# Calculating the path of a ball with spin moving across a table

A ping pong ball is rolling over a smooth (but not frictionless) table. During its travel, a clockwise spin is placed on the ball. The ball's path is changed to move to the right (in perspective from it's initial path). If the friction of the table is a variable $f$ that can be altered, how does one calculate the altered path of the ball after spin is applied?

A single parameter is not going to be sufficient to capture the effect you seek. Think of a spinning disk, which will have an extended contact area in the shape of a circle. The motion of any point of the disk can be decomposed into a velocity $v_\parallel$ in the direction of the bulk motion, and an orthogonal velocity $v_\perp$. If friction were independent of speed, then $v_\parallel$ and $v_\perp$ will be reduced at the same rate everywhere, leading to a net slowing of both the bulk motion and the spin but not resulting in any redirection of the bulk motion.

The reason spinning objects' paths are redirected is that they experience asymmetric friction/drag forces as different parts move at different relative speeds in the fixed environment. In other words, $\dot{v}_\perp$ does not average out to $0$ over the contact area.

In order to lay the framework, suppose the disk has a bulk linear velocity $v_\mathrm{lin}$, an angular frequency $\omega$, and a direction of travel $\theta$ measured in the usual sense with respect to an $xy$-grid on the (presumably inertial) surface. Label points on the disk with a distance $r$ from the center and an angle $\phi$ measured with respect to the same $xy$-grid.

With these definitions it is easy to see (especially if you make a sketch) that \begin{align} v_\parallel & = v_\mathrm{lin} - r\omega \sin(\phi-\theta) \\ v_\perp & = r\omega \cos(\phi-\theta). \end{align} These components have directions \begin{align} \hat{v}_\parallel & = \cos(\theta) \hat{x} + \sin(\theta) \hat{y} \\ \hat{v}_\perp & = -\sin(\theta) \hat{x} + \cos(\theta) \hat{y}, \end{align} and so the overall velocity is $$\vec{v} = v_\parallel \hat{v}_\parallel + v_\perp \hat{v}_\perp.$$ Let $$v = \sqrt{v_\parallel^2 + v_\perp^2}$$ be the speed and define $\hat{v} = \vec{v} / \lvert \vec{v} \rvert = \vec{v}/v$.

The object is a rigid body, and so we can integrate forces over the whole contact area to find the net results. (Without rigidity, each point would have its own dynamics coupled to those of its neighbors, making for a much more complicated problem.) Let $f(v)$ be the magnitude of the friction force per unit area as a function of relative speed. The direction will simply be $-\hat{v}$, since friction can only work exactly opposite the direction of motion at a point. Then the total force on the disk $D$ is $$\vec{F} = - \int\limits_D f(v) \hat{v} \ \mathrm{d}A.$$ At the same time, there will be a torque per unit area given by $-f(v) \vec{r} \times \hat{v}$, where $\vec{r}$ is the vector with magnitude $r$ pointing in the direction $\phi$. The net torque on the disk will be $$\vec{\tau} = -\int\limits_D f(v) \vec{r} \times \hat{v} \ \mathrm{d}A.$$

Finally, we can connect the forces with the change in the variables of interest. The linear velocity $$\vec{v}_\mathrm{lin} = v_\mathrm{lin} \left(\cos(\theta) \hat{x} + \sin(\theta) \hat{y}\right)$$ will evolve according to $$\dot{\vec{v}}_\mathrm{lin} = \frac{1}{m} \vec{F},$$ where $m$ is the mass of the disk. The disk will also spin down according to $$\dot{\omega} = \frac{1}{I} \tau,$$ where $I$ is the moment of inertia and $\tau = \lvert \vec{\tau} \rvert$.

As you can see, this can be complicated in general. Even some rather simple forms for $f(v)$ lead to intractable equations if you want analytic solutions. My recommendation is to simply numerically evolve the system under the guidance of the equations given.

The fact that the original question was for a ball rather than a disk changes little. The ball must have a non-vanishing contact area if it is to curve at all. The only trick is to remember to use the proper formulas for $m$ and $I$, and possibly to introduce $r$-dependence into $f$ to account for unequal weight per unit area over the contact surface.