# 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. Commented May 2, 2016 at 16:16
• Commented May 2, 2016 at 16:55

## 2 Answers

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.

You seem to apply a scalar mass moment of inertia to a vector rotation. If you are careful with the coordinate transformations and vector components things will start to make more sense to you.

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 $[R]$.

$$K = \frac{1}{2} \vec{\omega}_{body} \cdot [I_{body}] \vec{\omega}_{body}$$

• Thank you for the answer and I agree and understand: the energy of the system doesn't change by rotating the wheel counter to its axis. That said, I have updated my question to better clarify what I'm asking: if you get a chance to take a look, I would be most appreciative. Commented May 4, 2016 at 14:51

If the rotational velocity (for a given rotational inertia) does not change, then the rotational kinetic energy (a scaler) does not change. In rotation, work equals torque times angle (which can be expressed as a dot product). The torque vector which causes a precession is in the direction of the change in the angular momentum. The vector representing the angle of precession is at right angles to the change in angular momentum. That torque does no work (and causes no change in kinetic energy). Consider a top precessing on a table: No work is being done by gravity or the table.