## Determination of the Angular Velocity

When the abdominal wheel rotates on the horizontal plane, two points of the wheel have zero velocity: the point O representing one end of one of the holders and the point A on the outer edge of the disk that is in immediate contact with the floor. As both O and A belong to the same rigid object:

In the coordinate system E defined in Fig. 9.2, the displacement vector r can be written as r = (L2 + R2)(1/2) (cos 0 e1 + sin 0 e2) = (L2 + R2)(1/2) n in which the unit vector n is in the horizantal plane and in radial direction. Thus, we have

(«1 e1 + «2 e2 + «3 e3) X (cos 0 e1 + sin 0 e2) = 0

That means that either the angular velocity is zero or it is in the direction of n:

m = «o n where the scalar quantity «o is yet to be determined.

Let us next relate the velocity of center of mass (point C) to the velocity of point A and to the angular velocity:

Solving this equation for «o we find:

m = - [(vo/(R cos a)] (cos 0 e1 + sin 0 e2) «o = -[(»o/(R cos a)]

We could also express m in the coordinate system B that is associated with the abdominal wheel. Let b1 be the unit vector along the vector connecting point O to point C and let b3 be perpendicular to it in the plane created by e3 and n. The unit vector b2 is then given by the equality b2 = b3 X b1. It can be shown that b1 = cos a (cos 0 e1 + sin 0 e2) + sin a e3

Note that the reference frame B moves with the axis of the disk but does not rotate with the disk. In terms of these unit vectors, angular velocity m is expressed as m = (Vo/R) (-bi + tan a b3) (9.35) 