Question

A body of mass $m$ slides down an incline and reaches the bottom with a velocity $V$. If the same mass were in the form of a ring which rolls down the same incline, the velocity of its centre of mass at bottom of the plane is:

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

Answer 1

Answer: (C)

$\frac{V}{\sqrt{2}}$

Let $h$ the height of the point on the inclined plane at which the body starts to slide to reach the bottom.

For the body of mass $m$:

The initial kinetic energy of the body is zero.

Using work-energy theorem, $W=\frac{1}{2}m{V}^2$

$\therefore$ $mgh=\frac{1}{2}m{V}^2$

We get $gh=\frac{V^2}{2}$ ....(1)

For a ring of mass $m$ rolling down the plane:

The initial kinetic energy of the ring is zero.

Using work-energy theorem, $W=K.E_f$

$\therefore$ $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2$

Or $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\left(m{r}^2\right)w^2$

Or $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2$ (as $v=rw$)

Or $mgh=m{v}^2$

Or $v^2=gh$

Or $v^2=\frac{V^2}{2}$ (using 1)

$⟹$ $v=\frac{V}{\sqrt{2}}$