System of two wheels of different sizes with an axle through their centers

[Aurelius120]

Order, as in which you do first.
Place a die with 1 facing up, 2 towards you, 3 to the right.
If you rotate it 90° around the left-right (3-4) axis, top (1) away from you, then 90° about the vertical axis, right side away from you, the net result is a rotation about a long diagonal. The faces will now be:
1 left
2 top
3 back
If you had done those two rotations in the other order you would have
1 back
2 right
3 bottom

For infinitesimal rotations we need to switch to a sphere. If we draw little arrows on the surface to represent infinitesimal rotations, we can see that the net of two tiny rotations at right angles is almost the same as a rotation along the hypotenuse. So now they add like vectors.

Thanks
So for Option-B, the value of $\omega$ about axis through COM and perpendicular to axle is
$\dfrac{\omega}{5}\cos\theta$ and the angular momentum is calculated
Then the velocity of the COM of wheel will be
$$\dfrac{\omega}{5}\cos\theta \ \hat{n_1} \times \dfrac{-4l}{5} \ \hat{n_2}$$ and $$\dfrac{\omega}{5}\cos\theta \ \hat{n_1} \times \dfrac{+l}{5} \ \hat{n_2}$$
But this is different from the velocities obtained about z-axis?
Which are $\dfrac{\omega}{5}l\cos\theta$ and $\dfrac{\omega}{5}2l\cos\theta$

Note:
$\hat n_1$ is along perpendicular to axle
$\hat n_2$ is along axle

 

[haruspex]

$$\dfrac{\omega}{5}\cos\theta \ \hat{n_1} \times \dfrac{-4l}{5} \ \hat{n_2}$$

Where does the -4 come from?

 

[Aurelius120]

Where does the -4 come from?

Because it is measured from COM of system (9l/5) and COM of 1st wheel (5l/5) is on left so position vector is negative?
Even without that the magnitudes of velocity are different

 

[haruspex]

Because it is measured from COM of system

Then what you are calculating is the velocity relative to the COM. Whereas

$\dfrac{\omega}{5}l\cos\theta$ and $\dfrac{\omega}{5}2l\cos\theta$

are velocities relative to the ground frame.
You cannot expect them to be the same.

 

[Aurelius120]

Then what you are calculating is the velocity relative to the COM. Whereas

That should be zero if we assume COM to be moving because all points on axle have same velocity wrt ground.

are velocities relative to the ground frame.
You cannot expect them to be the same.

But didn't we assume COM to be at rest for calculating angular momentum?
Or as Orodruin calls it "COM Instantaneous Rest Frame"
So they be same?

 

[Orodruin]

That should be zero if we assume COM to be moving because all points on axle have same velocity wrt ground.

No they don’t. The velocity of a point on the axle relative to the ground is directly proportional to the distance from the point O as the axle is precessing around the z-axis.

 

[haruspex]

That should be zero if we assume COM to be moving because all points on axle have same velocity wrt ground.

Not so. A point on the axle at r from the z axis is moving at $\frac 15\omega r$ relative to the ground.

But didn't we assume COM to be at rest for calculating angular momentum?

Question B asks for angular momentum about the COM, and for that purpose it is fine to use velocity relative to that.

 

[Aurelius120]

No they don’t. The velocity of a point on the axle relative to the ground is directly proportional to the distance from the point O as the axle is precessing around the z-axis.

Not so. A point on the axle at r from the z axis is moving at $\frac 15\omega r$ relative to the ground.

How on Earth did I miss that? That was stupid of me.

Question B asks for angular momentum about the COM, and for that purpose it is fine to use velocity relative to that.

Ah! So we use velocity relative to COM
Since then I had been taking COM at rest to mean COM at rest wrt origin and wheels moving with their ground frame velocities.