A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is $\sqrt{\frac{x}{5}}$ where $\mathrm{x}=$ _____

Question

Aicharya · Accepted Answer

Applying Mechanical Energy conservation :
$\mathrm{k}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{k}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}$
$\begin{aligned}
& \Rightarrow 0+\mathrm{Mgh}=\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right)+0 \\ & \Rightarrow \mathrm{~V}=\sqrt{\frac{2 \mathrm{gh}}{1+\frac{\mathrm{k}^2}{\mathrm{R}^2}}}
\end{aligned}$
So Ratio of velocities
$\frac{\mathrm{V}_{\text {Ring }}}{\mathrm{V}_{\text {solids sphere }}}=\sqrt{\frac{1+\frac{2}{5}}{1+1}}=\sqrt{\frac{7}{10}}$
$\mathrm{x}=3.5$ Rounding off $\mathrm{x}=4$

A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without…

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