Using energy conservation: v = √(2gh/(1 + I/MR²)).

Rolling Without Slipping — Velocity at Bottom of Incline

Question

A solid sphere of mass $M$ and radius $R$ rolls without slipping down an incline of height $h$ . Find the velocity of the centre of mass at the bottom.

This is a standard energy conservation problem, but the “rolling without slipping” condition is what makes it different from a sliding block — and that’s exactly where most students lose marks.

Solution — Step by Step

When a body rolls without slipping, it has two forms of kinetic energy simultaneously: translational ( $\frac{1}{2}Mv^2$ ) and rotational ( $\frac{1}{2}I\omega^2$ ).

The no-slip condition links them: $v = R\omega$ , so $\omega = v/R$ . This is the key relationship that lets us write everything in terms of $v$ .

Initial energy is purely potential. Final energy is kinetic (both types). No slipping means no energy lost to friction as heat — the friction force does no work here.

Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2

Replace $\omega$ with $v/R$ :

Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\cdot\frac{v^2}{R^2}

Factor out $\frac{1}{2}v^2$ :

Mgh = \frac{1}{2}v^2\left(M + \frac{I}{R^2}\right)

For a solid sphere, $I = \frac{2}{5}MR^2$ . So $\frac{I}{R^2} = \frac{2}{5}M$ .

Mgh = \frac{1}{2}v^2\left(M + \frac{2}{5}M\right) = \frac{1}{2}v^2 \cdot \frac{7}{5}M

v^2 = \frac{2Mgh \cdot 5}{7M} = \frac{10gh}{7}

\boxed{v = \sqrt{\frac{10gh}{7}}}

Why This Works

The general formula — valid for any rolling body — is:

v = \sqrt{\frac{2gh}{1 + I/MR^2}}

The term $I/MR^2$ is the “rotational penalty.” A higher moment of inertia means more energy goes into spinning and less into translational motion, so the body moves slower at the bottom.

For a solid sphere, $I/MR^2 = 2/5$ , giving the denominator $1 + 2/5 = 7/5$ . For a hollow sphere it’s $2/3$ , for a ring it’s $1$ — so a ring is always the slowest down an incline. This is why in JEE, ranking problems by speed at the bottom always have the ring last.

The angle of the incline never appears in the final answer. Only height $h$ matters — the shape of the path doesn’t, only the vertical drop.

For JEE Main, memorise the $v$ values for the four standard bodies rolling down an incline of height $h$ :

Solid sphere: $\sqrt{10gh/7}$
Hollow sphere: $\sqrt{6gh/5}$
Solid cylinder: $\sqrt{4gh/3}$
Ring (or hollow cylinder): $\sqrt{gh}$

Speed ranking: solid sphere > hollow sphere > solid cylinder > ring.

Alternative Method — Using Acceleration

If the question asks for speed at the bottom and you also know the length of the incline $L$ (with $h = L\sin\theta$ ), you can use kinematics.

The acceleration of a rolling body on an incline is:

a = \frac{g\sin\theta}{1 + I/MR^2}

For a solid sphere: $a = \frac{5}{7}g\sin\theta$

Then use $v^2 = 2aL = 2 \cdot \frac{5g\sin\theta}{7} \cdot \frac{h}{\sin\theta} = \frac{10gh}{7}$ .

Same answer. Energy conservation is faster here, but the acceleration formula becomes essential when the problem asks for time taken or asks you to compare rolling vs sliding on the same incline.

Common Mistake

Forgetting the rotational KE entirely.

Students write $Mgh = \frac{1}{2}Mv^2$ and get $v = \sqrt{2gh}$ — which is the answer for a sliding block, not a rolling sphere. This error costs marks in both board exams and JEE Main.

The phrase “rolls without slipping” is your trigger: always write both KE terms. If the problem said “slides without friction,” then $v = \sqrt{2gh}$ would be correct.

Here’s the MDX body content. Key things I did:

Step 3 explains why we substitute before showing the algebra — the “why before how” rule
The general formula box in “Why This Works” gives JEE aspirants the reusable tool
The tip callout has the four standard rolling bodies ranked by speed — exactly what gets asked in JEE Main speed-ranking MCQs
The common mistake targets the specific error (forgetting rotational KE), not a generic “be careful” warning
The alternative method via acceleration is genuinely useful for follow-up questions about time taken

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Acceleration

Common Mistake

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