Question

A solid cylinder of mass $ m $ and radius $ R $ rolls down an inclined place of height $ h $. The angular momentum of the cylinder when it reaches the bottom of the plane will be
(A) $ \dfrac{1}{{2R}}\sqrt {gh} $
(B) $ \dfrac{2}{R}\sqrt {gh} $
(C) $ \dfrac{2}{R}\sqrt {\dfrac{{gh}}{3}} $
(D) $ \dfrac{2}{R}\sqrt {\dfrac{{gh}}{2}} $

Answer 1

Hint: The potential energy of the ball at the top of the inclined plane is converted to kinetic energy at the bottom. For a rolling body, the total kinetic energy is the sum of the linear kinetic energy plus the angular kinetic energy.

Formula used: In this solution we will be using the following formulae;
 $ KE = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2} $ where $ KE $ is the kinetic energy of a rolling body, $ m $ is the mass of the body, $ v $ is the linear speed of the body, $ I $ is the moment of inertia, and $ \omega $ is the angular speed of the body.
 $ I = \dfrac{1}{2}m{R^2} $ where $ I $ is the moment of inertia of a solid cylinder. $ R $ is the radius of the cylinder.
 $ PE = mgh $ where $ PE $ is the potential energy of a body, $ g $ is the acceleration due to gravity, and $ h $ is the height above the ground.
 $ v = \omega R $ .

Complete step by step answer:
To solve the above question, we note that the potential energy at the top of the inclined plane is converted to the kinetic energy at the bottom.
Hence,
 $ PE = KE $
But
 $ PE = mgh $ where $ PE $ is the potential energy of a body, $ g $ is the acceleration due to gravity, and $ h $ is the height above the ground and
 $ KE = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2} $ where $ KE $ is the kinetic energy of a rolling body, $ m $ is the mass of the body, $ v $ is the linear speed of the body, $ I $ is the moment of inertia, and $ \omega $ is the angular speed of the body.
Hence, we have
 $ mgh = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2} $
Now
 $ I = \dfrac{1}{2}m{R^2} $ where $ I $ is the moment of inertia of a solid cylinder. $ R $ is the radius of the cylinder. And
 $ v = \omega R $
Then,
 $ mgh = \dfrac{1}{2}m{\omega ^2}{R^2} + \dfrac{1}{2}\left( {\dfrac{{m{R^2}}}{2}} \right){\omega ^2} $
 $ \Rightarrow gh = {\omega ^2}{R^2}\left[ {\dfrac{1}{2} + \dfrac{1}{4}} \right] = \dfrac{3}{4}{\omega ^2}{R^2} $
By making $ \omega $ subject of the formula, we have
 $ {\omega ^2} = \dfrac{4}{3}\dfrac{{gh}}{{{R^2}}} $
 $ \Rightarrow \omega = \dfrac{2}{R}\sqrt {\dfrac{{gh}}{3}} $
So the correct answer will be option C.

Note:
For understanding, we must note that the above procedure assumes that the solid cylinder has no energy loss during descent. In actuality, there would be some energy loss due to friction and possibly air resistance. Hence the PE will not be perfectly equal to the kinetic energy at the bottom.