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:
(A) $ v $
(B) $ \sqrt 2 v $
(C) $ \dfrac{v}{{\sqrt 2 }} $
(D) $ \left( {\sqrt {\dfrac{2}{5}} } \right)v $

Answer 1

Hint: Only translational motion will take place in the first case. Thus we apply the work-energy theorem here. Rotational and translational motion will take place in the second case.

Formula used: The formulae used in the solution are given here.
Moment of inertia of the mass $ I = m{r^2} $ where mass is $ m $ and velocity is $ v $ .

Complete Step by Step Solution
It has been given that, a body of mass $ m $ slides down an incline and reaches the bottom with a velocity $ v $ . Let us assume that the $ r $ is the radius of the ring and $ m $ is its mass then,
Moment of inertia of the mass $ I = m{r^2} $ .....(i)
The total energy at the bottom of the inclined plane is given by the sum of transnational kinetic energy and rotational kinetic energy.
Total energy at the top of the plane = Potential energy.
Potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.
Now, as the mass/ring slides/rolls down the inclined plane then potential energy gets converted into kinetic energy.
In the first case when it is not a ring. If $ h $ is the height of the plane then,
 $ mgh = \dfrac{1}{2}m{v^2} $ where $ g $ is the acceleration due to gravity and $ v $ is the velocity of the mass.
 $ h = \dfrac{{{v^2}}}{{2gh}} $ ……….(ii)
Now, when the mass $ m $ is in the form of a ring and rolls down the incline.
According to the second case if $ v' $ is the velocity at the bottom of the plane then,
 $ mgh = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2} $ where $ \omega $ is the angular acceleration.
As we already know, $ v' = r \times \omega $ . Thus, we may write,
 $ mgh = \dfrac{1}{2}mv{'^2} + \dfrac{1}{2}m{r^2} \times {\left( {\dfrac{{v'}}{r}} \right)^2} $
Simplifying this, we get, $ mgh = mv{'^2} $
 $ \Rightarrow h = \dfrac{{v{'^2}}}{g} $ ……..(iii)
Comparing (ii) and (iii),
 $ \dfrac{{{v^2}}}{{2g}} = \dfrac{{v{'^2}}}{g} $
 $ \Rightarrow v{'^2} = \dfrac{{{v^2}}}{2} $
Thus, the velocity at the bottom of the plane is $ v' = \dfrac{v}{{\sqrt 2 }} $ .
Hence, the correct answer is Option C.

Note
The Work-Energy theorem is very useful in analysing situations where a rigid body moves under several forces. As we know that a rigid body cannot store potential energy in its lattice due to rigid structure, it can only possess kinetic energy. Thus the work done by any force acting on a rigid body is equal to the change in its kinetic energy. This is the basis of the work energy equation for rigid bodies.