Bouncing in an Elevator

Ignoring the simulation entirely.

First lets observe that the elevator increasing in acceleration results in the same relative moment between the elevator and the ball as if gravity were increasing at a rate of $\epsilon$

Then let's look at the energy the ball has at any given point.

$\displaystyle E = KE + PE$

$\displaystyle E = \frac12\,m\,\dot h^2 + m\,(g+ \epsilon\,t)\,h$

Normally, the kinetic energy and potential energy changes cancel out, but in this case we have an additional term:

$\displaystyle \frac{dE}{dt} = \,m\,\dot h\,\ddot h + m\,(g+ \epsilon\,t)\,\dot h + m\, \epsilon \,h$

Since $\ddot h = -(g+\epsilon\,t)$ the first two terms cancel as usual and we get:

$\displaystyle \frac{dE}{dt} = m\, \epsilon \,h$

If we have the energy then we can find the peak height:

$\displaystyle h_{peak} = \frac{E}{m\,(g+\epsilon\,t)}$

Looking at the time average of a parabola, the average height is exactly $\frac23$ of the peak height. When looking at these cubic curves it's a little off but accurate to 1 ppm. Thus:

$\displaystyle h_{ave}=\frac23 h_{peak}$

Now if we use this $h_{ave}$ value in our energy change equation we can create a solvable differential equation:

$\displaystyle \frac{dE}{dt} = m\, \epsilon \,h_{ave}$

$\displaystyle \frac{dE}{dt} = m\, \epsilon \,\frac23 h_{peak}$

$\displaystyle \frac{dE}{dt} = \epsilon \,\frac23 \frac{E}{g+\epsilon\,t}$

$\displaystyle \int \frac1E dE = \int \frac23 \frac{\epsilon}{g+\epsilon\,t} dt$

$\displaystyle \ln(E) = \frac23 \, \ln(C (g+\epsilon\, t) )$

$\displaystyle E = C (g+\epsilon\, t)^{\frac23}$

We know our initial energy:

$\displaystyle E_0 = m\,g\,h_0$

So we can find out constant C:

$\displaystyle E = m\,g\,h_0 \ \left( \frac{ g+\epsilon\, t } {g} \right) ^{\frac23}$

Since we have our energy we can now find our peak height:

$\displaystyle h_{peak} = h_0\, \frac{g}{g+\epsilon\,t} \, \left( \frac{ g+\epsilon\, t } {g} \right) ^{\frac23}$

$\displaystyle h_{peak} = h_0\, \left( \frac{ g+\epsilon\, t } {g} \right) ^{-\frac13}$

Since we know $\epsilon \,t = \frac12 \, g$ we can now plug that in to get the final answer:

$\displaystyle h_{peak} = h_0\, \left( \frac32 \right) ^{-\frac13} = 0.87358 h_0$

Elevator floor position changes as $x_{elev}\left(t\right)=x_{e} + v_{e} \cdot t + a_{e}\cdot\frac{t^2}{2}+ \epsilon\cdot\frac{t^3}{6}$ ball position is $x_{ball}\left(t\right)=x_b + v_b \cdot t-g\cdot\frac{t^2}{2}$ ball height relative to the elevator floor is $x_{ball}\left(t\right) - x_{elev}\left(t\right) = v_{b}\cdot t-g\cdot\frac{t^2}{2} - v_{e}\cdot t- a_{e}\cdot\frac{t^2}{2} - \epsilon\cdot\frac{t^2}{6}$ next bounce occurs when height is zero $t\cdot\left(\left( v_{b} - v_{e} \right) - \frac{\left(g+a_{e}\right)}{2} \cdot t-\frac{\epsilon}{6} \cdot t^2 \right)=0$ or $t_\text{next bounce} = \frac{-\frac{\left(g+a_{e}\right)}{2} + \sqrt{\frac{\left(g+a_{e}\right)^2}{4} + 4\cdot\frac{\epsilon}{6}\cdot\left( v_{b} - v_{e} \right)}}{2\cdot\frac{\epsilon}{6}}$

so we need to complete update function:

#calculate time of next bounce
t_next = 3 / (2 * epsilon) * (-(g + state["aElev"]) + math.sqrt((g + state["aElev"])**2 + 8 * epsilon * (state["vBall"] - state["vElev"]) / 3)) 

#calculate velocity and acceleration before next bounce
vBall_New = state["vBall"] - g * t_next
vElev_New = state["vElev"] + state["aElev"] * t_next + 0.5 * epsilon * (t_next ** 2)
aElev_New = state["aElev"] + epsilon * t_next

and simulation outputs answer:

The max height is 0.873609

This problem has an elegant solution using the adiabatic invariant, which should work as long as $\epsilon$ is small. First, using conservation of energy, we find an equation for the relationship between the maximum displacement and maximum momentum.

$E = \frac{ p_{0}^2 }{ 2 m } = m g x_0$

$x_0 = \frac{ p_{0}^2 }{ 2 m^2 g }$ , $p_0 = \sqrt{ 2 m^2 g x_0 }$

Then we can use the adiabatic invariant (area enclosed in the phase diagram is constant under adiabatic changes) as a second relationship between these two quantities.

$A = \oint p\,dx$

$A \propto x_0 \cdot p_0$

Combining these we can eliminate momentum and see how maximum displacement is affected by g, Earth's gravitational constant.

$A \propto g^{1/2} x_0^{3/2}$

$x_0 \propto g^{-1/3}$

In this problem, increasing acceleration to $0.5$ g has the same effect as increasing g by a factor of 1.5 so

$g \mapsto ( \frac{3}{2} ) g$

$p_0 \mapsto (\frac{3}{2})^{1/3} p_0$

$x_0 \mapsto (\frac{2}{3})^{1/3} x_0 \approx 0.874 x_0$