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(A) Consider the circular ring formed by the rod of length $L$ and mass $M$. Let the radius of this circle be $r$. The circumference is $2 \pi r = L$,

Vedclass · Accepted Answer

zero

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(A) Consider the circular ring formed by the rod of length $L$ and mass $M$. Let the radius of this circle be $r$. The circumference is $2 \pi r = L$, so $r = \frac{L}{2 \pi}$.Consider two diametrically opposite small mass segments $dM_1$ and $dM_2$ on the ring.The gravitational force exerted by segment $dM_1$ on mass $m$ at the centre is $F_1 = \frac{G m dM_1}{r^2}$ directed towards $dM_1$.The gravitational force exerted by the diametrically opposite segment $dM_2$ on mass $m$ at the centre is $F_2 = \frac{G m dM_2}{r^2}$ directed towards $dM_2$.Since the segments are diametrically opposite, the forces $F_1$ and $F_2$ are equal in magnitude and opposite in direction $(F_1 = -F_2)$.Therefore, the net force due to these two segments is $F_1 + F_2 = 0$.By symmetry, every mass segment on the ring has a corresponding diametrically opposite segment that exerts an equal and opposite gravitational force on the mass $m$ at the centre.Summing these forces over the entire ring, the net gravitational force acting on the mass $m$ at the centre is zero.

$A$ thin rod of length $L$ is bent in the form of a circle. Its mass is $M$. What force will act on a mass $m$ placed at the centre of this circle? $(G = \text{universal gravitational constant})$

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