Question

A ring has mass $M$, radius $R$. A point mass $m$ is placed at a distance $x$ on the axial line as shown in the figure. Find $x$, so that force experienced is maximum

• A. $\frac{R}{3\epsilon }$
• B. $\frac{R}{2}$
• C. $\frac{R}{\sqrt{2}}$
• D. $\frac{R}{\sqrt{3}}$

Answer 1

Answer: (C)

$\frac{R}{\sqrt{2}}$

All point on the circumference of the ring are at distance $\sqrt{R^2+x^2}$from the center O of the ring.
Force on the mass m at P : $F=\frac{GMm}{R^2+x^2}\times \left(2\cos \theta \right)$ where $\cos \theta =\frac{x}{\sqrt{R^2+x^2}}$
Due to symmetry, vertical components of the forces from two symmetrical elements cancel out, the horizontal components add up, hence we get a factor of 2.
$\therefore F=\frac{GMm}{R^2+x^2}\times 2\frac{x}{\sqrt{R^2+x^2}}$
When F is maximum, $\frac{dF}{dx}=0$
$\Rightarrow (R^2+x^2{)}^{-\frac{3}{2}}+x(-\frac{3}{2}(R^2+x^2{)}^{-\frac{3}{2}-1}\times 2x=0$
$\Rightarrow (R^2+x^2)=3x^2$
$\Rightarrow x=\frac{R}{\sqrt{2}}$