Question

A circular disc rolls down an inclined plane. The ratio of rotational kinetic energy to total kinetic energy is
A.$\dfrac{1}{2}$
B.$\dfrac{1}{3}$
C. $\dfrac{2}{3}$
D. $\dfrac{3}{4}$

Answer 1

Hint: The total energy is the sum of translational and rotational kinetic energy. Ratio is calculated by dividing rotational and total kinetic energy.

Step by step answer: A circular disc is rolling down the inclined plane i.e. without slipping. When the body (circular disc) is rolling down the body, it has both rotational and translational kinetic energy i.e., rotational energy as the body is rotating while coming down the inclined plane and translational energy because it is coming straight from the inclined plane.
Let the mass of the circular disc be m, velocity be v and angular velocity (the velocity possessed by a body during rotation) be ω and radius of the disc be r.
The moment of inertia (I) is defined as the sum of the products of the masses of all the particles of the body and square of the distances from the axis of rotation i.e. $I = \dfrac{1}{2}m{r^2}$ [since the disc is rotating about an axis passing through the centre]
Rotational Kinetic energy (R)=$\dfrac{1}{2}I{\omega ^2}$ [I is moment of inertia]
 $\Rightarrow$ $\dfrac{1}{2}\left( {\dfrac{1}{2} \times m{r^2}} \right){\omega ^2} = \dfrac{1}{4}\left( {m{r^2}{\omega ^2}} \right)$
The moment of inertia is defined as the sum of the products of the masses of all the particles of the body and square of the distances from the axis of rotation.
Translational kinetic energy (T)= $\dfrac{1}{2}\left( {m{v^2}} \right)$
 $\Rightarrow$ $\dfrac{1}{2}m{\left( {r\omega } \right)^2}$ $\left[ {v = r\omega } \right]$
 $\Rightarrow$ $\dfrac{1}{2}\left( {m{r^2}{\omega ^2}} \right)$
The total energy = $R + T$
 $\Rightarrow$ $\dfrac{1}{4}\left( {m{r^2}{\omega ^2}} \right) + \dfrac{1}{2}\left( {m{r^2}{\omega ^2}} \right) = \left( {\dfrac{1}{4} + \dfrac{1}{2}} \right)\left( {m{r^2}{\omega ^2}} \right)$
$\therefore $ $\dfrac{3}{4}\left( {m{r^2}{\omega ^2}} \right)$
Ratio of rotational and total kinetic energy = $\dfrac{{\dfrac{1}{4}m{r^2}{\omega ^2}}}{{\dfrac{3}{4}m{r^2}{\omega ^2}}}$ = $\dfrac{1}{3}$

Therefore, option B is correct.

Note: When a body rolls down an inclined plane, both translational and kinetic energy contributes to the motion. So, the total energy is being calculated by adding both these energies.