30.6: Ball Rolling on Inclined Rotating Plane

Figure $\PageIndex{1}$

We’ll take unit vectors $\begin{equation}\widehat{\vec{z}} \text { pointing vertically up, } \widehat{\vec{i}}\end{equation}$ perpendicularly up from the plane, the angle between these two unit vectors being α. (We will need a set of orthogonal unit vectors $\widehat{\vec{i}}, \widehat{\vec{j}}, \widehat{\vec{k}}$, not fixed in the plane, but appropriately oriented, with $\widehat{\vec{k}}$ horizontal.) The vector to the center of the sphere (radius $a$, mass $m$ ) from an origin on the axis of rotation, at a point a above the plane, is $\vec{r}$. The contact reaction force of the plane on the sphere is $\vec{R}$.

The equations of motion are:

$$\begin{equation} m \ddot{\vec{r}}=\vec{R}-m g \widehat{\vec{z}}, \quad I \dot{\vec{\Omega}}=-a \widehat{\vec{i}} \times \vec{R} \end{equation}$$

and the equation of rolling contact is $\dot{\vec{r}} - a \vec{\Omega} \times \widehat{\vec{i}}=\widehat{\omega i} \times \vec{r}$.

First, we eliminate $\vec{R}$ from the equations of motion to give

$$\begin{equation} \dot{\vec{\Omega}}=(a m / I)(\ddot{\vec{r}}+g \widehat{\vec{z}}) \times \widehat{\vec{i}} \end{equation}$$

Note that $\dot{\vec{\Omega}} \cdot \widehat{\vec{i}}=0$, so the spin in the direction normal to the plane is constant, $\vec{\Omega} \cdot \widehat{\vec{i}}=n$ say. (Both forces on the sphere have zero torque about this axis.)

Integrating,

$$\begin{equation} \vec{\Omega}+\text { const. }=(m a / I)(\dot{\vec{r}}+g t\widehat{\vec{z}}) \times \widehat{\vec{i}} \end{equation}$$

Now eliminate $\vec{\Omega}$ by multiplying both sides by $\times \vec{i}$ and using the equation of rolling contact

$$\begin{equation} \dot{\vec{r}}-a \vec{\Omega} \times \widehat{\vec{i}}=\omega \widehat{\vec{i}} \times \vec{r} \end{equation}$$

to find:

$$\begin{equation} \left(m a^{2} / I\right)[(\dot{\vec{r}}+g t \widehat{\vec{z}}) \times \widehat{\vec{i}}] \times \widehat{\vec{i}}=a \vec{\Omega} \times \widehat{\vec{i}}+\text { const. }=\dot{\vec{r}}-\omega \widehat{\vec{i}} \times \vec{r}+\text { const. } \end{equation}$$

then using $\begin{equation} (\dot{\vec{r}} \times \widehat{\vec{i}}) \times \widehat{\vec{i}}=-\dot{\vec{r}}, \; (\widehat{\vec{z}} \times \widehat{\vec{i}}) \times \widehat{\vec{i}}=-\widehat{\vec{j}} \end{equation}$, we find

$$\begin{equation} \dot{\vec{r}}\left(1+m a^{2} / I\right)+\left(m a^{2} / I\right) g t \widehat{\vec{j}} \sin \alpha+\text { const. }=\omega \widehat{vec{i}} \times \vec{r} \end{equation}$$

The constant is fixed by the initial position $\vec{r}_{0}$, giving finally

$$\begin{equation} \dot{\vec{r}}=\dfrac{\omega}{1+m a^{2} / I} \widehat{\vec{i}} \times\left[\left(\vec{r}-\vec{r}_{0}\right)+\dfrac{m a^{2} / I}{\omega} g t \widehat{\vec{k}} \sin \alpha\right] \end{equation}$$

The first term in the square brackets would give the same circular motion we found for the horizontal rotating plane, the second term adds a steady motion of the center of this circle, in a horizontal direction (not down the plane!) at constant speed $\left(m a^{2} / I \omega\right) g \sin \alpha$.

(This is identical to the motion of a charged particle in crossed electric and magnetic fields.)

Bottom line: the intuitive notion that a ball rolling on a rotating inclined turntable would tend to roll downhill is wrong! Recall that for a particle circling in a magnetic field, if an electric field is added perpendicular to the magnetic field, the particle moves in a cycloid at the same average electrical potential—it has no net movement in the direction of the electric field , only perpendicular to it. Our rolling ball follows an identical cycloidal path—keeping the same average gravitational potential.