JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Rotational Mechanics

A small object of uniform density rolls up a curved surface with an initial velocity

v

. It reaches up to a maximum height of

\frac{3 v ^{2}}{4 g}

with respect to the initial position. The object is

Answer: (D)

Solution

For a rolling object, the total kinetic energy is:

K E = \frac{1}{2} m v^{2} + \frac{1}{2} I_{c m} ω^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} I_{c m} (\frac{v}{R})^{2}

At maximum height, all kinetic energy converts to potential energy:

\frac{1}{2} m v^{2} + \frac{1}{2} I_{c m} \frac{v ^{2}}{R ^{2}} = m g h = m g \frac{3 v ^{2}}{4 g}

\frac{1}{2} m v^{2} (1 + \frac{I _{c m}}{m R ^{2}}) = \frac{3 m v ^{2}}{4}

1 + \frac{I _{c m}}{m R ^{2}} = \frac{3}{2}

\frac{I _{c m}}{m R ^{2}} = \frac{1}{2}

I_{c m} = \frac{1}{2} m R^{2}

This is the moment of inertia of a disc about its center.

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Single correct option