Ball rolling along a see-saw Suppose we have a see-saw which is 2 metres long, whose mid-point is connected to a fulcrum which is 0.5 metres tall. The see-saw has a mass of 1kg, and has uniform thickness and density. The see-saw starts with one side up, and hence has a slope of 30 degrees (or pi/6 radians).
You place a ball of mass $m$ on top of the see-saw. It starts rolling down the see-saw, but as it rolls, it starts pushing that side of the see-saw down. 
 
If the ball is too light, the see-saw will not swivel much and the ball will roll off the right hand side of the see-saw. If the ball is too heavy, the ball won't roll much, and instead will push the see-saw down quickly, and will end up rolling off the left hand side of the see saw. The question is: what must the mass of the ball be such that it gets stuck in a dynamic equilibrium.
That is, for the first phase (up until the ball reaches the fulcrum) the ball's gravitational potential energy is converted into kinetic energy both of the ball, and of the see-saw. When the ball is at the fulcrum, the see-saw is perfectly flat and the ball and see-saw have both reached their maximum velocity. In the second phase, the ball starts rolling up the right hand side of the see-saw.  The right side of the see saw is still swivelling up, although it's slowing down as the ball presses on it. The ball reaches the top of the see-saw, and then starts rolling down again; repeating the process.
NB Assume no friction or air resistance. Model the see-saw as being arbitrarily thin. Neglect the rotational energy of the ball (ie assume the ball is a point mass).
I have not attempted to solve this problem analytically (this would require differential equations, I'd imagine). I have used a spreadsheet model, though, to determine that the mass of the ball should be about 0.13 kg, in which case it reaches the fulcrum at around t=0.7 seconds (this was with a model in which time has a 'granularity' of 0.01 seconds).
Bonus points: Determine the curve that the ball traces. I think it should be a parabola.
 A: HINT
The system has two degree of freedom


*

*$\theta$ rotation of the see-saw, positive clockwise

*$s$ position of the ball along the see-saw, positive toward the right with origin at the fulcrum


then the equatons of the motion are


*

*$I\ddot \theta=ms\cos \theta$ with $I=I_0+ms^2$ rotational inertia of the system with respect to the fulcrum

*$m\ddot s=mg\sin \theta\implies \ddot s=g\sin \theta$


to be solved according to the initial conditions


*

*$s(0)=-1$

*$\dot s(0)=0$

*$\theta(0)=\frac{\pi}6$

*$\dot \theta(0)=0$


If we assume small rotation for the see-saw we have


*

*$I\ddot \theta=ms$ 

*$\ddot s=g\theta$
A: With
$m_{b}$ Ball mass
$J_{b}$ Ball inertia moment
$J_{s}$ See-saw inertia moment
$r_{b}$ Ball radius
$\theta$ See-saw angle
$\rho$ Ball distance from the fulcrum.
and using Lagrangians the modeling is straigthforward. The kinetic
energy can be represented as
$$
T=\frac{1}{2}J_{b}\left(\frac{\dot{\rho}}{r_{b}}\right)^{2}+\frac{1}{2}J_{s}\dot{\theta}^{2}+\frac{1}{2}m_{b}\left(\rho\dot{\theta}\right)^{2}+\frac{1}{2}m_{b}\dot{\rho}^{2}
$$
and the potential energy as
$$
V=-m_{b}g\rho\sin\theta
$$
the Lagrangian reads
$$
L=T-V
$$
so the Euler-Lagrange equations are obtained from
$$
\frac{\partial L}{\partial\phi}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{\phi}}\right)=0,\;\phi=\left\{ \theta,\rho\right\} 
$$
giving
$$
\left\{ \begin{array}{rcl}
\ddot{\theta} & = & \frac{m_{b}(g\cos\theta\rho-2\rho\dot{\rho}\dot{\theta)}}{J_{s}+m_{b}\rho^{2}}\\
\ddot{\rho} & = & \frac{m_{b}r_{b}^{2}(\rho\dot{\theta}^{2}+g\sin\theta)}{J_{b}+mr_{b}^{2}}
\end{array}\right. 
$$
This is an inherently unstable system so no dynamic equilibrium is
possible without external inputs.
