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30.6: Ball Rolling on Inclined Rotating Plane

  • Page ID
    30706
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    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 \(\vec{k}\) 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 \vec{r}=\vec{R}-m g \overrightarrow{\vec{z}}, \quad I \overrightarrow{\vec{\Omega}}=-a \overrightarrow{\vec{i}} \times \vec{R}
    \end{equation}

    and the equation of rolling contact is \(\vec{r}-a \vec{\Omega} \times \overrightarrow{\vec{i}}=\overrightarrow{\omega i} \times \vec{r}\).

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

    \begin{equation}
    \vec{\Omega}=(a m / I)(\vec{r}+g \vec{z}) \times \overrightarrow{\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, \(\vec{\Omega} \cdot \overrightarrow{\vec{i}}=n\)

    say. (Both forces on the sphere have zero torque about this axis.)

    Integrating,

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

    to find:

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

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

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

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

    \begin{equation}
    \dot{\vec{r}}=\frac{\omega}{1+m a^{2} / I} \overrightarrow{\vec{i}} \times\left[\left(\vec{r}-\vec{r}_{0}\right)+\frac{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.


    30.6: Ball Rolling on Inclined Rotating Plane is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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