JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Rotational Mechanics

A uniform solid cylinder of mass 'm' and radius 'r' rolls along an inclined rough plane of inclination

4 5^{\circ}

. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder axis will be :-

Answer: (C)

Solution

For a body of moment of inertia

I = k m r^{2}

rolling without slipping down an incline of angle

θ

, Newton's second law along the incline combined with the torque equation about the contact point gives

a = \frac{g sin θ}{1 + k}

For a uniform solid cylinder,

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

, so

k = \frac{1}{2}

.

With

θ = 4 5^{\circ}

sin 4 5^{\circ} = \frac{1}{2}

a = \frac{g / 2}{1 + \frac{1}{2}} = \frac{g / 2}{3/2} = \frac{2 g}{3 2} = \frac{2 g}{3}

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option