Question

A thin rod length L is bent to form a circle. Its mass is M. What force will act on the mass m placed at the center of the circle?

• A. $\frac{4{\pi }^{2}GMm}{L^2}$
• B. $\frac{GMm}{4{\pi }^2{L}^2}$
• C. $\frac{2\pi GMm}{L^2}$
• D. zero

Answer 1

Answer: (C)

$\frac{2\pi GMm}{L^2}$

REF.Image.

Express$d{F}_x:$

$D{F}_x=\frac{GmdM}{R^2}=\frac{GM\left(m\right)}{\pi (\frac{L}{\pi }{)}^2}d\theta cos\theta$

12

Integrate $d{F}_x$ over $\theta$ from $-\pi 2$ to $\pi 2:$

$F_x=\frac{\pi GMm}{L^2}{\int }_{-\pi /2}^{\pi /2}cos\theta d\theta =\frac{2\pi GMm}{L^2}$