How to Determine Reaction Forces on a Gyroscope at Points A and O?

[Like Tony Stark]

Homework Statement

An homogenous disk of radius $r$ and mass $m$ rotates about the point $O$ with $\omega_2$ and is restricted to move in a horizontal plane. It also rotates about the point $G$ with constant angular velocity $\omega_1$. Determine the reaction forces on $A$, $O$ and $C$. Consider that the bar of length $L$ has no mass.

Relevant Equations

$\Sigma M=I \dot\omega +\omega × I \omega$

I know that
$\vec{v_c}=(\omega_1;-\omega_2;0)×(L;-r;0)=0$
So $\omega_2=\frac{r\omega_1}{L}$

Then, using the system of coordinates shown in the picture and $\Sigma M_z$ I can find the reaction force in $C$.

But how can I find the reaction forces on $A$ and $O$? I mean, what system should I use to apply Euler equations?

 

[haruspex]

The question wording doesn't match the diagram well. Shouldn't it say that the axle of the disk "rotates about the point O with ω2 and is restricted to move in a horizontal plane"?
No mention is made of friction. You are assuming no slippage, but if there is friction that acts towards O, parallel to the tension in the rod, then it is unclear how to apportion the nett force in that direction between the two.
What did you get for the reaction at C?

For reaction at O, what is the tension in the rod?

 

[Like Tony Stark]

I agree with you, I think it should say "axle". The question says that the disk rolls without slipping.
I thought that there was no tension, I considered that the rod was there just to restrict the motion of the disk.

 

[haruspex]

I agree with you, I think it should say "axle". The question says that the disk rolls without slipping.
I thought that there was no tension, I considered that the rod was there just to restrict the motion of the disk.

If it is restricting the motion of the disk then it must be exerting a force.
What would the disk do if no rod?

 

[Like Tony Stark]

Using the fact that $\omega_1, \omega_2=constant$ and that the axis shown are the principal axis, I got:
$\Sigma M_z=\omega_1 \omega_2 (I_x-I_y)$
The only forces that produce torque in this case are the normal force on $C$ and weight, so we can get the value of the normal force
But this system that I've chosen doesn't allow me to find the reaction force on $A$ and $O$.

 

[haruspex]

Using the fact that $\omega_1, \omega_2=constant$ and that the axis shown are the principal axis, I got:
$\Sigma M_z=\omega_1 \omega_2 (I_x-I_y)$

And what are you equating M_z to in terms of the reaction force at C?

There's nothing to stop you taking moments about another axis.