Rolling Motion

A wheel rolling without slipping is a beautiful combination of translation (its CM moves forward) and rotation (it spins about its CM). At any instant, the contact point with the ground is momentarily at rest — and that's the secret of why rolling friction is tiny.

Concept

Pure rolling condition (no slipping): The contact point's instantaneous velocity is zero. For a wheel of radius $R$ with CM velocity $v$ and angular velocity $\omega$:

$v = R\omega$

Differentiating, the linear acceleration of the CM and the angular acceleration are linked by $a = R\alpha$.

Velocity decomposition. Every point on a rolling body has velocity = velocity of CM + velocity due to rotation about CM. Top of wheel moves at $2v$; bottom (contact point) has zero velocity.

Kinetic energy of a rolling body:

$K = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I_{\text{CM}}\omega^2 = \tfrac{1}{2}Mv^2\left(1 + \frac{I_{\text{CM}}}{MR^2}\right) = \tfrac{1}{2}Mv^2\left(1 + \frac{k^2}{R^2}\right)$

where $k$ is the radius of gyration about the CM.

Rolling on an incline (without slipping). A body of mass $M$, radius $R$, with $I_{\text{CM}} = \beta MR^2$ rolls down an incline of angle $\theta$.

$a = \frac{g\sin\theta}{1 + \beta}$

The acceleration is less than $g\sin\theta$ — some of the gravitational PE goes into rotational KE.

Common values of $\beta$: solid sphere $2/5$, hollow sphere $2/3$, disc $1/2$, ring $1$.

Derivation

Acceleration on incline. Forces along the incline: $Mg\sin\theta - f = Ma$, where $f$ is static friction (up the slope). Torque about CM: $fR = I_{\text{CM}}\alpha$. For rolling: $a = R\alpha$, so $\alpha = a/R$.

$fR = I_{\text{CM}}\frac{a}{R} \Rightarrow f = \frac{I_{\text{CM}} a}{R^2} = \beta M a$

Substituting in the translational equation:

$Mg\sin\theta - \beta Ma = Ma$

$g\sin\theta = a(1 + \beta) \Rightarrow a = \frac{g\sin\theta}{1+\beta}$

Speed at bottom of incline (height $h$). Using energy conservation,

$Mgh = \tfrac{1}{2}Mv^2(1+\beta)$

$v = \sqrt{\frac{2gh}{1+\beta}}$

Lower $\beta$ means higher speed at the bottom — solid sphere rolls faster than a ring or hollow sphere.

Worked Example

A solid sphere ($\beta = 2/5$) and a ring ($\beta = 1$) are released from the top of an incline of height 2 m, angle 30°. Find their speeds at the bottom, and which arrives first.

Solution:

Solid sphere: $v_{\text{sphere}} = \sqrt{2gh/(1+2/5)} = \sqrt{2 \cdot 10 \cdot 2 / 1.4} = \sqrt{40/1.4} = \sqrt{28.57} \approx 5.35 \, \text{m/s}$.

Ring: $v_{\text{ring}} = \sqrt{40/2} = \sqrt{20} \approx 4.47 \, \text{m/s}$.

Acceleration: sphere has $a = g\sin\theta/1.4 = 5/1.4 \approx 3.57$ m/s²; ring has $a = 5/2 = 2.5$ m/s². The sphere reaches the bottom first.

Common Confusions

• "Friction always opposes motion." Friction here acts up the incline (opposing the tendency of the contact point to slip backward) — but does no negative work because contact point is at rest in pure rolling.
• Static friction in rolling does no work. Since the contact point's velocity is zero, $W = \int \vec{F}\cdot d\vec{r}_{\text{contact}} = 0$. Energy is fully conserved in pure rolling.
• Heavier or larger ≠ faster. Acceleration of rolling depends only on $\beta = I/MR^2$, not on $M$ or $R$ individually.
• Top of wheel has speed $2v$, not $v$. That's why spokes appear to "bend" when a wheel is photographed.

Key Takeaways

• Pure rolling: $v = R\omega$ (no slipping).
• KE of rolling: $\tfrac{1}{2}Mv^2 (1 + k^2/R^2)$.
• Down an incline: $a = g\sin\theta / (1 + \beta)$ with $\beta = I/MR^2$.
• Lower $\beta$ ⇒ faster down the incline. Solid sphere beats ring.
• Static friction acts up the incline but does no work in pure rolling.