Question

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

The correct option is D zero

When the thin rod of mass $M$ and length $L$ is bent to form a circle, the mass $M$ gets distributed uniformly at the periphery of the circle.

The gravitational field produced by a mass $dm$ will get cancelled due to mass $dm$ located at diametrically opposite point, as shown in the image.

Hence, the gravitational field will be zero at the centre of the circle.i.e.,
$E_{centre}=0$

Thus, net force on the mass $m$ placed at centre will be,

$F_{net}=m{E}_{net}=m\times 0=0$


$\text{Alternative method}:$

The gravitational field due to a uniform circular ring having mass $M$ and radius $R$ at a distance $r$ on its axis is given by$$E=\frac{GMr}{(\sqrt{R^2+r^2}{)}^{3/2}}$$ At the centre of the ring, $r$ becomes zero.

Substituting $r=0$ in above equation, we get $E=0$.

Hence, force on the particle of mass $m$ is also zero.

Why this question? To make students understand the gravitational field is a vector quantity and the direction should be accounted while applying the principle of superposition.