Finding the precession of a gyro using Euler's equations of rotation

[Leo Liu]

Homework Statement

A gyroscope wheel consists of a uniform disk of mass M and radius R that is
spinning at a large angular rotation rate ωs. The gyroscope wheel is mounted onto
a ball-and-socket pivot by a rod of length D that has negligible mass, allowing the
gyroscope to precess over a wide range of directions. Constant gravitational
acceleration g acts downward. For this problem, ignore both friction and
nutational motion; i.e, assume the gyroscope only precesses uniformly. For all
parts, express your solution as a vector (magnitude and direction) with components
in the coordinate system shown above.

(a) [5 pts] Calculate the total angular momentum vector of the uniformly
precessing gyroscope in the orientation show above; i.e., the total of the spin and
precession angular momentum vectors.

Relevant Equations

Euler's equations

The rate of precession of this gyro $\Omega$ can be found by solving $\tau_1=DMg=I_s\omega_s\Omega$. But when I apply Euler's equations to this problem, it fails.
I first set the frame in the way shown in the diagram above.
Then I wrote the first equation:
$$\tau_1=\bcancel{I_1\dot\omega_1}+(I_3-I_2)\omega_2\omega_3$$
The first term on the right side of the equation is 0 because the question says that we should ignore nutation.
After applying perpendicular axis theorem and parallel axis theorem, we get
$$I_3=1/4MR^2$$
Therefore,
$$I_3-I_2=-1/4MR^2$$
The equation then becomes
$$Dg=1/4MR^2\omega_s\Omega$$
whose solution is different to the solution produced by the torque-angular momentum equation.
Could someone point out the mistake(s) in the solution which uses Euler's equations?
@etotheipi help me if you please.

 

[haruspex]

I refrained from responding because I only have part of the answer.
In the Euler form, the torque is about the centre of the disc, no? So no need for the parallel axis theorem.
But that still leaves a discrepancy: a magnitude of $\frac 12MR^2$ versus $\frac 14MR^2$. I can believe the sign should be reversed, making both positive.

 

[Leo Liu]

I refrained from responding because I only have part of the answer.
In the Euler form, the torque is about the centre of the disc, no? So no need for the parallel axis theorem.

Oh, I see where the problem is. I didn't quite understand the conditions under which Euler's equations hold so I made this mistake. Thank you!

That said, there are certainly some other mistakes in my work, as the coefficient is 1/4 instead of 1/2. Really weird.