The Rolling Race

The problem

A solid sphere, a solid disc and a thin hoop are released together from rest at the top of a 2.00 m ramp inclined at 30°. All roll without slipping. Which reaches the bottom first, and does it depend on their masses or radii?

Almost everyone's first instinct — that the heaviest, or the largest, wins — is wrong. So is the instinct that they tie.

Setting it up

Energy conservation from top to bottom, with the drop \(h = L\sin\theta\):

\[mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2\]

Rolling without slipping ties the two motions together, \(v = \omega R\). Write the moment of inertia as \(I = kmR^2\), where \(k\) is a pure number depending only on shape:

\[mgh = \tfrac12 mv^2 + \tfrac12 (kmR^2)\frac{v^2}{R^2} = \tfrac12 mv^2(1+k)\]

\[\boxed{\;v = \sqrt{\frac{2gh}{1+k}}\;},\qquad a = \frac{g\sin\theta}{1+k}\]

Both \(m\) and \(R\) have cancelled. Only \(k\) survives.

The answer

Body	\(k = I/mR^2\)	\(a\) (m/s²)	time (s)	\(v\) at bottom (m/s)
Frictionless block (slides)	0	4.905	0.903	4.429
Solid sphere	2/5	3.504	1.069	3.744
Solid disc / cylinder	1/2	3.270	1.106	3.617
Thin hoop	1	2.453	1.277	3.132

Sphere beats disc beats hoop, always, for any mass, any radius, any angle, any ramp length. And a frictionless sliding block beats all of them.

Why: the energy has to be shared

The ramp delivers the same \(mgh\) per unit mass to every body. Each must split it between translation (\(\frac12mv^2\)) and rotation (\(\frac12I\omega^2\)), and the split is fixed by \(k\):

\[\frac{K_{\rm rot}}{K_{\rm total}} = \frac{k}{1+k}\]

Hoop (\(k=1\)): half the energy goes into spin. Only half is left to move it down the ramp.
Disc (\(k=1/2\)): one third into spin.
Sphere (\(k=2/5\)): between a quarter and a third into spin.
Sliding block (\(k=0\)): none. It gets everything, and wins.

The physical reason \(k\) differs is where the mass sits. A hoop has all its mass at radius \(R\), so every gram must be spun up at the full rim speed. A sphere has most of its mass close to the axis, moving slowly, so spinning it is cheap. Mass distribution, not mass.

Why mass and radius cancel

Worth being explicit, because it surprises people twice.

Mass cancels for the same reason it does for a falling stone: gravity's pull and the inertia resisting it both scale with \(m\). Radius cancels because \(I \propto mR^2\) and the rolling constraint \(\omega = v/R\) brings in \(1/R^2\) — the two exactly undo each other.

So a marble and a bowling ball, released together, arrive together (both spheres). A bicycle wheel and a wedding ring also arrive together (both hoops) — and both lose to the marble.

Friction: the quiet requirement

Rolling needs static friction to supply the torque that spins the body up. Yet friction does no work here, because the contact point is instantaneously at rest — that is precisely what "rolling without slipping" means.

So friction is essential and free. If the ramp is too steep or too slippery the required static friction exceeds \(\mu_s N\), the body slips, kinetic friction takes over, energy is dissipated, and none of the above applies. The condition is

\[\tan\theta \le \mu_s\left(1 + \frac1k\right)\]

Note that a hoop tolerates the steepest ramp before slipping, and a sphere the shallowest — the opposite of the race order.

Common mistakes

Expecting the heaviest or largest to win. Both cancel exactly.
Forgetting the rotational term. Then every body would tie with the sliding block.
Treating friction as dissipative here. The contact point is at rest, so it does no work.
Using \(I\) about the wrong axis. \(k\) is defined about the centre of mass; the parallel-axis theorem appears only if you take torques about the contact point (which is a valid alternative route, and gives the same answer).

Moment of inertia · Rolling & angular momentum · Rotation · Conservation of energy · Friction & drag · Torque