Question

I am conducting an experiment to calculate the coefficient of rolling friction for two different objects (a handball and a tennis ball), and I need clarification as to what formula and information is required to calculate this value. I am thus far informed that it is a factor of speed, which ties in to the distance of deformation from the center of the ball (or wheel) to where the ball is no longer touching the surface (arc length). If there could be any clarifications to this, would be helpful. Thus far, I am laying the ball on a track, where the distance traveled is measured and and timed, a doggy door contraption is being used to apply force to the ball (so the force can be calculated if required to calculate coefficient of rolling friction), and the ball is on a track so it does not bob side to side. Masses, radius, arc length, and additional measurable components (with scale and ruler) are known. But what formula specifically and derivations can I use to calculate the coefficient of rolling friction? Use your own values for demonstration purposes (your own mass, arc length, radius, etc).

Answer 1

for a ball is a hollow spherer, the moment of inertia is given by
I = 2mr^2/3 [ where m is mass of ball, r is its radius]

now, for a ball rolling down an incline
let angle of incline be theta , and friction on the bottommost point of the ball be f
so, now from force balance

mg*sin(theta) - f = m*a [ where a is linear acceleration of the ball]
from moment balance
f*r = I*alpha ( where alpha is angular acceleration of the ball)

for pure rolling alpha = a/r
hence

f*r = I*a/r , f = I*a/r^2
hence

g*sin(theta) - 2a/3 = a
a = 3*g*sin(theta)/5

now for coefficient of friction
f = kN
N = mg*cos(theta)
hence
f = kmg*cos(theta)

so,
f = 2ma/3 = kmg*sin(theta)
hence
k = 2(3g*sin(theta))/5*3g*sin(theta)
k = 2/5

so we can find friction coefficient (dynamic) like this