A point particle is held on the axis of a ring of mass $m$ and radius $r$ at a distance $r$ from its centre…
A point particle is held on the axis of a ring of mass $m$ and radius $r$ at a distance $r$ from its centre $C$. When released, it reaches $C$ under the gravitational attraction of the ring. Its speed at $C$ will be
$\sqrt{\frac{2 G m}{r}(\sqrt{2}-1}$
$\sqrt{\frac{G m}{r}}$
$\sqrt{\frac{2 G m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$
$\sqrt{\frac{2 G m}{r}}$
Solution
Let ' $M$ ' be the mass of the particle
Now, $\mathrm{E}_{\text {initial }}=\mathrm{E}_{\text {final }}$
i.e. $\frac{\mathrm{GMm}}{\sqrt{2} r}+0=\frac{\mathrm{GM} m}{r}+\frac{1}{2} M \mathrm{~V}^2$
$
\begin{aligned}
& \text { or, } \frac{1}{2} M V^2=\frac{G M m}{r}\left[1-\frac{1}{\sqrt{2}}\right] \\
& \Rightarrow \frac{1}{2} V^2=\frac{G m}{r}\left[1-\frac{1}{\sqrt{2}}\right] \\
& \text { or, } V=\sqrt{\frac{2 G m}{r}\left(1-\frac{1}{\sqrt{2}}\right)}
\end{aligned}
$