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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?

A diagram of the problem

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?

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  • $\begingroup$ 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. $\endgroup$ Commented Dec 10, 2022 at 13:47
  • $\begingroup$ @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 $\Delta L_1 = L_2 \sin(\Delta \theta) = L_2 \Delta \theta = (2I\omega)(\Omega \Delta t)$. $\endgroup$ Commented Dec 10, 2022 at 17:34
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    $\begingroup$ 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. $\endgroup$
    – JAlex
    Commented Dec 12, 2022 at 19:37
  • $\begingroup$ 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. $\endgroup$ Commented Dec 12, 2022 at 21:02

1 Answer 1

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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

enter image description here

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*}

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

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