# Using Euler equations to solve for torque

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.

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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?

• Is external torque required to rotate a body around a principal axis? Nov 18 '18 at 18:50
• 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 $I_{yx}$? Nov 18 '18 at 18:54
• 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 $\vec{\omega}$. Nov 18 '18 at 19:11
• @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. Nov 19 '18 at 8:30
• Exactly. See my full answer below. Nov 19 '18 at 12:13

## 1 Answer

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.