Figure 1 shows the displacement of point P due to angular velocity w. The displacement due to a small amount of rotation (dq) is denoted as dr in the figure. During the rotation point P follows a circular path the center of which is located at point Q. The radius of this circular path can be described as
where n = the unit vector of angular velocity w. For a small dq, the displacement vector is perpendicular to both vectors n and r, thus,
where u = the unit vector of displacement vector dr. The length of displacement vector dr is equal to the length of the arc confined by points P and P':
Therefore, the displacement vector can be expressed as
where dq = the angular displacement vector. In other words, the displacement due to rotation about an axis of rotation is equal to the cross product of the angular displacement vector and the position vector.
Now, let's consider a slightly more complex situation. Figure 2 shows a rotating reference frame, the X'Y'Z' system, which rotates at an angular velocity of w. The inertial reference frame, the XYZ system, and the rotating frame share the origin together. The displacement of point P to P' is denoted as dr in the figure. Displacement vector dr can be split into two elements: drR and dr'. Vector drR is the displacement due to the rotation of the point with the rotating frame while dr' is the remainder of the displacement. Thus,
Since drR is due to the rotation of the point with the rotating frame, this displacement component won't be noticed within the rotating frame. Rather, dr' is the displacement of point P observed in the rotating frame and is in fact the displacement of point P relative to the rotating frame.
Substituting  into , one can obtain
Thus, the velocity of a point observed in the inertial frame is the sum of the velocity due to the rotation of the rotating frame and the velocity observed in the rotating frame.
One can rewrite  to
where rot = vector observed in the rotating frame.  can be generalized for an arbitrary vector V as
 is very useful in computing mechanical quantities of a rigid body since all points in the body rotates with the local reference frame fixed to the body. Let's apply  to computation of the torque acting on a rigid body. From  of Inertia Tensor, the angular momentum of a rigid body about its CM is
since the inertia tensor of the body changes as the body rotates in the inertial frame's perspective. Computing the time-derivative of the inertia tensor is not a simple job to do. Now, applying  instead, one can obtain
and  can be further reduced to
since the inertia tensor observed in the rotating frame (ICM) does not change as the body rotates. ar shown in  is the angular acceleration of the body relative to the rotating frame. Applying  to angular velocity:
that is, the angular acceleration observed in the fixed frame is equal to that observed in the rotating frame of the body. Substituting  into , one can obtain
Assuming the axes of the rotating reference frame fixed to the body are principal axes, from  of Principal Axes,  reduces to
 is the so-called Euler's equations for the motion of a rigid body. See Joint Torque for an example of using .
© Young-Hoo Kwon, 1998-