Question

A particle starts from rest at a distance R from the centre and along the axis of a fixed ring of radius R & mass M. Its velocity at the centre of the ring is:

• A. $\sqrt{\frac{\sqrt{2}GM}{R}}$
• B. $\sqrt{\frac{2GM}{R}}$
• C. $\sqrt{\left(2\sqrt{2}\right)\frac{GM}{R}}$
• D. $\sqrt{(2-\sqrt{2})\frac{GM}{R}}$

Answer 1

The correct option is C $\sqrt{\left(2\sqrt{2}\right)\frac{GM}{R}}$

Applying conservation of energy

$v=\frac{-G\times M}{\sqrt{x^2+R^2}}$

Potential energy$=v\times m=\frac{-GmM}{\sqrt{x^2+R^2}}$

Initial kinetic energy$r_i=0$

Potential energy at center$=\frac{-GMm}{R}\{x=0\}$

$\frac{-GMm}{\sqrt{x^2+R^2}}+0=\frac{GMm}{R}+\frac{1}{2}m{v}^2$

$\frac{1}{2}m{v}^2=\frac{GMm}{R}-\frac{GMm}{\sqrt{x^2+R^2}}=GMm\{\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}}\}$

$v^2=\frac{2GMm}{m}\{\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}}\}$

$v^2=\sqrt{(2-\frac{2}{\sqrt{2}})\left(\frac{GM}{R}\right)}=\sqrt{\left(2\sqrt{2}\right)\frac{GM}{R}}$