Apologies for the basic question but between the vectors and the spinning, I'm getting confused.
If I have a wheel with moment of inertia $I$ spinning at some rate $\vec\omega$ around an axis, what equation describes how much work is done to rotate the axle from one orientation to another? Yes, this is the standard "why doesn't a top fall down" problem, but I can't remember enough of my basic physics or google up the right equation to solve this. I'm trying to answer this basic question without precession before I complicate things with precession.
I know that the rotational energy in the system is $\frac{1}{2}I|\vec{\omega}|^2$, where $\vec \omega$ is the vector of rotation. So if we assume that $\vec\omega$ is initially rotation about the X-axis $\vec\omega_1 = \omega \vec i + 0\vec j + 0\vec k$ and then we apply a force to the axle to rotate the wheel 90 degrees so that it's rotating at the same rate completely in the Z plane $\vec\omega_2 = 0\vec i + 0\vec j + \omega\vec k$, is the change in energy (and thus the work done) simply:
\begin{aligned} \text{Work}&=\text{(Initial Energy) - (Final Energy)}\\ &=\frac{1}{2}I\omega_x^2 - \frac{1}{2} I\omega_z^2\\ &=\frac{1}{2}I \left(\omega_x^2-\omega_y^2\right)\\ &=\frac{1}{2}I(-2\omega^2)\\ &=-I\omega^2 \end{aligned}
Please let me know and thanks in advance.
Updated for clarity:
Thanks for the answers and the patience as I tune the question for additional clarity.
The first ~30 seconds of this video describe the question I'm asking: https://www.youtube.com/watch?v=_cMatPVUg-8&feature=youtu.be .
As you increase the weight of the mass attached to the end of the axel, how do we describe mathematically the "force" of wheel's resistance to being turned? I realize this is similar to friction in that it is not a real force that does work (as was pointed out, the energy of the system in preserved), but clearly a spinning wheel provides more resistance than a non-spinning one and a faster spinning wheel more than a slowly spinning one.
Does that help? Thanks in advance.
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