Question

A disc is rolling on the inclined plane, what is the ratio of its rotational $KE$ to the total $KE$ ?

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

Answer 1

Answer: (A)

The rotational kinetic energy of the disc is
$K_{rot}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }^2$
$=\frac{1}{4}M{R}^2{\omega }^2$
The translational kinetic energy is
$K_{trass}=\frac{1}{2}M{v}_{CM}^2$
Where, $v_{cM}$ is the linear velocity of its centre of mass
Now, $v_{cM}=R\omega$ (for pure rolling)
Therefore, $K_{\text{trans }}=\frac{1}{2}M{R}^2{\omega }^2$
Thus, $K_{\text{total }}=\frac{1}{4}M{R}^2{\omega }^2+\frac{1}{2}M{R}^2{\omega }^2$
$\frac{3}{4}M{R}^2{\omega }^2$
$\therefore \frac{K_{\text{rot }}}{K_{\text{total }}}=\frac{\frac{1}{4}M{R}^2{\omega }^2}{\frac{3}{4}M{R}^2{\omega }^2}$
$=\frac{1}{3}$