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Q. A circular disc rolls down an inclined plane. The ratio of the rotational kinetic energy to the total kinetic energy is

UPSEEUPSEE 2013

Solution:

Rotational kinetic energy, $K_{R}=\frac{1}{2} / \omega^{2}$
$K_{R}=\frac{1}{2} \times \frac{M R^{2}}{2} \omega^{2}=\frac{1}{4} M v^{2}$
$(\because \vee=R \omega)$
Translational kinetic energy
$K_{T}=\frac{1}{2} \,M v^{2}$
Total kinetic energy $=K_{T}+K_{R}$
$=\frac{1}{2} M v^{2}+\frac{1}{4} M v^{2}$
$=\frac{3}{4} M v^{2}$
$\therefore \frac{\text { Rotational kinetic energy }}{\text { Total kinetic energy }}$
$=\frac{\frac{1}{4} M v^{2}}{\frac{3}{4} M v^{2}}$
$=\frac{1}{3}$