A thin rod of mass M and length L is bent into semicircle as shown in figure. What is gravitational force on a particle with mass m at the centre of curvature?

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$\frac{G M m}{L^{2}}$

$\frac{π G M m}{L^{2}}$

$\frac{2 π G M m}{L^{2}}$

$\frac{2 G M m}{L^{2}}$

answer is B.

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Detailed Solution

Consider an element of rod of length $d l$ shown in figure and treat it as a small particle of mass (M/L) $d l$ situated at a distance R from P. Then gravitational force due to the element on the particle will be
$\begin{array}{l} d F = \frac{G m (M / L) (R d θ)}{R^{2}} a l o n g O P \\ [a s d l = R θ] \end{array}$

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So the components of this force along x and y axes will be
$\begin{array}{l} d F_{x} = d F \cos θ = \frac{G m M \cos θ d θ}{L R} \\ d F_{x} = d \sin θ = \frac{G m M \sin θ d θ}{L R} \end{array}$
So that
$\begin{array}{l} F_{x} = \frac{G m M}{L R} \int_{0}^{π} \cos θ d θ = \frac{G m M}{L R} = 0 \\ F_{y} = \frac{G m M}{L R} \int_{0}^{π} \sin θ d θ = \frac{G m M}{L R} \\ = \frac{2 π G m M}{L^{2}} [a s R = \frac{L}{π}] \end{array}$
$F = \sqrt{f_{x}^{2} + F_{y}^{2}} = F_{y} = \frac{2 π G m M}{L^{2}} (a s F_{x} = z e r o)$