Question

A thin rod of length 'L' is bent in the form of a circle. Its mass is 'M'. What force will act on mass 'm' placed at the centre of this circle? (G = constant of gravitation)

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

Hint:

Logic Tip: The gravitational field at the exact center of ANY uniform spherically or circularly symmetric mass distribution (like a ring, spherical shell, or solid sphere) is always zero due to perfect spatial cancellation. No calculation is needed if you recognize the symmetry!

Answer 1

The Correct Option is A

Concept:
The gravitational force is a vector quantity. The principle of superposition states that the net gravitational force on a particle is the vector sum of the forces exerted by all other individual mass elements. For a highly symmetric continuous mass distribution like a uniform circular ring, we can use symmetry to determine the net force at the geometric center.
Step 1: Consider a differential mass element on the ring.
Imagine a very small segment of the circular ring with mass $dM_1$. The gravitational force $d\vec{F}_1$ exerted by this small mass on the mass $m$ at the center is directed towards $dM_1$. $$|d\vec{F}_1| = \frac{G \cdot m \cdot dM_1}{r^2}$$ where $r$ is the radius of the circle ($r = \frac{L}{2\pi}$).
Step 2: Identify the diametrically opposite mass element.
For every mass element $dM_1$ on the ring, there exists an identical mass element $dM_2$ (where $dM_1 = dM_2$) located diametrically opposite to it on the circle. The force $d\vec{F}_2$ exerted by $dM_2$ on the central mass $m$ will have the same magnitude as $d\vec{F}_1$ but will point in the exact opposite direction. $$|d\vec{F}_2| = \frac{G \cdot m \cdot dM_2}{r^2}$$ Because $dM_1 = dM_2$, the magnitudes of the forces are equal: $|d\vec{F}_1| = |d\vec{F}_2|$.
Step 3: Apply the superposition principle.
Since $d\vec{F}_1$ and $d\vec{F}_2$ are equal in magnitude and opposite in direction, they cancel each other out perfectly: $$d\vec{F}_1 + d\vec{F}_2 = 0$$ By integrating around the entire ring, every infinitesimal mass element pairs up with its diametrically opposite partner, resulting in a complete cancellation of all force vectors. Thus, the net gravitational force acting on the mass $m$ at the center is exactly zero.