Rotational Motion: Tricky Questions Solved (7)

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Question

A uniform solid sphere of mass MM and radius RR rolls without slipping down a smooth incline of angle θ\theta. Find the linear acceleration of its centre of mass. Compare with a hollow sphere of the same mass and radius.

Solution — Step by Step

Three forces act: gravity MgMg down, normal NN perpendicular to incline, and friction ff up the incline. Friction is what enables rolling without slipping.

Taking down the incline as positive:

Mgsinθf=MaMg\sin\theta - f = Ma

Friction is the only force producing torque about the centre. With τ=fR\tau = fR and τ=Iα\tau = I\alpha:

fR=IαfR = I\alpha

For rolling without slipping, a=Rαa = R\alpha, so α=a/R\alpha = a/R. Then f=Ia/R2f = Ia/R^2.

Substituting into the force equation:

MgsinθIaR2=Ma    a=gsinθ1+I/(MR2)Mg\sin\theta - \tfrac{Ia}{R^2} = Ma \implies a = \tfrac{g\sin\theta}{1 + I/(MR^2)}

For a solid sphere, I=25MR2I = \tfrac{2}{5}MR^2, so I/(MR2)=2/5I/(MR^2) = 2/5:

asolid=gsinθ1+2/5=5gsinθ7a_{\text{solid}} = \tfrac{g\sin\theta}{1 + 2/5} = \tfrac{5g\sin\theta}{7}

For a hollow sphere, I=23MR2I = \tfrac{2}{3}MR^2:

ahollow=gsinθ1+2/3=3gsinθ5a_{\text{hollow}} = \tfrac{g\sin\theta}{1 + 2/3} = \tfrac{3g\sin\theta}{5}

570.714\tfrac{5}{7} \approx 0.714 vs 35=0.6\tfrac{3}{5} = 0.6. The solid sphere accelerates faster because more of its mass is closer to the axis, so less rotational kinetic energy is “wasted” — more energy goes into translation.

Final answer: asolid=5gsinθ7a_{\text{solid}} = \mathbf{\tfrac{5g\sin\theta}{7}}, ahollow=3gsinθ5a_{\text{hollow}} = \mathbf{\tfrac{3g\sin\theta}{5}}.

Why This Works

The factor 1+I/(MR2)1 + I/(MR^2) tells us how “rotation-heavy” a body is. A point particle with I=0I = 0 would accelerate at gsinθg\sin\theta — the full slide value. Any rolling body is slower because some of the gravitational energy converts to rotational KE.

Memorise these factors: ring 12\tfrac{1}{2}, disc 23\tfrac{2}{3}, solid sphere 57\tfrac{5}{7}, hollow sphere 35\tfrac{3}{5}. They appear in every rolling question in JEE.

Alternative Method

Energy conservation gives the same result for speed after rolling distance LL. Using MgLsinθ=12Mv2+12Iω2Mg L\sin\theta = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2 and v=Rωv = R\omega, we get v2=2gLsinθ1+I/MR2v^2 = \tfrac{2gL\sin\theta}{1 + I/MR^2}, which combined with v2=2aLv^2 = 2aL recovers our answer.

Common Mistake

Many students forget that friction does no work during pure rolling. The point of contact is instantaneously at rest, so the displacement of the friction force is zero. Energy is conserved entirely between gravitational PE, translational KE and rotational KE — friction is just the constraint that links them.

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