(8 A)
that is to say, the angular velocity about each axis is proportional to the sine of the angle between the other two.
Demonstration.—From C draw CF perpendicular to OA, and CG perpendicular to OE
| Then CF = 2 × | area EC | , |
| CE |
| and CG = 2 × | area ECO | ; |
| OE |
∴ CG : CF :: CE = OD : OE.
Let vc denote the linear velocity of the point C. Then
vc = α · CF = γ · CG
∴ γ : α :: CF : CG :: OE : OD,
which is one part of the solution above stated. From E draw EH perpendicular to OB, and EK to OA. Then it can be shown as before that