In preparation for an exam, I'm revisiting old exam questions. This one seems neat, but also quite complicated:
A soccer ball with Radius $R=11cm$ is inflated at a pressure of $P =9 \times 10^4 Pa$, then dropped from a height of $0.1m$ (distance floor to lowest part of ball) onto a hard floor and bounces off elastically.
Question: Find approximate expressions for:
- Surface area of ball in contact with floor
- Amount of time the ball is in contact with the floor
- Peak force exerted on the floor
if the mass is $0.42 kg$.
My attempt at a solution: Assume the ball is filled with an ideal gas and that the process is adiabatic. Assume that the deformation leads to a simple spherical segment, i.e. a ball where one part is cut off flat. This gives an expression of the volume $V$ in terms of the height of the "center" of the ball, $h$ as $$V = \frac{11}{6}\pi R^3 + \frac{\pi}{6}h^3$$ The surface area then is simply $$A = \pi (R^2 - h^2)$$.
Next: The ball has potential energy $mgh_0$, which is completely converted to internal energy at the peak point of the motion. The internal energy of an ideal gas is $$U = 3/2 N kT = 3/2 PV = 3/2 P_0 V_0^\gamma V^{1-\gamma}$$ where $\gamma$ is the adiabatic coefficient (1.4 for air).
The change in $U$ due to a changing $V$ then comes completely from the initial potential energy $E$. Some algebra and some sensible binomic approximations then yield a simple expression for the surface area of $$A = \frac{32 mgh}{9 V_0 P_0 (\gamma - 1)}$$.
Using $F = PA$ then allows me to calculate the peak force.
But what about the contact time? My initial guess was to approximate $F(t)$ as a triangle curve that goes from 0 to $F_{max}$ and back to $0$ and then use that $F$ equals change in momentum over time, $dp = Fdt$, which in this simple case would mean the total change in momentum, $\delta p$, would be equal to $1/2 F_{max} \delta t$. I can calculate the initial momentum from the initial potential energy, and since the process is elastic, the change in momentum is (minus) two times that value. Then I know everything to calculate $\delta t$.
I am, however, unsure about whether I can apply this law about momentum at all, since I am not talking about a simple point of mass here, rather about a bunch of gas molecules confined to a certain volume. This is also why my standard approach to mechanics questions, i.e. Lagrangian mechanics, doesn't seem to work: A simple coordinate describing the entire process would be $h$, but what is the kinetic term in terms of $h$?
EDIT I just realized that my formula for the spherical cap is wrong, and a bit more complicated than I wanted: If $h$ is the distance from the center of the sphere to the base of the cap, the volume becomes $$V = \frac{2}{3} R^3 - hR^2 + \frac{h^3}{3}$$
If I keep the pressure constant, the work needed for a volume change is $P \Delta V$, so we can equate: $$\Delta V = E_0 / P$$ where $E_0 = mgh$ is the initial potential energy of the ball.
The only problem then is that I have a third-order equation for $h$ which, I feel, is too complicated to solve. But let's see... let $d = R - h$, then we get $V == Rd^2 - d^3/3$ and now we assume that $d \ll R$ so that we have $V \approx Rd^2$.
Next, we need the surface area, $A = \pi (R^2 - h^2) = \pi (R^2 - (R-d)^2) \approx 2\pi R d$ where we again use $R \gg d$.
Plugging in some numbers gives $A \approx 44 cm^2$ which amounts to a radius of the spherical cap of approx $3.7cm$.
The peak force $F_{max}$ is just $AP \approx 401 N$.
Assuming that the force grows linearly with time until $F_{max}$ is reached and then drops linearly to $0$, the total change in momentum over the contact time $\Delta t$ is $\Delta p = 1/2 F_{max} \delta T$. The change in momentum is two times the initial momentum, so it is $$\Delta p = 2\sqrt{2mE_0} = 1/2 * 2\pi \sqrt{E_0 R P} \Delta t,$$ which we can solve for $\Delta t$ to obtain $\Delta t \approx 9 ms$.
That sounds reasonable to me.
