A uniform circular disc of radius R and mass M is rolling without slipping on a horizontal surface with a linear velocity v . What is its total kinetic energy?

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TestHub · Accepted Answer

The total kinetic energy of a body rolling without slipping is the sum of its translational kinetic energy and rotational kinetic energy: $K_{t o t a l} = K_{t r an s} + K_{ro t}$ . Translational kinetic energy is $K_{t r an s} = \frac{1}{2} M v^{2}$ .

Rotational kinetic energy is $K_{ro t} = \frac{1}{2} I ω^{2}$ . For a uniform circular disc, $I = \frac{1}{2} M R^{2}$ . For rolling without slipping, $v = R ω$ , so $ω = v / R$ . Substituting these values, $K_{ro t} = \frac{1}{2} (\frac{1}{2} M R^{2}) (\frac{v}{R})^{2} = \frac{1}{4} M R^{2} \frac{v ^{2}}{R ^{2}} = \frac{1}{4} M v^{2}$ . Therefore, $K_{t o t a l} = \frac{1}{2} M v^{2} + \frac{1}{4} M v^{2} = \frac{2}{4} M v^{2} + \frac{1}{4} M v^{2} = \frac{3}{4} M v^{2}$ .

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JEE Main 2027 (Class 12+)