Understanding work with rotational momentum/moment of inertia

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.

• I took the liberty of editing your question to make the math look better. Please make sure I didn't mess up. If you don't like my edit you can revert it by clicking on the edit button. – AccidentalFourierTransform May 2 '16 at 16:16
• – BowlOfRed May 2 '16 at 16:55

Kinetic energy is $$K = \frac{1}{2} \vec{\omega} \cdot [I] \vec{\omega}$$ when the 3×3 mass moment of inertia matrix $[I]$ is expressed in world coordinates. Remember $$[I] = [R] [I_{body}] [R]^\top$$ is how body inertias is transformed into world inertias.
If the body is rotated from spinning around the x-axis to the z-axis, then the energy is conserved. To find that out resolve the rotation into body coordinates $$\vec{\omega}_{body} = [R]^\top \vec{\omega}$$ you will find the kinetic energy expression does not depend on the rotation matrix $[R]$.
$$K = \frac{1}{2} \vec{\omega}_{body} \cdot [I_{body}] \vec{\omega}_{body}$$