Question

A solid spherical ball is rolling without slipping on an inclined plane. The fraction of its total kinetic energy associated with rotation is

• A. $\frac{2}{5}$
• B. $\frac{2}{7}$
• C. $\frac{3}{5}$
• D. $\frac{3}{7}$

Answer 1

The correct option is B $\frac{2}{7}$

Given ball is rolling without slipping, hence $v=\omega R$
where, $v=$ linear velocity of the ball
$\omega =$ angular velocity of the ball
Total kinetic energy = Translational Kinetic Energy + Rotational Kinetic Energy
$T.K.E=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
Here, $I_{com}=\frac{2}{5}m{R}^2$
$T.K.E=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right)\times {\left(\frac{v}{R}\right)}^2$
$T.K.E=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2$
$T.K.E=\frac{7}{10}m{v}^2$
Rotational $KE=\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}\left(\frac{2}{5}m{v}^2\right)\times {\left(\frac{v}{R}\right)}^2=\frac{1}{5}m{v}^2$
$\therefore \frac{\text{Rotational K.E}}{\text{Total K.E}}=\frac{\frac{1}{5}m{v}^2}{\frac{7}{10}m{v}^2}=\frac{2}{7}$