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What conditions must be met for a ball to roll perfectly down an incline without slipping? A mathematically rigorous definition, please.

I honestly don't know where to begin with answering this problem.

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do a google rolling dynamics. You will get pretty comprehensive search results.. – Vineet Menon Feb 21 '12 at 4:55
Comment to the question (v1): Can one assume that the incline is a plane, or could it be a curved surface? In the latter case there would be additional restrictions to secure that the ball doesn't loose contact with the surface. – Qmechanic Feb 21 '12 at 11:07
That's... a very good point, Qmechanic. I don't think I'm told specifically that the incline is a plane. – badreferences Feb 21 '12 at 17:00
If you like this question you may also enjoy reading this question. – Qmechanic Mar 22 '12 at 12:29

The formula is $$\mu_s \geq \frac{g\tan\theta}{1+\frac{k^2}{r^2}}$$ where $\mu_s$ is static friction coefficient for the ball-incline interface. $\theta$ is the angle of the incline, and $k$ is the radius of gyration of the ball (for a solid uniform spherical ball, $k=R\sqrt{\frac{2}{5}}$). R is the radius of the ball. If you have a more complicated body, R will be the radius of the circular surface that is rolling (This comes into place if you have a spool rolling down an incline). This formula is only applicable when the center of mass of the body is at the center of the rolling circle.

Where did I get this formula? I first assumed the friction to be $f$. Now, I calculated the acceleration using Newton's laws, and I similarly calculated angular acceleration through torque. Using $a=\alpha r$, I got a value for $f$. Now I just had to set its upper bound, i.e $\mu_s N$ (N is normal reaction force, denoted by R by some).

If you do not understand the explanation, read up a bit on rolling dynamics as @Vineet suggested.

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I understand. Thank you very much; you make it seem obvious in retrospect. – badreferences Feb 21 '12 at 16:55

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