Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?

Hint – In this problem, you have to first apply the concept of conservation of energy. First calculate the potential energy of the three bodies and the kinetic energy of the three bodies and then compare them. Use this comparison to find out the velocity of the three bodies.

Step By Step Answer:
Potential energy of the three bodies $(PE) = mgh$

Here,

$m = $ Mass of bodies,
$g = 9.8m{s^{ - 1}}$ (Acceleration due to gravity)
$h = $ Height of the bodies from the ground

When the bodies reach the ground, all this potential energy will be converted into kinetic energy. The kinetic energy will be of two types –

Kinetic energy due to the translation motion of the bodies is

${(KE)_{Tr}} = \dfrac{1}{2}m{v^2}$(Equation 1)

Kinetic energy due to the rotational motion of the body is

${(KE)_{Ro}} = \dfrac{1}{2}I{w^2}$

We know that,

$I = m{k^2}$ ($k = $Radius of gyration)
And $\omega = \dfrac{v}{R}$ ($R = $radius of the three bodies)

Substitute the values of $I$ and$\omega $ in equation 2

$ \Rightarrow {(KE)_{Ro}} = \dfrac{1}{2}m{k^2}{\left( {\dfrac{v}{R}} \right)^2}$
$ \Rightarrow {(KE)_{Ro}} = \dfrac{1}{2}\dfrac{{m{k^2}{v^2}}}{{{R^2}}}$(Equation 3)
To get the total kinetic energy we add equation $2$ and $3$

${(KE)_{To}} = {(KE)_{Tr}} + {(KE)_{Ro}}$
$ \Rightarrow {(KE)_{To}} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}\dfrac{{m{k^2}{v^2}}}{{{R^2}}}$
$ \Rightarrow {(KE)_{To}} = \dfrac{1}{2}m{v^2}\left( {1 + \dfrac{{{k^2}}}{{{R^2}}}} \right)$

We also know by law of conservation of energy

$(PE) = {(KE)_{To}}$
$ \Rightarrow mgh = \dfrac{1}{2}m{v^2}\left( {1 + \dfrac{{{k^2}}}{{{R^2}}}} \right)$
$ \Rightarrow {v^2} = \dfrac{{2gh}}{{\left( {1 + \dfrac{{{k^2}}}{{{R^2}}}} \right)}}$
$ \Rightarrow v = \sqrt {\dfrac{{2gh}}{{\left( {1 + \dfrac{{{k^2}}}{{{R^2}}}} \right)}}} $
For a ring, ${k^2} = {R^2}$
$ \Rightarrow v = \sqrt {gh} $

For a solid cylinder, ${k^2} = \dfrac{{{R^2}}}{2}$
$ \Rightarrow v = \sqrt {\dfrac{{2gh}}{{1 + \dfrac{1}{2}}}} = \sqrt {\dfrac{{4gh}}{3}} = \sqrt {1.33gh} $

For a solid sphere,${k^2} = \dfrac{{2{R^2}}}{5}$
$ \Rightarrow v = \sqrt {\dfrac{{2gh}}{{1 + \dfrac{2}{5}}}} = \sqrt {\dfrac{{10gh}}{7}} = 1.43$

So in between a ring, a cylinder and a solid sphere of same radii, the ring has the least velocity at the point it reaches the ground and the solid sphere has the most velocity at the point it reaches the ground.

Note – For finding out the rotational kinetic energy of the bodies, we used the concept of moment of inertia. Moment of inertia is a quantity that expresses the tendency of a body to resist angular acceleration. It is the product of mass of every particle in the body with the square of its distance which is measured from the axis of rotation.