Question

A disc on a table whose ratio of its K.E. of rotation to the total K.E. is
A. $2/5$
B. $1/3$
C. $5/6$
D. $2/3$

Answer 1

Rotational kinetic energy is the kinetic energy possessed by the body when it is rotating about its axis. Total kinetic energy is the sum of rotational kinetic energy and translational kinetic energy. Recall the expressions for translational kinetic energy and rotational kinetic energy to answer this question.

Formula used:
${E_{Total}} = {E_T} + {E_r}$
$\therefore {E_T} = \dfrac{1}{2}m{v^2},{E_r} = \dfrac{1}{2}I{w^2}$
Here, m is the mass, v is the velocity, I is the moment of inertia and $\omega$ is the angular velocity.

Complete step by step answer:
We know that the Kinetic energy is the energy possessed by a body when it is in motion.
Translational kinetic energy is energy possessed by a body at the time of translational motion. Translational kinetic energy of a body is given by,
${E_T} = \dfrac{1}{2}m{v^2}$. . . . . (1)
Where,
${E_T} =$Translational kinetic energy
$m$ is mass of the body
$v$ is radius

Rotational kinetic energy is the energy possessed by a body when it is in rotational motion.
Rotational kinetic energy is given by
${E_r} = \dfrac{1}{2}I{w^2}$. . . . . (2)
Where, ${E_r}$ is rotational kinetic energy,
$I$ is moment of inertia,
$w$ is angular velocity of the body.

We know that the moment of inertia of a disc is given by,
$I = \dfrac{1}{2}m{r^2}$
Where, m is the mass and r is the radius.
By putting this value in equation (2), we get
${E_r} = \dfrac{1}{2}\left( {\dfrac{1}{2}m{r^2}} \right){w^2}$
$\Rightarrow{E_r} = \dfrac{1}{4}m{r^2}{w^2}$
$\Rightarrow{E_r} = \dfrac{1}{4}m{\left( {rw} \right)^2}$

But we know that, $v = rw$
${E_r} = \dfrac{1}{4}m{v^2}$. . . . . (3)
Now, the total kinetic energy of a body is the sum of translational kinetic energy and rotational kinetic energy. Therefore,
${E_{Total}} = {E_T} + {E_r}$
$\Rightarrow {E_{Total}} = \dfrac{1}{2}m{v^2} + \dfrac{1}{4}m{v^2}$(from equation (1) and (3))
By simplifying, we get
$\Rightarrow {E_{Total}} = \dfrac{1}{2}m{v^2}\left( {1 + \dfrac{1}{2}} \right)$
$\Rightarrow {E_{Total}} = \dfrac{1}{2}m{v^2}\left( {\dfrac{{2 + 1}}{2}} \right)$
$\Rightarrow {E_{Total}} = \dfrac{1}{2}m{v^2}\left( {\dfrac{3}{2}} \right)$
$\Rightarrow {E_{Total}} = \dfrac{3}{4}m{v^2}$
Therefore, the ratio of rotational kinetic energy to total kinetic energy can be written as,
$\dfrac{{{E_r}}}{{{E_{Total}}}} = \dfrac{{\dfrac{1}{4}m{v^2}}}{{\dfrac{3}{4}m{v^2}}}$(from equation 3 and 4)
On simplifying we get
$\therefore \dfrac{{{E_r}}}{{{E_{Total}}}} = \dfrac{1}{3}$

Therefore, from the above explanation the correct option is B.

Note: In translational motion the body moves without rotating about its axis. The block sliding on the inclined plane is the example of translational motion while the sphere rotating while sliding down is the example of rotational and translational motion simultaneously. Students should remember the moment of inertia of some of the often used objects to solve such types of questions.