thanks phyzguy and mfb.
I got a hold of Goldstein's book and looked at chapter 4. The book is advanced. I am slowly grasping the meaning of your replies. I understand that the rotation of an object is not an absolute concept. For example, the angular velocity vector is a free vector.
We start with a rigid body and a triad of unit vectors ( ##\hat{i},\hat{j},\hat{k}##) that moves with the rigid body itself. The triad has an origin ##\overrightarrow{O}=(x_O, y_O,z_O)## that can be located at any arbitrary point in space. The angular velocity vector ##\overrightarrow{\omega}## is a
free vector given by $$\omega_x = \frac {d \hat{j} }{dt } \cdot \hat{k}$$ $$\omega_y = \frac {d \hat{k} }{dt } \cdot \hat{i}$$ $$\omega_z = \frac {d \hat{i}}{dt } \cdot \hat{j}$$
- An arbitrary point on the rigid body has a position ##\overrightarrow{P}=(x_P, y_P, z_P)## relative to the external, fixed frame of reference (not relative to the triad).
- The triad has an origin ##\overrightarrow{O} =(x_O, y_O, z_O)## relative to the external, fixed frame of reference.
- The triad's origin ##O## has velocity ##\overrightarrow{v_0}## relative to the external, fixed frame of reference.
- An arbitrary point on the rigid body has a velocity ##\overrightarrow{v_P}## relative to the external, fixed frame of reference:
$$ \overrightarrow{v_P}= \overrightarrow{v_0}+ \overrightarrow{\omega} (\overrightarrow{P}-\overrightarrow{O})$$
So, it is clear that the arbitrary point instantaneous velocity ##\overrightarrow{v_P}## depends on the choice of the triad's origin ##\overrightarrow{O}## and the point rotates about an axis parallel to the vector ##\overrightarrow{\omega}## and passing this arbitrarily chosen point ##\overrightarrow{O}## with velocity ##\overrightarrow{\omega} (\overrightarrow{P}-\overrightarrow{O})##.
So, this means that there is no absolute rotation axis for an object that is physically rotating. Therefore, when we throw an object in the air and the object is said to "rotate" about the center of mass c.m. in a plane perpendicular to the ground, the rotation is not physically happening about the center of mass as much as it is happening about any other arbitrary point ##\overrightarrow{O} \neq c.m. ## as long as the point ##\overrightarrow{O}## is fixed (moves with the rigid body so an observer located at ##\overrightarrow{O}##sees all the rigid body's point having zero velocity) with the rigid body (the point can be internal or even external to the rigid body).
In general, the point ##\overrightarrow{O}## is suitably chosen to be equal to the center of mass ##c.m.## for
mathematical convenience.
Could you explain this mathematical convenience in more detail again? Does the total kinetic energy ##KE## of the rigid body have a simpler expression when ##\overrightarrow{O} = c.m.##? What about the angular momentum expression for the rigid body?
Thanks!
