A rolling ball does two things at once: its center moves, and it spins. So it has two kinds of kinetic energy, the ordinary kind and rotational kinetic energy. Adding both to energy conservation explains a famous race.
Why it works
Rolling without slipping means the point touching the ground is momentarily at rest, like the bottom of a tire tread. The center then moves the distance $R\Delta\theta$ while the object turns $\Delta\theta$, so $v = R\omega$. With $I = cmR^2$ (where $c$ depends on the shape), the total kinetic energy is $$K = \tfrac{1}{2}mv^2 + \tfrac{1}{2}(cmR^2)\left(\frac{v}{R}\right)^2 = \tfrac{1}{2}(1 + c)\,mv^2.$$ Static friction at the contact point does no work, because that point doesn't slide, so mechanical energy is conserved down a ramp: $mgh = \tfrac{1}{2}(1 + c)mv^2$.
Key idea
For rolling without slipping: $v = R\omega$ and $K = \tfrac{1}{2}(1 + c)mv^2$, with $c = \dfrac{I}{mR^2}$: hoop $1$, hollow sphere $\tfrac{2}{3}$, solid cylinder $\tfrac{1}{2}$, solid sphere $\tfrac{2}{5}$. From rest down a height $h$: $$v = \sqrt{\frac{2gh}{1 + c}}.$$ Mass and radius cancel; only the shape matters.
Worked examples
Worked example: Kinetic energy of a rolling can
A $2.0\ \text{kg}$ solid cylinder rolls at $3.0\ \text{m/s}$. Find its total kinetic energy.
- Moving: $\tfrac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}$.
- Spinning: $c = \tfrac{1}{2}$, so $\tfrac{1}{2} \times 9.0 = 4.5\ \text{J}$.
- Total: $13.5\ \text{J}$. One third of it is spinning energy.
Answer: $13.5\ \text{J}$
Worked example: Speed at the bottom
A solid ball rolls from rest down a ramp, dropping $1.4\ \text{m}$. How fast is it moving at the bottom?
- $$v = \sqrt{\dfrac{2gh}{1 + 2/5}} = \sqrt{\dfrac{2(9.8)(1.4)}{1.4}} = \sqrt{19.6} = 4.4\ \text{m/s}.$$
- A frictionless sliding block would reach $\sqrt{2(9.8)(1.4)} = 5.2\ \text{m/s}$. The ball is slower because $\tfrac{2}{7}$ of its energy goes into spinning.
Answer: $4.4\ \text{m/s}$
Worked example: Rolling uphill
A hoop rolls toward a hill at $2.0\ \text{m/s}$. How high does it roll before stopping?
- All its kinetic energy becomes potential energy: $\tfrac{1}{2}(1 + 1)mv^2 = mgh$.
- $h = \dfrac{2v^2}{2g} = \dfrac{(2.0)^2}{9.8} = 0.41\ \text{m}$, twice as high as a sliding block with the same speed would go.
Answer: $0.41\ \text{m}$
Common mistake
Thinking the heavier or larger ball wins a rolling race. In $v = \sqrt{2gh/(1 + c)}$ there is no $m$ and no $R$. A heavy solid ball and a light solid ball tie; a solid ball beats a hoop of any size. Check: if your answer for a rolling speed depends on the mass, a step has gone wrong (or something other than gravity is doing work).
The ramp's static friction is what makes the object spin, yet it takes no energy away: it only shares the energy between moving and spinning. A ball that skids, like a bowling ball just after release, does lose energy to kinetic friction until it starts to roll.