Question

A small particle of mass $m$ slides down on inclined plane and reaches the bottom with linear velocity $V$. If the same mass is in the form of ring and rolls without slipping down the same inclined plane, its velocity will be

• A. $V$
• B. $\sqrt{2}V$
• C. $\frac{V}{\sqrt{2}}$
• D. 2V

Answer 1

The correct option is C $\frac{V}{\sqrt{2}}$

So we can apply conservation of mechanical energy (kinetic energy = potential energy).
In first case,
$\frac{m{V}^2}{2}=mgh\Rightarrow V=\sqrt{2gh}$.
In second case, let velocity of ring be $V^{\prime }$ then
$\frac{m{\left(V\text{'}\right)}^2}{2}+\frac{I}{2}\times {\left(\frac{V\text{'}}{r}\right)}^2=mgh$
where, $I=m{r}^2$
$\frac{m{\left(V\text{'}\right)}^2}{2}+\frac{m{\left(V\text{'}\right)}^2}{2}=mgh$
$\Rightarrow {\left(V\text{'}\right)}^2=gh\Rightarrow V\text{'}=\sqrt{gh}$
Therefore,
$V\text{'}=\frac{V}{\sqrt{2}}$