Question

A ring has radius $R$ and mass $m_1=m\text{kg}$ which is distributed uniformly over its circumference. A highly dense particle of mass, $m_2=2m\text{kg}$ is placed at rest on the axis of the ring at a distance $x_0=\sqrt{3}R$ from the centre. Neglecting all other forces, except mutual gravitational interaction between the ring and particle. Calculate the speed of the ring at the instant when the particle is at the centre of the ring.

• A. $\sqrt{\frac{2G}{3R}}$
• B. $\sqrt{\frac{G}{3R}}$
• C. $\sqrt{\frac{G}{R}}$
• D. $\sqrt{\frac{4G}{3R}}$

Answer 1

The correct option is D $\sqrt{\frac{4G}{3R}}$

Since, there are no external forces, so on applying conservation of linear momentum,
$$m_1{v}_1=m_2{v}_2...\left(1\right)$$ Where, $v_1$ is the velocity of ring and $v_2$ is the velocity of particle.

Also, there is no work done by external forces, so on applying conservation of mechanical energy,

$P.E_i+K.E_i=P.E_f+K.E_f$

Substituting the values,

$-\left(\frac{G{m}_1}{\sqrt{R^2+x_0^2}}\right)m_2+0=-\frac{G{m}_1{m}_2}{R}+\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

$\Rightarrow G{m}_1{m}_2[\frac{1}{R}-\frac{1}{\sqrt{R^2+x_0^2}}]=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

$\Rightarrow G{m}_1[\frac{1}{R}-\frac{1}{\sqrt{R^2+x_0^2}}]=\frac{1}{2}\left(\frac{m_1}{m_2}\right)v_1^2+\frac{1}{2}{v}_2^2$

Using equation $\left(1\right)$,

$\Rightarrow G{m}_1[\frac{1}{R}-\frac{1}{\sqrt{R^2+x_0^2}}]=\frac{1}{2}\times \left(\frac{m_1}{m_2}\right)v_1^2+\frac{1}{2}{\left(\frac{m_1{v}_1}{m_2}\right)}^2$


$\Rightarrow G{m}_1[\frac{1}{R}-\frac{1}{\sqrt{R^2+x_0^2}}]=\frac{1}{2}\times \frac{m_1}{m_2}[1+\frac{m_1}{m_2}]\times v_1^2$

$\Rightarrow v_1=\sqrt{2G{m}_2\frac{(\frac{1}{R}-\frac{1}{\sqrt{R^2+x_0^2}})}{(1+\frac{m_1}{m_2})}}$

Substituting $m_1=m$, $m_2=2m$ and $x_0=\sqrt{3}R$

$\Rightarrow v_1=\sqrt{2G\left(2m\right)\frac{(\frac{1}{R}-\frac{1}{\sqrt{R^2+(\sqrt{3}R{)}^2}})}{(1+\frac{m}{2m})}}$

$\Rightarrow v_1=\sqrt{\frac{8G}{3}(\frac{1}{R}-\frac{1}{\sqrt{R^2+3R^2}})}$

$\Rightarrow v_1=\sqrt{\frac{4G}{3R}}$

Hence, option (d) is the correct answer.

Key Concept: The gravitational potential due to a ring on its axis is given by $$V=-\frac{GM}{\sqrt{R^2+r^2}}$$Why this question: To make students apply multiple concepts in a problem. This problem the various conceptes like gravitational potential due to a ring, conservation of linear momentum and conservation of mechanical energy.