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 Here given the cylinder is rolling without slipping down on an inclined plane. We have to find the speed of its centre of mass when the cylinder reaches the bottom. This problem can be solved by using the conservation of energy according to which when the body reaches the bottom its total potential energy is converted into kinetic energy we will obtain the velocity of the center of mass of the cylinder.
Formula used:
Conservation of total energy
$P.E = K.{E_{rot.}} + K.{E_{trans.}}$
where $K.{E_{rot.}}$= kinetic energy due to the rotation
$K.{E_{trans.}}$= kinetic energy due to translation

Complete Step by step solution
Here our cylinder is solid and it consists of potential energy when it is at rest. As our body starts rolling its potential energy starts converting into kinetic energy. The motion of the cylinder will involve two types of kinetic energy one due to motion in a straight line which is known as translational kinetic energy and another due to circular motion which is known as rotational kinetic energy.
Hence according to total conservation of energy, we can deduce that
$P.E = K.{E_{rot.}} + K.{E_{trans.}}$
where potential energy $P.E = mgh$
Kinetic energy due to translation $K.{E_{trans.}} = \dfrac{1}{2}M{v^2}$,
Kinetic energy due to the rotation $K.{E_{rot.}} = \dfrac{1}{2}I{\omega ^2}$
Now the equation can be given as
$Mgh = \dfrac{1}{2}I{\omega ^2} + \dfrac{1}{2}M{v^2}$ --------------- Equation $(1)$
where $h$= height of an inclined plane
$v$= translational velocity
$\omega $= rotational velocity
$I$= moment of inertia
$M$=mass of the cylinder
Now as our cylinder is rolling without slipping hence our rotation is pure as a result $v = \omega R$, where $R$is the radius of the cylinder.
Also moment of inertia $I = \dfrac{{M{R^2}}}{2}$
Substituting the values of $I$and $v$ into the equation $(1)$ we get
$Mgh = \dfrac{1}{2}{\dfrac{{MR}}{2}^2} \times \dfrac{{{v^2}}}{{{R^2}}} + \dfrac{1}{2}M{v^2}$
$ \Rightarrow Mgh = \dfrac{M}{4} \times {v^2} + \dfrac{1}{2}M{v^2}$
Now eliminating the mass also from both side
$gh = \dfrac{1}{4}{v^2} + \dfrac{1}{2}{v^2}$
$ \Rightarrow gh = \dfrac{3}{4}{v^2}$
Rearranging the terms by transposition
$ \Rightarrow {v^2} = \dfrac{4}{3}gh$
$\therefore v = \sqrt {\dfrac{4}{3}gh} $
Hence the velocity of the Centre of the mass of cylinder rolling without slipping is $v = \sqrt {\dfrac{4}{3}gh} $.

Hence option (C) is the correct answer.

Note One should be aware of the difference between rotational kinetic energy and translational kinetic energy is that translational occurs due to straight-line motion while rotational kinetic energy occurs due to circular rotation.