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.
Best Answer
The problem gets very simple if you consider only small deformations $d = R - h$ and calculate the force $F=PA$ to first order in $d$. To this order, $A = 2 \pi R d$ and $F = (P_0 2 \pi R) d$, where $P_0$ is the pressure inside the undisturbed ball (note that changes in pressure arising from volume changes only contribute terms of higher order in $d$).
In this approximation the force on the ball is essentially an elastic restoring force $F = - k d$ , with 'spring constant' $ k = 2 \pi R P_0 \approx 6 \times 10^4 N/m $.
The contact time is, then, one-half the oscillator period: $t_{contact} = \pi \sqrt{m/k} \approx 8\, ms$ (a very reasonable result). Note that, because of linearity, the contact time is independent of the velocity of the ball as it hits the ground.
From conservation of energy, the maximum deformation of a ball dropped from a height $H$ is $ d_{max} = \sqrt{2 m g H/k}$. The peak force is $ F_{max} = k d_{max} = \sqrt{2 m g k H}$.