Nonuniform circular motion due to gravity.

[3dot]

The question I encountered asked at what angle would a ball rolling off a cylindrical incline leave the surface of the incline. This problem i solved. But it got me thinking, how would I describe the motion of that ball in terms of time?
To generalize the problem I started like this. A ball starts rolling from some point on a circular incline with some initial speed.
We draw 2 lines. A line through a point on the circular incline and the circle's center. And a vertical line through the center (parallel to the direction of gravity). Let the angle between these lines be θ. Then the tangential acceleration, which changes speed is:

dv/dt=gsinθ
(dv/ds)(ds/dt)=gsinθ (s being the distance traveled,R-radius, s=Rθ, ds=Rdθ)
vdv=gRsinθdθ
v²-v₀²=2gR(cosθ₀-cosθ)
ds/dt=±√(2gR)×√(cosθ₀+v₀²/(2gR)-cosθ)

Here you dt=$$\frac{a}{\sqrt{b-cos\theta}}$$dθ where both a and b are constants, integrating this is where I get stuck. replacing tan(theta/2)=x and using the identity cos theta = 1-tan^2(x/2)/sec^2(x/2) does seem to help until you need to integrate 1 over square root of a quartic.
Can someone offer help with this integral?

 

[Gib Z]

Hello 3dot, and welcome to Physicsforums!

Unfortunately, the solution to that integral is somewhat complicated and can not be solved in terms of elementary functions. The solution, if you want it, is

$$t = \frac{2a\sqrt{\frac{b - \cos x}{b-1}} \cdot F\left[\frac{x}{2}, \frac{-2}{(b-1)}\right]}{ \sqrt{b - \cos x}} + C$$

where F(x|m) is the elliptic integral of the first kind, and C is an arbitrary constant.

You can find some information of That Elliptic Integral here:

http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html

Hopefully that helps.

 

[3dot]

Thanks, I now know what I should read about.