Question

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

• A.
• B.
• C.
• D. zero

Answer 1

The correct option is A

Considering an element of rod of lenght dl as shown in fugure and treating it as a point of mass (M/L)

dl situated at a distance R from P, the gravitational force due to this element on the particle will be

dF = $\frac{Gm\frac{M}{L}Rd\theta }{R^2}$ along OP [as dl = rd $\theta$]

So the component of this force along x and y axes will be

d $F_x$ = dF cos $\theta$ = $\frac{GmMsin\theta d\theta }{LR}$

d $F_y$ = dF sin $\theta$ = $\frac{GmMsin\theta d\theta }{LR}$

So that $F_x$ = $\frac{GmM}{LR}$ $\underset{0}{\overset{\pi }{\int }}$ cos $\theta$ d $\theta$ =$\frac{GmM}{LR}$ $[sin\theta$ ${]}_0^{\pi }$ = 0

and $F_y$ = $\frac{GmM}{LR}$ $\underset{0}{\overset{\pi }{\int }}$ cos $\theta$ d $\theta$ =$\frac{GmM}{LR}$ $[-cos\theta$ ${]}_0^{\pi }$ = 0

= $\frac{2\pi GmM}{L^2}$

So, F = $\sqrt{{F_x}^2+{F_y}^2}$ = $F_y$ = $\frac{2\pi GmM}{L^2}$ [as $F_x$] is zero