A cylinder rolls down an incline without slipping — compare with sliding

Question

A solid cylinder and a block (both of mass $M$ ) are released from the top of an incline of angle $\theta$ and height $h$ . The cylinder rolls without slipping; the block slides without friction. Find the acceleration and speed at the bottom for each. Which reaches the bottom first?

(JEE Advanced 2021, similar pattern)

Solution — Step by Step

For the frictionless sliding block, all PE converts to translational KE:

Mgh = \frac{1}{2}Mv_{\text{slide}}^2

v_{\text{slide}} = \sqrt{2gh}

Acceleration along the incline: $a_{\text{slide}} = g\sin\theta$

For the rolling cylinder, PE converts to both translational and rotational KE:

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

For a solid cylinder: $I = \frac{MR^2}{2}$ and $v = R\omega$ (pure rolling).

Mgh = \frac{1}{2}Mv^2 + \frac{1}{2} \cdot \frac{MR^2}{2} \cdot \frac{v^2}{R^2} = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2

v_{\text{roll}} = \sqrt{\frac{4gh}{3}}

Using the formula for rolling on an incline:

a_{\text{roll}} = \frac{g\sin\theta}{1 + k^2/R^2}

For solid cylinder, $k^2/R^2 = 1/2$ :

a_{\text{roll}} = \frac{g\sin\theta}{1 + 1/2} = \frac{2g\sin\theta}{3}

Quantity	Sliding block	Rolling cylinder
Acceleration	$g\sin\theta$	$\frac{2}{3}g\sin\theta$
Speed at bottom	$\sqrt{2gh}$	$\sqrt{4gh/3}$

Since $a_{\text{slide}} > a_{\text{roll}}$ , the sliding block reaches the bottom first.

\boxed{v_{\text{slide}} = \sqrt{2gh}, \quad v_{\text{roll}} = \sqrt{4gh/3}, \quad \text{Block wins}}

Why This Works

The rolling cylinder is slower because some of its energy goes into spinning. Of the total PE, $\frac{2}{3}$ goes to translation and $\frac{1}{3}$ goes to rotation. The block puts all its energy into translation.

The ratio depends on $k^2/R^2$ — the geometry of the rolling body. A hollow cylinder ( $k^2/R^2 = 1$ ) would be even slower: $v = \sqrt{gh}$ . A solid sphere ( $k^2/R^2 = 2/5$ ) would be faster than the cylinder: $v = \sqrt{10gh/7}$ . The ranking from fastest to slowest: block > solid sphere > solid cylinder > hollow sphere > hollow cylinder/ring.

Alternative Method — Force analysis for rolling

Net force along incline: $Mg\sin\theta - f = Ma$ (translation)

Torque about centre: $fR = I\alpha = \frac{MR^2}{2} \cdot \frac{a}{R} = \frac{MRa}{2}$

So $f = Ma/2$ . Substituting: $Mg\sin\theta - Ma/2 = Ma$ , giving $a = \frac{2g\sin\theta}{3}$ .

The $k^2/R^2$ ratios are tested repeatedly in JEE. Memorise: ring = 1, hollow cylinder = 1, solid sphere = 2/5, solid cylinder = disc = 1/2, hollow sphere = 2/3. The one with the smallest ratio reaches the bottom first among rollers. This is independent of mass and radius — a powerful result.

Common Mistake

Students often assume friction does work during pure rolling and subtract it from the energy equation. In pure rolling, the contact point has zero velocity, so friction does zero work — it only redirects energy from translation to rotation. You can use energy conservation directly ( $Mgh = \frac{1}{2}Mv^2(1 + k^2/R^2)$ ) without worrying about friction’s work. Friction is needed for the force analysis but not for the energy method.

Question

Solution — Step by Step

Why This Works

Alternative Method — Force analysis for rolling

Common Mistake

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