Question

A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$ . What is the speed of its centre of mass when the cylinder reaches its bottom –
(A) $\sqrt {2gh}$
(B) $\sqrt {\dfrac{3}{4}gh}$
(C) $\sqrt {\dfrac{4}{3}gh}$
(D) $\sqrt {4gh}$

Answer 1

Hint : To solve this question, we need to use the work energy theorem between the top and the bottom of the inclined plane. As the cylinder is rolling as well as translating so the kinetic energy will be the sum of rotational and translational.

Formula used: The formulae used for solving this question are given by,
$K = \dfrac{1}{2}m{v^2}$ , here $K$ is the kinetic energy of a body of mass $m$ moving with a velocity of $v$ .
$K = \dfrac{1}{2}I{\omega ^2}$ here $K$ is the rotational kinetic energy of a body which is rotating with an angular velocity of $\omega$ , and $I$ is the moment of inertia.
$I = \dfrac{1}{2}m{R^2}$ , here $I$ is the moment of inertia of a solid cylinder of mass $m$ and radius $R$ about its natural axis.

Complete step by step answer
Let the speed of the centre of mass of the cylinder and the angular velocity at the bottom of the inclined plane be $v$ and $\omega$ respectively. As the cylinder moves under the influence of the gravitational force only, which is a conservative force, its total mechanical energy remains conserved. So on applying the work energy theorem, we get
$W = {K_2} - {K_1}$ ……………………….(1)
As the only force acting on the cylinder is the gravitational force, so the work done is equal to the negative of the change in its potential energy, that is,
$W = - \Delta U$
$\Rightarrow W = - \left( { - Mgh} \right) = Mgh$ ……………………….(2)
Now, kinetic energy at the top of the plane is equal to zero, that is,
${K_1} = 0$
The kinetic energy at the bottom of the plane is rotational plus translational. So it is given by
${K_2} = \dfrac{1}{2}I{\omega ^2} + \dfrac{1}{2}M{v^2}$ ……………………….(3)
As the solid cylinder rolls about its natural axis, so its moment of inertia is
$I = \dfrac{1}{2}M{R^2}$ ……………………….(4)
Also, as it is rolling without slipping, so we have
$v = \omega R$
$\Rightarrow \omega = \dfrac{v}{R}$ ……………………….(5)
Substituting (4) and (5) in (3) we get
${K_2} = \dfrac{1}{2}\left( {\dfrac{1}{2}M{R^2}} \right){\left( {\dfrac{v}{R}} \right)^2} + \dfrac{1}{2}M{v^2}$
$\Rightarrow {K_2} = \dfrac{3}{4}M{v^2}$ ……………………….(6)
Substituting (2), (3) and (6) in (1) we have
$Mgh = \dfrac{3}{4}M{v^2} - 0$
$\Rightarrow {v^2} = \dfrac{4}{3}gh$
Finally taking square root both the sides, we get
$v = \sqrt {\dfrac{4}{3}gh}$
Thus, the velocity of the cylinder at the bottom of the inclined plane is equal to $\sqrt {\dfrac{4}{3}gh}$ .
Hence, the correct answer is option C.

Note
In this question, the length of the inclined plane is just the extra information and is not required in the solution. It would have been required if friction would be present between the cylinder and the plane.