Rotational Motion: PYQ Walkthrough (6)

hard 2 min read

Question

A solid sphere of mass MM and radius RR rolls without slipping down an incline of angle θ\theta. Find its acceleration along the incline. (JEE Main 2024 Shift 1 pattern.)

Solution — Step by Step

Along the incline, two forces act on the sphere: MgsinθMg\sin\theta down the slope and friction ff up the slope (friction must point up because the sphere tends to slide down).

Mgsinθf=MaMg\sin\theta - f = Ma

Friction provides the torque about the centre of mass. Normal force and gravity pass through the centre, so they contribute nothing.

fR=Iα=25MR2αfR = I\alpha = \frac{2}{5}MR^2 \cdot \alpha

For pure rolling, a=Rαa = R\alpha, so α=a/R\alpha = a/R.

From the torque equation: f=25Maf = \frac{2}{5}Ma.

Substituting into the translation equation:

Mgsinθ25Ma=MaMg\sin\theta - \frac{2}{5}Ma = Ma

Mgsinθ=75MaMg\sin\theta = \frac{7}{5}Ma

a=5gsinθ7a = \frac{5g\sin\theta}{7}

Final answer: a=5gsinθ7a = \dfrac{5g\sin\theta}{7}

Why This Works

The factor 57\frac{5}{7} comes directly from the moment of inertia of a solid sphere (25MR2\frac{2}{5}MR^2). For a hollow sphere (23MR2\frac{2}{3}MR^2), the same algebra gives a=3gsinθ5a = \frac{3g\sin\theta}{5}. For a ring or hollow cylinder, a=gsinθ2a = \frac{g\sin\theta}{2}.

The pattern: a=gsinθ1+I/MR2a = \dfrac{g\sin\theta}{1 + I/MR^2}. Memorise this — it cracks every “rolling down an incline” question instantly.

Alternative Method

Energy conservation gives velocity, not acceleration directly. But for time-of-descent questions, energy works faster:

Mgh=12Mv2+12Iω2=12Mv2(1+IMR2)Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2\left(1 + \frac{I}{MR^2}\right)

So v2=2gh1+I/MR2v^2 = \dfrac{2gh}{1 + I/MR^2}.

Common Mistake

Students forget to include rotational kinetic energy and write Mgh=12Mv2Mgh = \frac{1}{2}Mv^2. This gives a=gsinθa = g\sin\theta — the answer for a sliding (not rolling) block. JEE loves this trap because both answers look “clean”.

Rolling acceleration is always less than gsinθg\sin\theta because part of the gravitational PE goes into spinning, not translating. If your answer comes out larger than gsinθg\sin\theta, you have a sign error.

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