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Consider a wheel with the center point O of the wheel is fixed and rotating freely around an axis passing through O. Assume that it is spinning in the horizontal plane and that there is no force applied to it. Now, suddenly we apply a force $\vec F$ to it as shown in the figure. enter image description here

From experiment, due to the gyroscope effect, the wheel will start rotating about x axis, since the $\vec F$ force will take effect 90 degrees ahead.

But the equations do not show that. We have the Euler's equations of motion $M_x = I_x\dot \omega_x - (I_y - I_z)\omega_y\omega_z$, $M_y = I_y\dot \omega_y - (I_z - I_x)\omega_z\omega_x$ and $M_z = I_z\dot \omega_z - (I_x - I_y)\omega_x\omega_y$.

We also have at the instant applying force $\vec F$ that, $$\omega_x = \omega_y = 0,$$ $$M_x = M_z = 0, \text{and}$$ $$M_y = -F \cdot R,$$ where R is the radius of the wheel.

Then Euler's equations are then, $$0 = I_x\dot \omega_x,$$ $$-F \cdot R = I_y\dot \omega_y \text{ and}$$ $$0 = I_z\dot \omega_z.$$

This implies that $\dot \omega_x = 0$. This result does not co-incide with the experimental result that $\dot \omega_x \neq 0$.

What am I missing?

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You didn't miss anything. In your question you analyze the dynamics correctly at the instant $t=0$. If you use the same argument, after the application of the force $F$ over the time interval $dt$ you will see that $$-F.R = I_y\dot \omega_y \implies \omega_y = -F.R \frac {dt}{I_y},$$ and then from the Euler equations $$M_x = I_x\dot \omega_x - (I_y - I_z)\omega_y\omega_z \implies \dot \omega_x = - \frac {(I_y - I_z)}{I_x} \omega_z F.R \frac {dt}{I_y} \neq 0.$$

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