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Q.
A disc is rolling on the inclined plane, what is the ratio of its rotational $KE$ to the total $KE$ ?
System of Particles and Rotational Motion
Solution:
The rotational kinetic energy of the disc is
$K _{ rot } =\frac{1}{2} I \omega^{2}=\frac{1}{2}\left(\frac{1}{2} MR ^{2}\right) \omega^{2}$
$=\frac{1}{4} MR ^{2} \omega^{2}$
The translational kinetic energy is
$K _{ trass }=\frac{1}{2} Mv _{ CM }^{2}$
Where, $v_{cM}$ is the linear velocity of its centre of mass
Now, $v_{c M}=R \omega$ (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} $