Rolling Without Slipping showed with energy that a hoop loses a race down a ramp to a solid ball. Newton's Second Law for Rotation lets us find the forces and the acceleration directly, and see what static friction is doing.
Why it works
Take an object of mass $M$, radius $R$ and $I = cMR^2$ rolling down a ramp at angle $\theta$. Along the ramp: $$Mg\sin\theta - f = Ma.$$ About the center, only friction has a lever arm: $$fR = I\alpha = cMR^2\alpha.$$ Rolling without slipping links them: $a = R\alpha$. So $f = cMa$, and substituting, $Mg\sin\theta = Ma(1 + c)$.
Key idea
$$a = \dfrac{g\sin\theta}{1 + I/(MR^2)}, \qquad f = \dfrac{I}{R^2}a.$$ Mass and radius cancel; only the shape factor $c = I/(MR^2)$ matters: solid ball 0.4, solid disk 0.5, hollow ball 0.67, hoop 1. Static friction must supply $f$, so rolling needs $\mu_s \ge \dfrac{c}{1 + c}\tan\theta$.
Worked example: A ball on a ramp
A solid ball rolls without slipping down a $30^\circ$ ramp. Find its acceleration and the least $\mu_s$ that allows rolling.
- $a = \dfrac{9.8\sin 30^\circ}{1 + 0.4} = \dfrac{4.9}{1.4} = 3.5\ \text{m/s}^2$ (sliding without friction it would be $4.9\ \text{m/s}^2$).
- Friction needed: $f = 0.4Ma = 0.4(3.5)M = 1.4M$ (in newtons, with $M$ in kilograms).
- $\mu_s \ge \dfrac{f}{Mg\cos\theta} = \dfrac{0.4}{1.4}\tan 30^\circ = 0.165$.
Answer: $a = 3.5\ \text{m/s}^2$; $\mu_s \ge 0.165$.
Worked example: Speed at the bottom
A hoop and a solid ball each roll from rest down a ramp, dropping $1.0\ \text{m}$. Find each one's speed at the bottom.
- Energy: $Mgh = \tfrac{1}{2}Mv^2(1 + c)$, so $v = \sqrt{\dfrac{2gh}{1 + c}}$.
- Hoop ($c = 1$): $v = \sqrt{9.8} = 3.13\ \text{m/s}$.
- Solid ball ($c = 0.4$): $v = \sqrt{19.6/1.4} = 3.74\ \text{m/s}$. The hoop puts half its energy into spinning; the ball only 2/7.
Answer: Hoop $3.13\ \text{m/s}$; ball $3.74\ \text{m/s}$.
Why don't mass and radius matter? A bigger or heavier object of the same shape has more weight pulling it down the ramp, but proportionally more inertia, both for moving and for spinning. Only the shape, through $I/(MR^2)$, decides how the energy splits between moving and spinning.
Common mistake
Thinking static friction drains energy from a rolling object, so energy conservation fails. The contact point of a rolling object is momentarily at rest ($v_{\text{contact}} = v - R\omega = 0$), so friction acting there does no work. It moves energy from "moving" to "spinning" without losing any. Check: the energy method above, which ignores friction, gives the same answer as the force method, which includes it. (Kinetic friction during skidding is different; see From Skidding to Rolling.)