Question

A disc is rolling without slipping on a straight surface. The ratio of its translational kinetic energy to its total kinetic energy is
A. $\dfrac{2}{3}$
B. $\dfrac{1}{3}$
C. $\dfrac{2}{5}$
D. $\dfrac{3}{5}$

Answer 1

Hint: Here the disc has both translational kinetic energy due to translational motion and rotational kinetic energy due to rotational motion. So, the total kinetic energy will be the sum of these two kinetic energies. If we divide the kinetic energy due to translation with the total kinetic energy, we will get the answer.

Complete answer:
It is given that a disc rolls without slipping on a straight surface.
We know that whenever there is rotational motion involved there will be rotational kinetic energy.
And due to translational motion, it also has translational kinetic energy.
The translational kinetic energy is given by the equation
$ \Rightarrow $${K_T} = \dfrac{1}{2}m{v^2}$ ……………………..(1)
Where, $m$ is the mass and $v$ is the linear velocity,
In the rotational motion, rotational kinetic energy is given as
$ \Rightarrow $${K_R} = \dfrac{1}{2}I{\omega ^2}$ …………….(2)
Where, $I$ is the moment of inertia and $\omega $ is the angular velocity.
Moment of inertia is the rotational analogue of mass and angular velocity is the rotational analogue of linear velocity.
The relationship between angular velocity $\omega $ and linear velocity $v$ is given as
$ \Rightarrow $$\omega = \dfrac{v}{R}$
We know the moment of inertia of a disc is given as
$ \Rightarrow $ $I = \dfrac{1}{2}m{R^2}$
Where, $m$ is the mass and $R$ is the radius of the disc.
By substituting $I$ and $\omega $ in equation (2) we get
$ \Rightarrow $${K_R} = \dfrac{1}{2}\left( {\dfrac{1}{2}m{R^2}} \right)\dfrac{{{v^2}}}{{{R^2}}}$
$ \Rightarrow {K_R} = \dfrac{1}{4}m{v^2}$
Now total kinetic energy, $K$ can be written as sum of kinetic energy due to rotational and kinetic energy due to translation. That is
$ \Rightarrow $$K = {K_T} + {K_R}$
On substituting the values, we get
$ \Rightarrow $$K = \dfrac{1}{2}m{v^2} + \dfrac{1}{4}m{v^2}$
$ \Rightarrow K = \dfrac{{2m{v^2} + m{v^2}}}{4}$
$\therefore K = \dfrac{{3m{v^2}}}{4}$
We need to find the ratio of translational kinetic energy to the total kinetic energy.
Thus
$ \Rightarrow $$\dfrac{{{K_T}}}{K} = \dfrac{{\dfrac{1}{2}m{v^2}}}{{\dfrac{3}{4}m{v^2}}}$
$ \Rightarrow \dfrac{{{K_T}}}{K} = \dfrac{2}{3}$

So, the correct answer is option A.

Note:
Here in this question it is given that the disc rolls without slipping. So, we considered the friction to be negligible. Friction is the resistance offered to the motion so due to friction some of the kinetic energy is converted into heat. Thus, the total kinetic energy will be less than the value that we got when friction is acting on the disc.