The ring always loses: what one rolling-race question taught me about rotational motion

Last month's Tuesday mock had a question I still think about. A ring, a disc, a solid sphere and a thin hollow sphere are released together from the top of the same incline. All of them roll without slipping. Which one reaches the bottom first?

I picked the ring. My reasoning, if you can call it that: all the mass is at the rim, so it has more "oomph" once it gets going. It was wrong, and not by a little. The ring is dead last.

I play a lot of chess, and there's a mistake I make there too. I think I'm attacking, but I've actually spent my tempo on a move that looks active and does nothing. The ring does exactly that. It is busy, it is spinning like crazy, and all that busy-ness is energy that is not going into moving forward.

So let me write down the one idea that fixed this for me.

Energy has two places to go

A body rolling down loses potential energy $mgh$. That energy has to be shared between moving forward and spinning:

$mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2$

For rolling without slipping, $v = \omega R$. Write $I = mk^2$ (k is the radius of gyration) and the whole thing collapses into

$mgh = \tfrac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)$

Only the first term makes you faster down the slope. The second term is the "tax" for spinning. A ring has $k = R$, so $k^2/R^2 = 1$ and half its energy goes into spinning. A solid sphere has $I = \frac{2}{5}mR^2$, so the ratio is $2/5$ and only $2/7$ of its energy is spent on rotation.

Rearranging, the acceleration along the incline is

$a = \frac{g\sin\theta}{1 + k^2/R^2}$

Body	$I$ about centre	$k^2/R^2$	$a$
Solid sphere	$\frac{2}{5}mR^2$	2/5	$\frac{5}{7}g\sin\theta$
Disc / solid cylinder	$\frac{1}{2}mR^2$	1/2	$\frac{2}{3}g\sin\theta$
Thin hollow sphere	$\frac{2}{3}mR^2$	2/3	$\frac{3}{5}g\sin\theta$
Ring	$mR^2$	1	$\frac{1}{2}g\sin\theta$

Order of arrival: solid sphere, disc, hollow sphere, ring. Notice that mass and radius are nowhere in the final formula. A huge ring and a tiny ring tie. That was my second mistake in the test, I kept wanting the heavier body to win. Mass cancels everywhere in gravity-driven motion and I forgot it the moment a rotation showed up.

"But friction opposes motion!"

This one tripped me for a whole year. If friction acts on the rolling body, doesn't that waste energy?

For pure rolling, no. The point touching the ground is instantaneously at rest, so static friction there does no work. What friction does is supply the torque that makes the body spin up while it speeds up:

$\tau = I\alpha, \qquad \tau = f R$

Take the torque about the centre. Gravity gives no torque about the centre of mass, so friction is the only thing that can make it rotate. Without friction (a perfectly smooth incline) the bodies just slide, nothing rotates, and all four tie with $a = g\sin\theta$.

How much friction is needed? From $f = mg\sin\theta - ma$ you get

$f = mg\sin\theta\cdot\frac{k^2/R^2}{1 + k^2/R^2}$

and since $N = mg\cos\theta$, the condition for rolling is $\mu \ge \tan\theta\cdot\frac{k^2/R^2}{1+k^2/R^2}$. Ring needs $\frac{1}{2}\tan\theta$, solid sphere only $\frac{2}{7}\tan\theta$. So the ring is also the one most likely to start slipping on a rough-ish surface.

The moment of inertia sheet I keep in my geometry box

• Ring (axis through centre, perpendicular to plane): $MR^2$
• Disc: $\frac{1}{2}MR^2$
• Solid sphere: $\frac{2}{5}MR^2$, thin spherical shell: $\frac{2}{3}MR^2$
• Rod, axis through centre and perpendicular: $\frac{ML^2}{12}$; about one end: $\frac{ML^2}{3}$

Then two theorems that I mixed up for a long time.

Parallel axis: $I = I_{cm} + Md^2$. Works for any body, but the two axes must be parallel and one must pass through the centre of mass.

Perpendicular axis: $I_z = I_x + I_y$. Works only for flat, planar bodies (laminas). I once used it on a solid sphere in a test and got a number so wrong that even I knew something was off.

Angular momentum, with the skater everyone draws

$L = I\omega$, and it stays constant if the net external torque is zero. A skater pulls her arms in, $I$ drops, so $\omega$ goes up. What I never saw in school: kinetic energy goes up too, because $KE = L^2/2I$ and $L$ is fixed. The extra energy comes from the work her muscles do pulling the arms inward. Energy is not appearing from nowhere.

What I actually do now

Before I touch a formula in any rotation question I write one line: where does the energy go? Sometimes it's a bar divided into "translation" and "rotation", nothing fancy. It forces me to notice if a spinning body is involved at all, and that alone has saved me from the ring trap twice since.

I also did the plank experiment because I was annoyed. A wooden plank against a pile of books, a steel marble, a solid wooden cylinder I found in a drawer, and a tin of chickpeas. The marble won every time, which felt like a small personal victory over my old self. The tin of chickpeas came last-ish, and I'm still not sure how to explain it cleanly, so I'm not going to pretend I can.

If you're in class 11 and rotational motion feels like a different subject from the rest of mechanics, it isn't. It's the same energy accounting you already know, with one extra bucket.