First slide

Rolling motion

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 ring at the bottom would have been:

Moderate

A

$\sqrt{2} v$
B

$v$
C

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

$v / \sqrt{2}$

Solution

For a sliding body of mass m,
$v_{body} = \sqrt{2 g h} = v . . . . (i)$
For a rolling ring of same mass m,
$\begin{matrix} v_{ring} = \sqrt{\frac{2 g h}{β}} \\ = \sqrt{\frac{2 g h}{1 + \frac{I}{m R^{2}}}} = \sqrt{\frac{2 g h}{1 + \frac{m R^{2}}{m R^{2}}}} = \sqrt{\frac{2 g h}{2}} = \sqrt{g h} = \frac{v}{\sqrt{2}} \end{matrix}$

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