# Rolling motion

A special case of rigid body motion is rolling without slipping on a stationary ground surface. This is defined by motion where the point of contact with the ground has zero velocity, so it matches the ground velocity and is not slipping.

Rolling without slipping on stationary ground surfaces.

$\text{Contact point P has zero velocity:} \quad \vec{v}_P = 0$

Rolling without slipping means by definition that the contacting points have the same velocity. For a stationary ground the velocity is zero, so the contact point on the body must also have zero velocity.

It is helpful to think about the motion of the body in two ways:

1. The body rotates about the moving center $C$.
2. The body rotates about the instantaneous center at the contact point $P$.

These two ways of visualizing the motion can be seen on the figure below.

 Movement: flat big flat rock inside 2-circle inside 3-circle inside rock outside 1-circle outside 2-circle outside rock

Center $C$ Body $P$ velocity $P$ acceleration
none none
$\vec{v}_C$ (trans.) $\vec{a}_C$ (trans.)
$\vec{\omega} \times \vec{r}_{CP}$ (rot.) $\vec{\alpha} \times \vec{r}_{CP}$ (ang.)
$\vec{\omega} \times (\vec{\omega} \times \vec{r}_{CP})$ (cent.)
$\vec{v}_P$ (total) $\vec{a}_P$ (total)

Velocity and acceleration of points on a rigid body undergoing different rolling motions.

## Rolling on a 2D flat surface

The most common and also the simplest form of rolling occurs on a flat surface.

Geometry and variables for rolling without slipping on a flat surface.

While rolling, the velocity and acceleration are directly connected to the angular velocity and angular acceleration, as shown by the next equations.

Center velocity and acceleration while rolling on a flat surface.

\begin{aligned} \vec{v}_C &= r \omega \,\hat{e}_t \\ \vec{a}_C &= r \alpha \,\hat{e}_t \end{aligned}

We begin by observing that the sign conventions in Figure #rko-ff mean that $\vec\omega = -\omega\,\hat{e}_b$. Now rolling without slipping means the contact point $A$ must instantaneously have zero velocity, so using #rkg-er gives: \begin{aligned} \vec{v}_C &= \vec{v}_A + \vec{\omega} \times \vec{r}_{AC} \\ &= (-\omega \,\hat{e}_b) \times r \,\hat{e}_n \\ &= r \omega \,\hat{e}_t. \end{aligned} Because the surface is flat, the tangential basis vector $\hat{e}_t$ is constant, and the radius $r$ is also constant. Differentiating the velocity expression thus results in: \begin{aligned} \vec{v}_C &= r \omega \,\hat{e}_t \\ \dot{\vec{v}}_C &= r \dot\omega \,\hat{e}_t \\ \vec{a}_C &= r \alpha \,\hat{e}_t. \end{aligned}

Another way to express the connection between angular and linear velocity and acceleration is via the distance $s$ traveled by the rolling body:

Distance-angular relationships for rolling on a flat surface.

\begin{aligned} s &= r \theta \\ \dot{s} &= r \omega \\ \ddot{s} &= r \alpha \end{aligned}

The contact point $P$ and center $C$ are offset by the constant vector $r\,\hat{e}_n$, so $\dot{s} = v_P = v_C = r\omega$, from #rko-ef. Because $r$ is constant, differentiating the velocity expression gives $\ddot{s} = r \dot\omega = r \alpha$, while integrating with zero initial displacement gives $s = r \theta$.

While rolling, the contact point will have zero velocity but will have a centripetal acceleration towards the rolling center:

Contact point $P$ velocity and acceleration while rolling on a flat surface.

\begin{aligned} \vec{v}_P &= 0 \\ \vec{a}_P &= \omega^2 \,\vec{r}_{PC} \end{aligned}

By definition of non-slip rolling contact, the point of contact $P$ has zero velocity. The acceleration can be computed from the center $C$ with #rkg-e2: \begin{aligned} \vec{a}_P &= \vec{a}_C + \vec{\alpha} \times \vec{r}_{CP} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{CP}) \\ &= \alpha r \,\hat{e}_t + (-\alpha\,\hat{e}_b) \times (-r\,\hat{e}_n) + (-\omega\,\hat{e}_b) \times \Big((-\omega\,\hat{e}_b) \times (-r\,\hat{e}_n)\Big) \\ &= \alpha r \,\hat{e}_t - \alpha r \,\hat{e}_t + \omega^2 r \,\hat{e}_n \\ &= \omega^2 \,\vec{r}_{PC}. \end{aligned}

## Rolling on a 2D curved surface

When a circular rigid body rolls without slipping on a surface which is itself curved, the radius of curvature of the surface affects the acceleration (but not velocity) of points on the rolling body.

Geometry and variables for rolling without slipping on a curved surface. The left diagram shows rolling on the inside of the curve, while the right diagram is rolling on the outside of the curve.

Because there are two different geometries for rolling on a curved surface (inside and outside), there are different sign conventions and variable definitions in the two cases, as listed below.

Geometric quantities for rolling on a curved surface.

\begin{aligned} \left.\begin{aligned} R &= \rho - r \\ \vec{\omega} &= -\omega \,\hat{e}_b \\ \vec{\alpha} &= -\alpha \,\hat{e}_b \end{aligned}\right\} & \text{ when rolling on the inside of a curved surface} \\[1em] \left.\begin{aligned} R &= \rho + r \\ \vec{\omega} &= \omega \,\hat{e}_b \\ \vec{\alpha} &= \alpha \,\hat{e}_b \end{aligned}\right\} & \text{ when rolling on the outside of a curved surface} \end{aligned}

These formulas are all choices of sign conventions for $\omega$ and $\alpha$, and definitions of $R$. Figure #rko-fc shows the appropriate geometry. Note that $\omega$ and $\alpha$ are defined with positive values corresponding to motion in the tangential direction.

Warning: Radii of curvature $\rho$ and $R$ may not be constant.

The radius of curvature $\rho$ of the surface may be varying with position as the body rolls. If $\rho$ changes then this will also cause $R$ to change. These two variables will only be constant if the surface is in fact perfectly circular.

The center velocity for rolling on a curved surface is the same as for a flat surface, while the acceleration also depends on the radius of curvature of the surface:

Center velocity and acceleration while rolling on a curved surface.

\begin{aligned} \vec{v}_C &= r \omega \,\hat{e}_t \\ \vec{a}_C &= r \alpha \,\hat{e}_t + \frac{(r\omega)^2}{R} \,\hat{e}_n \end{aligned}

The two configurations shown in Figure #rko-eg must be considered individually. We will do the derivation for the case of rolling inside (the left diagram), as the rolling-outside case is very similar, but with different signs.

We begin with the point of contact $P$, which has zero velocity and so it is the instantaneous center of the body. Using #rkg-er with the expressions #rko-eg thus gives the velocity expression: \begin{aligned} \vec{v}_C &= \vec{v}_P + \vec{\omega} \times \vec{r}_{PC} \\ &= (-\omega\,\hat{e}_b) \times r\,\hat{e}_n \\ &= r \omega \,\hat{e}_t. \end{aligned} Differentiating this velocity and using #rkt-ed gives the acceleration: \begin{aligned} \vec{a}_C &= r \dot{\omega} \,\hat{e}_t + r \omega \,\dot{\hat{e}}_t \\ &= r \alpha \,\hat{e}_t + r \omega \frac{\dot{s}}{\rho} \,\hat{e}_n. \end{aligned} To replace $\dot{s}$ in this acceleration expression, we first consider the instantaneous center $M$, which moves along the surface with speed $\dot{s}$ and tangential acceleration $\ddot{s}$. Thus from #rkt-ev: \begin{aligned} \vec{v}_M &= \dot{s} \,\hat{e}_t. \end{aligned} We can now compute the velocity of $C$ from $M$, as follows. Note that we cannot use equation #rkg-er because $M$ and $C$ are not fixed to the same body ($M$ is not fixed to the body). Thus: \begin{aligned} \vec{r}_C &= \vec{r}_M + r\,\hat{e}_n \\ \vec{v}_C &= \vec{v}_M + r\,\dot{\hat{e}}_n \\ &= \dot{s}\,\hat{e}_t - r \frac{\dot{s}}{\rho} \,\dot{\hat{e}}_t \\ &= \frac{(\rho - r)}{\rho} \dot{s} \,\hat{e}_t \\ &= \frac{R}{\rho} \dot{s} \,\hat{e}_t. \end{aligned} Comparing this expression for $\vec{v}_C$ to the one above, we see that: \begin{aligned} r \omega &= \frac{R}{\rho} \dot{s} \\ \dot{s} &= \frac{\rho r}{R} \omega. \end{aligned} This expression for $\dot{s}$ can be substituted back into our earlier equation for $\vec{a}_C$, giving: \begin{aligned} \vec{a}_C &= r \alpha \,\hat{e}_t + r \omega \frac{1}{\rho} \left( \frac{\rho r}{R} \omega \right) \,\hat{e}_n \\ &= r \alpha \,\hat{e}_t + \frac{(r \omega)^2}{R} \,\hat{e}_n. \end{aligned} The case of rolling on the outside of the curved surface (the right diagram in Figure #rko-fc) is similar to the above but with the different sign conventions from #rko-eg. The final expressions in #rko-ec apply to rolling on both the inside and outside.

The angular velocity of the rolling rigid body can be related to the derivative $\dot{s}$ of the distance traveled on the surface as follows. In the special case of a perfectly circular surface, the angular acceleration also has a simple relationship with $\ddot{s}$.

Distance-angular relationships for rolling on a curved surface.

\begin{aligned} \omega &= \frac{R}{\rho r} \dot{s} \quad \text{on any curved surface} \\ \alpha &= \frac{R}{\rho r} \ddot{s} \quad \text{on a circular surface} \end{aligned}

See the derivation for #rko-ec, where the relationship between $\omega$ and $\dot{s}$ is obtained. If the surface is exactly circular then by definition $\rho$ and $R$ are constant, so differentiating the angular velocity expression immediately gives the angular acceleration equation.

Just as with rolling on a flat surface, when a rigid body rolls on a curved surface the contact point has zero velocity but nonzero acceleration, as shown below.

Contact point $P$ velocity and acceleration while rolling on a curved surface.

\begin{aligned} \vec{v}_P &= 0 \\ \vec{a}_P &= \frac{\rho}{R} \omega^2 \,\vec{r}_{PC} \end{aligned}

The derivations for rolling on the inside or outside of a surface are different. Here we show the rolling inside case, as the outside case is very similar.

By definition the contact point $P$ has zero velocity. The acceleration can be computed from the center $C$: \begin{aligned} \vec{a}_P &= \vec{a}_C + \vec{\alpha} \times \vec{r}_{CP} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{CP}) \\ &= r \alpha \,\hat{e}_t + \frac{(r\omega)^2}{R} \,\hat{e}_n + (-\alpha \,\hat{e}_b) \times (-r\,\hat{e}_n) + (-\omega \,\hat{e}_b) \times \Big( (-\omega \,\hat{e}_b) \times (-r\,\hat{e}_n) \Big) \\ &= r \alpha \,\hat{e}_t + \frac{(r\omega)^2}{R} \,\hat{e}_n - r \alpha \,\hat{e}_t + r \omega^2 \,\hat{e}_n \\ &= \frac{(r + R)}{R} r \omega^2 \,\hat{e}_n \\ &= \frac{\rho}{R} \omega^2 \,\vec{r}_{PC}. \end{aligned}