Question

A disc is rolling on an inclined plane without slipping then what fraction of its total energy will be in form of rotational kinetic energy :-

• A. $1:3$
• B. $1:2$
• C. $2:7$
• D. $2:5$

Answer 1

The correct option is C $2:7$

The moment of inertia of a sphere about its central axis and a disc of mass $M$ and radius $R$ as shown in figure above is

$I_{disc}=\frac{2}{5}M{R}^2$

Kinetic energy of the rotation of the disc, $K.E_{rotation}=\frac{1}{2}I{\omega }^2$ where $\omega$ is the angular velocity.

$⟹K.E_{rotation}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\frac{2}{5}M{R}^2{\omega }^2$

Now, angular velocity, $\omega =\frac{v}{R}$

where $v$ is the linear velocity of the sphere.

$⟹K.E_{rotation}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\frac{2}{5}M{R}^2\frac{v^2}{R^2}=\frac{1}{5}M{v}^2$

Kinetic energy of the linear motion $K.E_{linearmotion}=\frac{1}{2}M{v}^2$

Total kinetic energy $=K.E_{rotation}+K.E_{linearmotion}=\frac{1}{5}M{v}^2+\frac{1}{2}M{v}^2=\frac{7}{10}M{v}^2$

Now, the fraction of its total energy associated with rotation $=\frac{K.Eofrotation}{TotalK.E}$

$=\frac{\frac{1}{5}M{v}^2}{\frac{7}{10}M{v}^2}=\frac{2}{7}$

So the correct answer is $2:7$