First slide

Rolling motion

Question

A small object of uniform density rolls up a curved surface with an initial velocity ' $υ$ '. It reaches upto a maximum height of $\frac{3 υ^{2}}{4 g}$ with respect to the initial position. The object is

Moderate

A

hollow sphere
B

disc
C

ring
D

solid sphere

Solution

The kinetic energy of the rolling object is converted into potential energy at height
$h (= \frac{3 v^{2}}{4 g})$
So by the law of conservation of mechanical energy, we have
$\frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2} = M g h \frac{1}{2} M v^{2} + \frac{1}{2} I {(\frac{v}{R})}^{2} = M g (\frac{3 v^{2}}{4 g}) (∵ ω = \frac{v}{R}) \frac{1}{2} I \frac{v^{2}}{R^{2}} = \frac{3}{4} M v^{2} - \frac{1}{2} M v^{2} \frac{1}{2} I \frac{v^{2}}{R^{2}} = \frac{1}{4} M v^{2} o r I = \frac{1}{2} M R^{2}$
Hence, the object is disc.

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