Question

A body of mass $m$ slides down an smooth incline and reaches the bottom with a velocity $v$. Now smooth incline surface is made rough and the same mass was in the form of a ring which rolls down this incline, the velocity of the rind at the bottom would have been:

• A. $\sqrt{2v}$
• B. $v$
• C. $\left(\sqrt{\frac{2}{5}}\right)v$
• D. $v/\sqrt{2}$

Answer 1

The correct option is D $v/\sqrt{2}$

Case 1: Only translational motion will take place.

Work-energy theorem, $mgh=\frac{1}{2}m{v}^2$ .................(1)

Case 2: Rotational and translational motion will take place and $W_{friction}=0$

Work-energy theorem, $mgh=\frac{1}{2}m(v^{\prime }{)}^2+\frac{1}{2}I{w}^2$

$mgh=\frac{1}{2}m(v^{\prime }{)}^2+\frac{1}{2}\times m{R}^2{w}^2$

Also $v^{\prime }=Rw$

$mgh=\frac{1}{2}m(v^{\prime }{)}^2+\frac{1}{2}m(v^{\prime }{)}^2$ ...................(2)

Equating (1) & (2), $\frac{1}{2}m{v}^2=m(v^{\prime }{)}^2$

$⟹v^{\prime }=\frac{v}{\sqrt{2}}$