{LL/\pi}\]
\[ = \frac{2\pi G Mm}{L^2}\]
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