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I am trying to solve the torque needed to rotate a rectangular plate of sides $a$ and $b$, about a diagonal with constant angular velocity $\omega$.

Euler equations are given by,

$$ I_1\dot{\omega}_1 + (I_2 - I_3)\omega_2\omega_3 = N_1, $$

$$ I_2\dot{\omega}_2 + (I_3 - I_1)\omega_3\omega_1 = N_2, $$

$$ I_3\dot{\omega}_3 + (I_1 - I_2)\omega_1\omega_2 = N_3, $$

where $I_1$, $I_2$ and $I_3$ are the principal moments of inertia of the rigid body, $\omega_1$, $\omega_2$ and $\omega_3$ are the angular velocities around the axes of these moments of inertia, and $N_i$ denotes the external torque applied along the axis of $\omega_i$ and $i$ = 1,2,3.

$\hskip2in$image1

For this problem, suppose $a > b$. $$I_1 = \frac{M a^2}{12}, \quad I_2 = \frac{M b^2}{12}, \quad I_3 = \frac{M (a^2 + b^2)}{12}$$

$$\omega_1 = \frac{\omega b}{\sqrt{a^2+b^2}}, \quad \omega_2 = \frac{\omega a}{\sqrt{a^2+b^2}}, \quad \omega_3 = 0$$

We find from Euler equations that, $$N_1 = 0 = N_2, \text{ while } N_3 = \frac{Mab\omega^2}{12 (a^2+b^2)}(a^2 - b^2) $$.

But this implies if $a = b$, the torque needed is zero. How should we find this intuitive that it requires zero torque for a square hinged at opposite corners?

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    $\begingroup$ Is external torque required to rotate a body around a principal axis? $\endgroup$
    – eranreches
    Nov 18 '18 at 18:50
  • $\begingroup$ Right. We don't need an external torque to rotate the body around any of the principal axes. So, is it just that for a square, the inertia tensor is a diagonal matrix and for a rectangle, we should be considering non-zero $I_{xy}$ and $I_{yx}$? $\endgroup$ Nov 18 '18 at 18:54
  • $\begingroup$ Even in the rectangle case $I$ is diagonal. The non-zero torque is related to the fact that $\vec{L}$ and $\vec{\omega}$ don't point in the same direction. Therefore, $\vec{L}$ precesse around $\vec{\omega}$. $\endgroup$
    – eranreches
    Nov 18 '18 at 19:11
  • $\begingroup$ @eranreches Okay, now I get it. This torque is arising from $\vec{\omega} \times \vec{L}$ since they are not pointing in the same direction. And in the case of a square lamina, these vectors are parallel. $\endgroup$ Nov 19 '18 at 8:30
  • $\begingroup$ Exactly. See my full answer below. $\endgroup$
    – eranreches
    Nov 19 '18 at 12:13
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Let me expand my comments. In the Lab frame you can write

$$\dfrac{{\rm d}\vec{L}}{{\rm d}t}=\vec{\omega}\times\vec{L}$$

where $\vec{L}$ is the angular momentum and $\vec{\omega}$ is the angular velocity. Now since $\vec{L}=\hat{I}\vec{\omega}$ and $\vec{\tau}=\dfrac{{\rm d}\vec{L}}{{\rm d}t}$, one has

$$\vec{\tau}=\vec{\omega}\times\hat{I}\vec{\omega}$$

Therefore, the torque vanishes iff $\vec{\omega}\parallel\hat{I}\vec{\omega}$, in other words iff there exists $\lambda$ such that

$$\hat{I}\vec{\omega}=\lambda\vec{\omega}$$

i.e. the rotation is about one of the principal axes.

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