Two bodies, a ring and a solid cylinder of same material are rolling down without slipping on an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is $\frac{\sqrt{\mathrm{x}}}{2}$. Then, the value of x is …… .

Let us consider the case for ring first.

Rotational kinetic energy of ring, $\mathrm{K}=\frac{1}{2}{\mathrm{I\omega }}^2$

$\because$ Moment of inertia for ring about the point of contact,

$\begin{array}{l}{\mathrm{l}}_{\text{ring }}=2{\mathrm{mR}}^2\\ \mathrm{K}=\frac{1}{2}\left(2{\mathrm{mR}}^2\right){\mathrm{\omega }}^2\\ \mathrm{K}=\frac{1}{2}\left(2{\mathrm{mR}}^2\right)\frac{{\mathrm{v}}_{\mathrm{R}}^2}{{\mathrm{R}}^2} \left\{\because \mathrm{\omega }=\frac{{\mathrm{v}}_{\mathrm{R}}}{\mathrm{R}}\right\}\end{array}$

As, this is a case of pure rolling and the ring is rolling down, then by conservation of energy its potential energy will be equal to its kinetic energy at the bottom

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{PE}=\mathrm{mgh}=\frac{1}{2}\left(2{\mathrm{mR}}^2\right)\frac{{\mathrm{v}}_{\mathrm{R}}^2}{{\mathrm{R}}^2}\\ \Rightarrow {\mathrm{v}}_{\mathrm{R}}=\sqrt{\mathrm{gh}} ...\left(\mathrm{i}\right)\end{array}$

Now, consider the case of solid cylinder,

$\because$ Moment of inertia in this case as well as about the point of contact is given as

$\begin{array}{l}{\mathrm{I}}_{\text{cylinder }}=\frac{3}{2}{\mathrm{mR}}^2\\ \mathrm{mgh}=\frac{1}{2}{\mathrm{I}}_{\text{cylinder }}{\mathrm{\omega }}^2\\ \Rightarrow \mathrm{mgh}=\frac{1}{2}\left(\frac{3}{2}{\mathrm{mR}}^2\right)\cdot \frac{{\mathrm{v}}_{\mathrm{c}}^2}{{\mathrm{R}}^2} \left\{\because \mathrm{\omega }=\frac{\mathrm{v}}{\mathrm{R}}\right\}\\ \Rightarrow {\mathrm{v}}_{\mathrm{C}}=\sqrt{\frac{4\mathrm{gh}}{3}} ...\left(\mathrm{ii}\right)\end{array}$

On dividing Eq. (i) by Eq. (ii), we get

$\frac{{\mathrm{v}}_{\mathrm{R}}}{{\mathrm{v}}_{\mathrm{C}}}=\frac{\sqrt{3}}{2}$                    ...(iii)

As per question, the ratio of velocity of centre of mass at the bottom of inclined plane of the ring to that of cylinder is $\frac{\sqrt{\mathrm{x}}}{2}$. So, on comparing it with Eq. (iii), we can write x = 3.