Equation to Rolling sphere/ball

[Mickeysofine1972]

Hey Guys

I've been looking all over for a good way to work out how much a ball or sphere rotates when rolling any ideas?

I know that the circumference of a ball is:

2*PI*Radius

and that PI == 180degrees

but would the value in degrees relate to the units travelled when a sphere rolls?

i think i'm on the right track but can find any further maths info HELP!

Mike

[hyperair]

Supposing the ball rolls in a straight line, then the number of turns it makes, multiplied by the circumference equals to the distance it travels. But if it doesn't go in a straight line... then I'm not so sure.

[cb951303]

it's been a while since I took a dynamics lesson but if I remember correctly

angular speed (w) = 2 * PI * frequency (f) [rd/s]
so lets say if the ball rolls for t seconds then the rotating it does = w * t [rd]

[WW]

If a ball with radius R rolls in a straight line for distance D, then the total rotation is D/R radians, or (180/Pi)*D/R degrees.

[Lux Perpetua]

If anybody is interested in the case of a path that is not necessarily straight, then I believe the net rotation matrix R(s) of the sphere after rolling for a total distance of s can be found as the solution of the following differential equation: R'(s) = (1/r)A(s)R(s), R(0) = I (the identity matrix).r is the radius of the sphere, and A(s) is the matrix
[ 0 0 x'(s) ]
[ 0 0 y'(s) ]
[-x'(s) -y'(s) 0 ],where x(s) and y(s) are the coordinates of the path at length s. (Note: the parameter s doesn't actually need to be the distance traveled; you can reparametrize if convenient.)

If the path is a straight line, then A(s) is a constant matrix, and the differential equation can be solved exactly, giving you one version of the formulas of the previous posters. In general, I don't know if there is an explicit solution to that equation, but you could still approximate it.

I also tried to consider the generalization where the sphere rolls on a surface that isn't necessarily flat, but my solution to that turned out to be incorrect, so I'll leave that be.