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?(adsbygoogle = window.adsbygoogle || []).push({});

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

v^{2}-v_{0}^{2}=2gR(cosθ_{0}-cosθ)

ds/dt=±√(2gR)×√(cosθ_{0}+v_{0}^{2}/(2gR)-cosθ)

Here you dt=[tex]\frac{a}{\sqrt{b-cos\theta}}[/tex]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?

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# Nonuniform circular motion due to gravity.

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