/sci/ - Science & Math

>>16185206
If I'm not mistaken, this mostly reduces to a pendulum starting out inverted. If [math]\theta(t)[/math] is the angular distance between the bottom of the ball and the box corners (with [math]\theta(-\infty) = 0[/math]), we get
[math]
\dot{\theta}^2 = \frac{2g}{r}(1 - \cos \theta).
[/math]
This holds as long as the velocity of a box (which is [math]v = \partial_t(r \sin \theta)[/math]) is increasing: for [math]\dot{v} > 0[/math], the ball is pushing the boxes, [math]\dot{v} < 0[/math] would correspond to the ball pulling the boxes closer together. The separation point [math]\theta_1[/math] is thus at [math]\dot{v} = 0[/math], which you can solve for
[math]
\cos \theta_1 = \frac{2}{3}.
[/math]
Then you calculate the energy of the ball at separation time (after which it is in free fall), and eventually get a final velocity of
[math]
\frac{10}{3\sqrt{3}} \sqrt{gr}.
[/math]

Or something like that, I haven't really double-checked anything.