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

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?

• How sure are you that a random answer found on study.com is correct? I haven't checked the result myself, but I wouldn't necessarily trust it. Dec 10, 2022 at 13:47
• @MichaelSeifert Our TA posted a similar solution (same torque). The torque comes from looking at a differential time step and projecting the angular momentum from before the rotation onto the principal axes after the rotation. This gives $\Delta L_1 = L_2 \sin(\Delta \theta) = L_2 \Delta \theta = (2I\omega)(\Omega \Delta t)$. Dec 10, 2022 at 17:34
• Hint: The two wheels are not a rigid body. They are TWO rigid bodies each with its own motion. So you cannot just combine the motion and apply the EOM at the center. Dec 12, 2022 at 19:37
• Surely a rigid body is one for which every particle maintains a constant distance from every other particle, ensuring that the body's shape and size do not change. And surely the wheels and axle form such a body. Dec 12, 2022 at 21:02

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)

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*}

• I don't understand why the axle being massless makes $I_z=0$. Don't the wheels mass contribute to that moment of inertia? Dec 14, 2022 at 22:44
• @AlexWaese-Perlman , yes this is correct , I will edit my answer .
– Eli
Dec 15, 2022 at 8:53