| Groove Depths | 0 | .1 | .2 | .3 | .4 | .5 | .6 | .7 | .8 | .9 | 1.0 |
| Frictions | 0 | .01 | .02 | .03 | .05 | .07 | .09 | .12 | .15 | .18 | .21 |
To what is this friction due? Look at this diagram ([Fig. 2]), representing a transverse section of the groove and ball.
Fig. 2.
Transverse section groove and ball.
Is it not evident that the ball really rolls on two parallel lines in the groove somewhere between D and E, say the lines through c c perpendicular to the plane of the paper? This granted, it follows that points on the ball-surface touching the groove above c are going faster, while those touching below c are going slower than points touching at c. Hence, no wonder there is friction. The position of c c is such that the sum of the moments of friction above c c balances the sum of the moments of friction below c c. Take axes O X, O Y, as indicated; let the x of c c be a, and that of D D, b; put d s for an element of arc, and let A be the angle between the radius to d s and O Y. Then the friction on d s is proportional to d s cos A = d y, and its moment about c c is proportional to d y (x − a), or, d y (a − x), according as d s is above or below c c.
Therefore, ∫√1 − a2(x − a) dy0 = ∫√1 − b2 (a − x) dy√1 − a2
The ball’s radius being unity, the solution of the above equation is,—
a = 1⁄2(arc cos b√1 − b2 + b √1 − b2),
which determines a for all values of b; that is, determines the points c, c. It was stated above that d s was proportional to the friction upon itself. Of course, we meant that it was proportional so long as a remained constant. In terms of the unit given at the beginning of this discussion, the friction is ds2a √1 − a2, and the total friction upon the ball is therefore