First slide

Rolling motion

Question

A body of mass m slides down an incline and reaches the bottom with a velocity v . If 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:

Easy

A

v
B

$(\sqrt{2}) v$
C

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

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

Solution

$\begin{array}{l} From COME, \frac{1}{2} {mv}^{2} = mgh \\ and \frac{1}{2} {mv}^{2} (1 + \frac{K^{2}}{R^{2}}) = mgh \\ \Rightarrow \frac{1}{2} {mv}^{' 2} (1 + 1) = \frac{1}{2} {mv}^{2} \Rightarrow v^{'} = \frac{v}{\sqrt{2}} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App