Why don't Euler's formulas for torque apply to this problem?

Question

In my mechanics class we were assigned problem 9.44 from "Introduction to Classical Mechanics" by David Morin as homework. The problem and figure are below:

Two wheels of mass $m$ and moment of inertia $I$ are connected by a massless axle of length $l$, as shown in Fig. 9.61. The system rests on a frictionless surface, and the wheels rotate with frequency $\omega$ around the axle. Additionally, the whole system rotates with frequency $\Omega$ around the vertical axis through the center of the axle. What is the largest value of $\Omega$ for which both wheels stay on the ground?

Here is how I calculated torque with Euler's formulas:

Euler's torque formula says that $\tau_1 = I_1 \dot{\omega}_1 + (I_3 - I_2)\omega_2\omega_3$. If we let $I_1$ be the principal axis that initially points through the page, and $I_2$ be the principal axis that is parallel to the axle, we have $\tau_1 = (I_3 - I_2)\omega\Omega$. It's easy to see that $I_2 = 2I$, and we can calculate $I_3 = \frac{1}{2}ml^2 + I$ using axis theorems, so $\tau = (\frac{1}{2}ml^2-I)\omega\Omega$.

This torque is incorrect. The correct one (found here), uses $\tau = 2I\omega\Omega$, which is slightly different than the torque given by Euler's formulas. My question is, why don't Euler's formulas correctly calculate the torque in this system? And where does the value $\tau = 2I\omega\Omega$ come from?

Answer 1

The Euler equation is:

$$\mathbf\Theta\, {\dot{\vec{O}}_m}+\vec O_m\times (\mathbf\Theta\,\vec O_m)=\vec \tau$$

with

\begin{align*} &\vec{O}_m=\begin{bmatrix} 0 \\ 2\,\omega(t) \\ \Omega \\ \end{bmatrix}\quad, \mathbf\Theta=\begin{bmatrix} 0 & 0 & 0 \\ 0 & I & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \end{align*}

you obtain

\begin{align*} &\vec\tau=\begin{bmatrix} -2\,I\,\Omega\,\omega \\ I\,\dot{\omega} \\ 0 \\ \end{bmatrix} \end{align*}

notice that the inertia $~I_x~,I_y~,I_z$ of the axle are zero.(mass less)

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Edit

with the parallel axis theorem the inertia tensor, $~\mathbf\Theta_{xyz}~$ is:

\begin{align*} &\mathbf\Theta_{xyz}= \mathbf \Theta_{w}-m\,\left[\vec r_{wo}\right]_\times \,\left[\vec r_{wo}\right]_\times \end{align*} where m is the mass of the wheel and $~\mathbf \Theta_{w}~$ is the wheel inertia . $~\vec r_{wo}~$ is the vector from the center of the wheel to the coordinate system $~xyz~$.

thus \begin{align*} \mathbf\Theta_{xyz}= \begin{bmatrix} 0 & 0 & 0 \\ 0 & I & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}&\underbrace{-m\, \begin{bmatrix} 0 & 0 & -L/2 \\ 0 & 0 & 0 \\ L/2 & 0 & 0 \\ \end{bmatrix}\, \begin{bmatrix} 0 & 0 & -L/2 \\ 0 & 0 & 0 \\ L/2 & 0 & 0 \\ \end{bmatrix}}_{\rm left\, wheel}\\ &\underbrace{ -m\, \begin{bmatrix} 0 & 0 & L/2 \\ 0 & 0 & 0 \\ -L/2 & 0 & 0 \\ \end{bmatrix}\, \begin{bmatrix} 0 & 0 & L/2 \\ 0 & 0 & 0 \\ -L/2 & 0 & 0 \\ \end{bmatrix}}_{\rm right\, wheel}\\ &\mathbf\Theta_{xyz}= \begin{bmatrix} 1/2\,m\,L^2 & 0 & 0 \\ 0 & I & 0 \\ 0 & 0 & 1/2\,m\,L^2 \\ \end{bmatrix} \end{align*}

the result is now

\begin{align*} &\vec \tau= \begin{bmatrix} \tau_x \\ \tau_y \\ \tau_z \\ \end{bmatrix}= \begin{bmatrix} \Omega\,\omega\,(m\,L^2 -2\,I)\\ I\,\dot{\vec{\omega}} \\ 0 \\ \end{bmatrix} \end{align*}