First slide

Rolling motion

Question

A body of mass m slides down an incline (without friction) and reaches the bottom with a velocity v. If the same mass were in the form of a ring which rolls down(without slipping) this incline, the velocity of the ring at bottom would have been

Moderate

A

v
B

$\sqrt{2} v$
C

$\frac{v}{\sqrt{2}}$
D

$\sqrt{(\frac{2}{5}) v}$

Solution

v = $\sqrt{2 gh}$ ------(i)
For second case: $\frac{1}{2} {mv}^{' 2} + \frac{1}{2} {Iω}^{2} = mgh$
$(ω = \frac{v^{'}}{r}, I = {mr}^{2})$
$\Rightarrow v^{'} = \sqrt{gh} - - - - - - - (2)$
from (1) and (2) $v^{'} = \frac{v}{\sqrt{2}}$

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