JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Rotational Mechanics

A body rolls down an inclined plane without slipping. The kinetic energy of rotation is

50%

of its translational kinetic energy. The body is :

(A)Hollow cylinder
(B)Ring
(C)Solid sphere
(D)Solid cylinder✓ Correct

Answer: (D)

Solution

For rolling without slipping,

v = ω R

, and writing

I = m k^{2}

gives

\frac{K _{rot}}{K _{trans}} = \frac{\frac{1}{2} I ω ^{2}}{\frac{1}{2} m v ^{2}} = \frac{k ^{2}}{R ^{2}}

The rotational energy is half the translational energy, so

\frac{k ^{2}}{R ^{2}} = \frac{1}{2}

and

I = \frac{1}{2} m R^{2}

. That is the moment of inertia of a solid cylinder about its own axis, so the body is a solid cylinder.

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option

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