How to calculate/use horizontal coefficient of restitution value for basketball?

Question

I'm trying to run an experiment in which a basketball is dropped straight down with a certain angular speed and then undergoes a bounce. My goal is to create a relationship between the initial angular speed of the ball and its horizontal displacement between its first two bounces. After some research, here's what I've found:

$$v_{x_{2}}=\frac{(1-\alpha e_x)v_{x_{1}}+\alpha(1+e_x)R\omega_1}{1+\alpha}$$ Of course, if the ball is dropped straight down, the initial horizontal velocity should be zero, meaning one can drop the first term. All the same, I have two questions.

(1) Would it be incorrect to split $v_{x2}$ into $(t)(s_x)$ where t is the duration of time between the first and second bounce and $s_x$ is the horizontal displacement? For some reason this feels wrong, but I'm not sure.

(2) What do I do with horizontal coefficient of restitution ($e_x$)? How could I calculate/approximate this value? Would it be wrong, for instance, to assume energy conservation and just give it the value of 1?

Thanks for any and all suggestions and help!

Answer 1

I will take a stab at this with a few assumptions. I will assume that on the first bounce when the basketball comes in contact with the floor, it will continue spinning for a time. This will create a frictional force between the ball and the floor. This force will product a torque slowing the spin and will also linearly accelerate the ball in the horizontal direction.

I am also going to assume the ball remains in contact with the floor long enough to achieve pure rolling before leaving the floor. I believe this to be reasonable for a moderately rough floor or cement, but may not work for a very slippery floor.

Take the basketball to be a hollow sphere. The moment of inertia about a diameter is:

$I=\frac{2}{3} m R^2$

Call the friction force $f$, the equation for the angular velocity $\omega$ becomes:

$I\ \omega '(t)=-f\ R$

and if we take the initial angular velocity as $\omega 0$ we get

$\omega=\omega 0-\frac{3 ft}{2 m R}$

The equation for the horizontal velocity $v_x$ is:

$v_x'(t)=\frac{f}{m}$

For an initial horizontal velocity of $0$ we get

$v_x=\frac{ft}{m}$

The condition for pure rolling is:

$v_x=R\ \omega$

from which we can solve for $f t$.

$ft=\frac{2 m R \omega 0}{5}$

which gives the horizontal velocity after the first bounce as:

$v_x=\frac{2 R \omega 0}{5}$

You can also compute $\omega$ if you want. After the first bounce the ball's $v_x$ and $\omega$ match each other for pure rolling, so those values should remain constant, barring other external forces.

Vertical velocities at each bounce from the coefficient of restitution and their subsequent times of flight should be straightforward.

If contact is not long enough to achieve pure rolling, then more information concerning coefficient of friction and time of contact is necessary to work the problem by this method.