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

[Aurelius120]

Homework Statement

Two thin circular discs are rigidly fixed by a massless rigid rod passing through centers and laid on on a firm flat surface and set rolling without slipping

Relevant Equations

$$\vec L=\vec r\times m\vec v$$
$$\vec L= I\vec \omega$$

If I understand correctly :
The angular velocity vector has two components: one along $\text{-ve z-axis}$ and one along $\text{-ve x-axis}$

So the motion can be considered to be two rotations:(some animation might help)

• Rotation about $\text{z-axis}$ with angular speed $\omega\sin\theta$
• Rotation about $\text{x-axis}$ which is the Instantaneous Axis Of Rotation with angular speed $\omega \cos \theta$

Now the center of mass of system is at $\frac{9l}{5}$ from origin. So angular momentum about COM is found.

In order to find angular velocity about $\text{z-axis}$, value of $$\sin\theta=\frac{a}{\sqrt{l^2+a^2}}=\frac{1}{5}$$

So here is a doubt:

1. Why does the system rotate in a way where the radius is perpendicular to the axis of rotation? It's not given in the question so why the assumption?
2. Why do the wheels tilt? Why can't the wheels be perpendicular to the surface with an axle connecting their centers? How might that motion look? Does it destabilize the system rotation or is it because of the diagram given?
3. The angular momentum of COM is non-zero (though not what the option says) However if the COM is on the axis, it means it should have zero angular velocity and therefore zero angular momentum. This contradicts the fact that the COM has a velocity and therefore angular momentum is not zero?

 

[Lnewqban]

1. Consider that the system is similar to a cone rolling on its side on a flat horizontal surface.

2. For that to happen, rather than zero, the z value of point O must be equal to radius 2a and should be a support of such type that it allows the axle to rotate freely about the z axis and simultaneously about the axle itself. The smaller disk would never touch the flat surface, while the vertical bigger disk would roll over the flat surface describing a circle.

3. Not a contradiction, since, as you noted, there are two simultaneous rotations. The angular momentum is about the z axis. Try visualizing yourself as trying to start pushing a very heavy cone to make it roll on its side on a flat horizontal surface.

 

[Aurelius120]

Consider that the system is similar to a cone rolling on its side on a flat horizontal surface.

In a cone the discs are fixed in their postions but here they are free to choose any angle of tilt right depending on how we join them but still they behave with radius perpendicular to axis?

For that to happen, rather than zero, the z value of point O must be equal to radius 2a and should be a support of such type that it allows the axle to rotate freely about the z axis and simultaneously about the axle itself. The smaller disk would never touch the flat surface, while the vertical bigger disk would roll over the flat surface describing a circle

So keeping both discs perpendicular to and touching the surface, if we connect their centers by a rod (obviously not horizontal), it is not possible to rotate it?

Not a contradiction, since, as you noted, there are two simultaneous rotations. The angular momentum is about the z axis. Try visualizing yourself as trying to start pushing a very heavy cone to make it roll on its side on a flat horizontal surface.

True treating it as two simultaneous rotations gives the correct answer using $\vec L=m(\vec r \times \vec v)$
But if I treat it as a net rotation about the axis given in the figure then it should also give the same answer, right? but $\vec L=I\vec \omega$ fails here( because $\omega$ on axis is zero). The other formula will probably work methinks

 

[Lnewqban]

In a cone the discs are fixed in their positions but here they are free to choose any angle of tilt right depending on how we join them but still they behave with radius perpendicular to axis?

Let's say that the "two thin circular discs are rigidly fixed by a massless rigid rod passing through their centers" means two welded connections.

If those connections are made in such a way that the plane containing each disc is other than perpendicular to the rigid rod central axis, would the angle that the rod forms with the flat surface be constant while the system rotates around the z axis?
Would the location of point O remain put in place?

If not, the system CM would be oscillating up and down respect to the flat surface.
Also, the center of the described circle over the flat surface would "walk".

So keeping both discs perpendicular to and touching the surface, if we connect their centers by a rod (obviously not horizontal), it is not possible to rotate it?

It will be possible to rotate that system, but not as shown in the problem.
We would have a situation similar to the one described above.

True treating it as two simultaneous rotations gives the correct answer using $\vec L=m(\vec r \times \vec v)$
But if I treat it as a net rotation about the axis given in the figure then it should also give the same answer, right? but $\vec L=I\vec \omega$ fails here( because $\omega$ on axis is zero). The other formula will probably work methinks

I agree.
Angular momentum only about the z axis represents the product of the rotational inertia and rotational velocity of both thin discs (the rigid rod is massless).
In that case, we don't need the rotation about the x-axis to have the same angular momentum respect to the Z axis.

 

[Aurelius120]

Angular momentum only about the z axisrepresents the product of the rotational inertia and rotational velocity of both thin discs (the rigid rod is massless).

In that case, we don't need the rotation about the x-axis to have the same angular momentum respect to the Z axis.

But I was talking about the option which said:

Angular Momentum of COM about point O is $81ma^2\omega$

The COM is a point on the axis so its angular momentum about O should be zero and all my other arguments

True treating it as two simultaneous rotations gives the correct answer using L→=m(r→×v→)
But if I treat it as a net rotation about the axis given in the figure then it should also give the same answer, right? but L→=Iω→ fails here( because ω on axis is zero). The other formula will probably work methinks

 

[haruspex]

If I understand correctly :
The angular velocity vector has two components: one along $\text{-ve z-axis}$ and one along $\text{-ve x-axis}$

So the motion can be considered to be two rotations:(some animation might help)

• Rotation about $\text{z-axis}$ with angular speed $\omega\sin\theta$
• Rotation about $\text{x-axis}$ … with angular speed $\omega \cos \theta$

Wouldn’t that all be true if the axis of the discs remained fixed? The rolling contact implies that axis is rotating around the z axis.
It might be easier to work in a rotating frame so that the discs remain in one place.

 

[Aurelius120]

Wouldn’t that all be true if the axis of the discs remained fixed? The rolling contact implies that axis is rotating around the z axis.
It might be easier to work in a rotating frame so that the discs remain in one place.

Ok my bad. I worded it incorrectly. The x-y plane contains the Instantaneous Axis of Rotation.
x-axis being it for a moment
Unless of course we consider the coordinate axes to be rotating about z-axis as well

 

[haruspex]

Ok my bad. I worded it incorrectly. The x-y plane contains the Instantaneous Axis of Rotation.
x-axis being it for a moment
Unless of course we consider the coordinate axes to be rotating about z-axis as well

That wasn't my point.
Consider an upright wheel rolling around in a circle. Not only is it rotating about a horizontal axis through its centre, it is also rotating about a vertical axis through its centre.

 

[Aurelius120]

Wouldn’t that all be true if the axis of the discs remained fixed?

So how do I decide whether it is also rotating about z-axis? Is is simply the use of 'rolling without slipping' in the question? Will a cone rotate on a frictionless surface?

I don't think it can rotate about the axis in the diagram unless it rotates about z-axis as well

 

[haruspex]

So how do I decide whether it is also rotating about z-axis? Is is simply the use of 'rolling without slipping' in the question? Will a cone rotate on a frictionless surface?

I don't think it can rotate about the axis in the diagram unless it rotates about z-axis as well

You can write the rotation as a sum of a rotation about the instantaneous axis of the cone (as you have it) and a rotation of that axis about the z axis.

 

[Aurelius120]

You can write the rotation as a sum of a rotation about the instantaneous axis of the cone (as you have it) and a rotation of that axis about the z axis.

But you said that happens even when axis is fixed :

Wouldn’t that all be true if the axis of the discs remained fixed? The rolling contact implies that axis is rotating around the z axis.

So how to write?

 

[Nik_2213]

FWIW, this is the 'kinematics' of railway wheels on bogies etc with rigid axles.

 

[haruspex]

But you said that happens even when axis is fixed :

So how to write?

There are three components altogether.
In post #1, you took the rotation about the instantaneous position of the cone's axis and resolved it into a component about the x axis and a component about the z axis. Neither of those involves the length L.
What that misses is the rotation of the cone's axis about the z axis. This adds another rotation about the z axis, and does involve L.

 

[Aurelius120]

There are three components altogether.
In post #1, you took the rotation about the instantaneous position of the cone's axis and resolved it into a component about the x axis and a component about the z axis. Neither of those involves the length L.
What that misses is the rotation of the cone's axis about the z axis. This adds another rotation about the z axis, and does involve L.

So the net z-component of angular velocity is not equal to $\omega \sin \theta= \omega /5$ but the solution says option b and d are correct?

Option-b should be correct because relative angular velocity will still be $\omega$

 

[haruspex]

So the net z-component of angular velocity is not equal to $\omega \sin \theta= \omega /5$ but the solution says option b and d are correct?

Hmm.. I think that is wrong.
Consider the case where the assembly is sliding around the z axis with the same points constantly in contact with the ground, instead of rolling. According to that analysis it has no angular momentum about its mass centre, which is clearly false.

But I am not confident… I need a second opinion.
@TSny? @Orodruin? @jbriggs444 ?

 

[Orodruin]

The center of mass certainly rotates with angular velocity $\omega/5$ around the $z$-axis purely by geometrical consideration. The smaller wheel should roll around a circle of radius $\sqrt{\ell^2 + a^2} = 5a$ and the wheel itself has radius $a$. It will therefore go around once every 5 revolutions, meaning $\omega/5$ is the angular velocity of the CoM.

 

[haruspex]

The center of mass certainly rotates with angular velocity $\omega/5$ around the $z$-axis purely by geometrical consideration. The smaller wheel should roll around a circle of radius $\sqrt{\ell^2 + a^2} = 5a$ and the wheel itself has radius $a$. It will therefore go around once every 5 revolutions, meaning $\omega/5$ is the angular velocity of the CoM.

Ok, I start to see where my confusion is, but it wasn't part d that I meant to query - I agree with that.
And maybe I misunderstood the OP's analysis in post #1. I read that as decomposing the rotation about the cone's axis as a rotation about the x axis plus a rotation ($\omega/5$) about the z axis. But that does not consider the rotation of the cone's axis about the z axis, also $\omega/5$.
Rereading it, maybe what I read as the first of those was intended as the second, in which case it is the first that is missing. E.g., if the cone's axis were fixed (so slipping, not rolling) there would still be a z axis $\omega/5$ component of the angular velocity.

So part b is the one I am worried about. The given answer would be true for that slipping case. Shouldn't the precession of the axis change the magnitude of the angular momentum? (In fact, reduce it, I think.)

 

[Orodruin]

So part b is the one I am worried about. The given answer would be true for that slipping case. Shouldn't the precession of the axis change the magnitude of the angular momentum? (In fact, reduce it, I think.)

Yes.

 

[Aurelius120]

Yes.

But if the COM is on the axle/axis , its angular velocity about the axle should be zero. And since the entire system(including COM is rotating about z-axis), the only velocity (of the discs) in the COM frame should be $\omega$ about axle
Then the angular momentum should be $$I \omega =\left( \frac{ma^2}{2}+\frac{4m(2a)^2}{2} \right) \omega$$

 

[Aurelius120]

And maybe I misunderstood the OP's analysis in post #1. I read that as decomposing the rotation about the cone's axis as a rotation about the x axis plus a rotation ($\omega/5$) about the z axis. But that does not consider the rotation of the cone's axis about the z axis, also $\omega/5$.
Rereading it, maybe what I read as the first of those was intended as the second, in which case it is the first that is missing. E.g., if the cone's axis were fixed (so slipping, not rolling) there would still be a z axis $\omega/5$ component of the angular velocity.

My analysis was wrong. I imagined it correctly as a pure rotation about point of contact (substitute for rolling) and z-axis but for whatever reason decided it was because of the two components of $\omega$
So my equation was $$\vec \omega_{net}=-\omega (\cos \theta \hat i + \sin \theta \hat k)$$
Rather than $$\vec \omega_{net}=\vec \omega+ \text{ angular velocity about z-axis }$$

And that's why I thought angular momentum of COM will be zero about origin

 

[Orodruin]

But if the COM is on the axle/axis , its angular velocity about the axle should be zero. And since the entire system(including COM is rotating about z-axis), the only velocity in the COM frame should be $\omega$ about axle
Then the angular momentum should be $$I \omega =\left( \frac{ma^2}{2}+\frac{4m(2a)^2}{2} \right) \omega$$

Angular velocity does not depend on relative position as angular momentum does. The angular velocity is what it is regardless of frame.

 

[Orodruin]

Angular velocity does not depend on relative position as angular momentum does. The angular velocity is what it is regardless of frame.

To put that into context. If you factor out the CoM motion, the axle will still precess around the z-axis. The answer given in (b) is the angular momentum of the assembly if it only spins about the axis without precessing - but as @haruspex points out: it does precess.

 

[Aurelius120]

Angular velocity does not depend on relative position as angular momentum does. The angular velocity is what it is regardless of frame.

But it should depend on frame of reference? I mean when we calculate angular velocity, it is relative to Earth. We don't take rotation and revolution of Earth into consideration.

Similarly relative to the COM the angular velocity of the discs should be $\vec \omega$ about the axle.

 

[Orodruin]

But it should depend on frame of reference?

Not really no.

I mean when we calculate angular velocity, it is relative to Earth. We don't take rotation and revolution of Earth into consideration.

That’s because it is so small that it usually does not matter for the purposes of say a spinning top. When you decompose the angular momentum into the CoM angular momentum and the angular momentum about the CoM, these are both computed in the same inertial frame.

If you want to use a non-inertial frame you will have additional terms.

 

[Aurelius120]

@Orodruin So if I were to sit on the axle, I will still be able to see the wheels rotating about z-axis? This seems a bit a counterintuitive

So to calculate angular momentum about COM do we take COM stationary or COM rotating about z-axis?

It is so much easier with linear velocity and linear momentum

 

[haruspex]

So to calculate angular momentum about COM do we take COM stationary or COM rotating about z-axis?

For the angular momentum about the COM, take the COM as stationary but consider all rotations about it. That means both the rotation of the cone about its axis and the rotation of that axis about a vertical axis.

 

[Aurelius120]

For the angular momentum about the COM, take the COM as stationary but consider all rotations about it. That means both the rotation of the cone about its axis and the rotation of that axis about a vertical axis.

So for the purpose of this question, the option is wrong
But I can't calculate angular momentum about COM now

 

[Aurelius120]

For the angular momentum about the COM, take the COM as stationary but consider all rotations about it. That means both the rotation of the cone about its axis and the rotation of that axis about a vertical axis.

Since about the COM and parallel to z-axis angular velocity of the wheels will be in opposite sense, does it mean angular momentum is substracted

 

[haruspex]

Since about the COM and parallel to z-axis angular velocity of the wheels will be in opposite sense, does it mean angular momentum is substracted

Yes, it reduces the overall magnitude.
It is probably easiest if you resolve the angular velocities into components along the cone axis and normal to it. (Otherwise you will have angular momentum of rotations of discs about a skew axis to deal with.)
For the rotation about the cone axis, that's trivial. Try resolving the rotation of the cone axis about a vertical axis through the COM that way.

 

[Aurelius120]

Yes, it reduces the overall magnitude.
It is probably easiest if you resolve the angular velocities into components along the cone axis and normal to it. (Otherwise you will have angular momentum of rotations of discs about a skew axis to deal with.)
For the rotation about the cone axis, that's trivial. Try resolving the rotation of the cone axis about a vertical axis through the COM that way.

Plz Check this
This is my attempt at Option-B:
About axle :
Angular Momentum: $L_1=I\omega= \dfrac{17ma^2\omega}{2}$

About axis perpendicular to axle:
Velocity of COM of 1st wheel about z-axis=$\dfrac{\omega}{5} l \cos\theta = \omega_1×\dfrac{4l}{5}$
Velocity of COM of 2nd wheel about z-axis= $\omega_2×\dfrac{l}{5}= \dfrac{\omega}{5} × 2l \cos \theta$
Angular Momentum: $L_2=I_1\omega_1 - I_2\omega_2$

So $$L_1=\frac{17ma^2\omega}{2}$$
$$L_2=\left(\dfrac{ma^2}{4}+\dfrac{16ml^2}{25}\right)\omega_1-\left(\dfrac{4m(2a)^2}{4}+\dfrac{4ml^2}{25}\right)\omega_2$$
$$L_{COM}=\sqrt{L_1^2+L_2^2}$$

 

[Orodruin]

You do not need to find the angular momentum to conclude that (b) is false. You only need to note that the value given in (b) is the angular momentum it would have if you did not account for the precession. Doing that, the angular momentum will clearly be different.

 

[Orodruin]

t is probably easiest if you resolve the angular velocities into components along the cone axis and normal to it.

I would just compute the moment of inertia tensor for the system and multiply by the angular velocity …

 

[Aurelius120]

Please Check this as well
This is my attempt at Option-A:

Angular Momentum about axle and passing through origin: $L_3=I\omega=\dfrac{17ma^2\omega}{2}$

About an axis perpendicular to axle and through origin:

Velocity of COM of 1st wheel= $\omega_3 ×l=\dfrac{\omega}{5}× l\cos\theta$
Velocity of COM of 2nd wheel=$\omega_4 ×2l=\dfrac{\omega}{5}×2l\cos\theta$
Angular momentum: $L_4=I_3\omega_3+I_4\omega_4$
$$L_3=I\omega=\dfrac{17ma^2\omega}{2}$$ $$L_4=\left(\dfrac{ma^2}{4}+ml^2\right)\omega_3 + \left(\dfrac{4m(2a)^2}{4}+4m(2l)^2\right)\omega_4$$

z-component of angular momentum about origin=
$$L_z=L_4\cos\theta-L_3\sin\theta$$

 

[Aurelius120]

You do not need to find the angular momentum to conclude that (b) is false. You only need to note that the value given in (b) is the angular momentum it would have if you did not account for the precession. Doing that, the angular momentum will clearly be different.

Yes I know that. I was just trying to compute it for the sake of it.
Also I can't use that for Option-A

 

[TSny]

Please Check this as well
This is my attempt at Option-A:

Angular Momentum about axle and passing through origin: $L_3=I\omega=\dfrac{17ma^2\omega}{2}$

You need to use the net angular velocity component along the axle. The precession angular velocity $\omega/5$ has a component along the axle that needs to be included here.

$$L_4=\left(\dfrac{ma^2}{4}+ml^2\right)\omega_3 + \left(\dfrac{4m(2a)^2}{4}+4m(2l)^2\right)\omega_4$$

I think this is correct if you are taking $\omega_3 = \omega_4 = \dfrac{\omega}{5}\cos\theta$.

z-component of angular momentum about origin=
$$L_z=L_4\cos\theta-L_3\sin\theta$$

Looks good.

 

[Aurelius120]

@TSny Can you also check my attempt at Option B in post 30? Thank You

 

[haruspex]

About axis perpendicular to axle:
Velocity of COM of 1st wheel about z-axis=$\dfrac{\omega}{5} l \cos\theta = \omega_1×\dfrac{4l}{5}$
Velocity of COM of 2nd wheel about z-axis= $\omega_2×\dfrac{l}{5}= \dfrac{\omega}{5} × 2l \cos \theta$

I do not understand what you are doing there. For each disc you have a rotation about the z axis, but you need to

• resolve that into a rotation about the axle of the cone and and a perpendicular rotation
• find the moment of inertia of the disc about axes parallel to those through the mass centre of the assembly.

 

[Aurelius120]

I do not understand what you are doing there. For each disc you have a rotation about the z axis, but you need to

• resolve that into a rotation about the axle of the cone and and a perpendicular rotation
• find the moment of inertia of the disc about axes parallel to those through the mass centre of the assembly.

I was going like this
Since linear velocity is same irrespective of axis
I equated the product of angular velocity and perpendicular distance from axis in each case
As in
$$\omega_z × r_z= \omega_{COM}×r_{COM}$$

 

[TSny]

Plz Check this
This is my attempt at Option-B

About axle :
Angular Momentum: $L_1=I\omega= \dfrac{17ma^2\omega}{2}$

You need to include the component of the precessional angular velocity along the axle. That is, the moment of inertia about the axle should be multiplied by the component of the net angular velocity along the axle. See post #26 of @haruspex .

About axis perpendicular to axle:
Velocity of COM of 1st wheel about z-axis=$\dfrac{\omega}{5} l \cos\theta = \omega_1×\dfrac{4l}{5}$
Velocity of COM of 2nd wheel about z-axis= $\omega_2×\dfrac{l}{5}= \dfrac{\omega}{5} × 2l \cos \theta$
Angular Momentum: $L_2=I_1\omega_1 - I_2\omega_2$

So $$L_1=\frac{17ma^2\omega}{2}$$
$$L_2=\left(\dfrac{ma^2}{4}+\dfrac{16ml^2}{25}\right)\omega_1-\left(\dfrac{4m(2a)^2}{4}+\dfrac{4ml^2}{25}\right)\omega_2$$

I get a bit confused with the meaning of your subscripts "1" and "2".

In your expression for $L_2$, I don't understand why you distinguish between $\omega _1$ and $\omega_2$. Also, why the subtraction?

For $L_2$ you need to find the total moment of inertia about the perpendicular axis through the center of mass and multiply by the component of the net angular velocity along this axis.

$$L_{COM}=\sqrt{L_1^2+L_2^2}$$

Ok.

 

[Aurelius120]

I get a bit confused with the meaning of your subscripts "1" and "2".

That represents the angular velocity of wheels 1 and 2 about the axis perpendicular to axle.

In your expression for $L_2$, I don't understand why you distinguish between $\omega _1$ and $\omega_2$. Also, why the subtraction?

The wheels are rotating about z-axis in say ACW sense.
About an axis passing in between the wheels, but perpendicular to axle the rotation of wheels will be in opposite sense to each other, right?

For $L_2$ you need to find the total moment of inertia about the perpendicular axis through the center of mass and multiply by the component of the net angular velocity along this axis.

What is confusing is that angular velocity of both wheels about COM perpendicilar to z-axis can't be same, right?
Because linear velocity is independent of axis and is different for either wheel as is their distance from COM
For the wheels,
$$\omega_{COM}×r_{COM}= \omega_z×r_z$$

 

[TSny]

The wheels are rotating about z-axis in say ACW sense.
About an axis passing in between the wheels, but perpendicular to axle the rotation of wheels will be in opposite sense to each other, right?
What is confusing is that angular velocity of both wheels about COM perpendicilar to z-axis can't be same, right?

Consider a dumbbell rotating about the axis as shown. Both balls have the same angular velocity $\omega$ in the same direction. The linear velocities of the two balls will be in opposite directions with different magnitudes.

 

[Aurelius120]

Consider a dumbbell rotating about the axis as shown. Both balls have the same angular velocity $\omega$ in the same direction. The linear velocities of the two balls will be in opposite directions with different magnitudes.

View attachment 347426

But what if the balls were rotating about an outside axis. Both will have same direction of linear velocity. However when the that system is analysed about an axis passing in-between the balls, the linear velocities being kn the same direction will make angular velocity in opposite directions.

 

[TSny]

But what if the balls were rotating about an outside axis. Both will have same direction of linear velocity. However when the that system is analysed about an axis passing in-between the balls, the linear velocities being kn the same direction will make angular velocity in opposite directions.

Suppose the dumbbell rotates about an "outside" axis. Both balls will have the same angular speed $\omega$. Their linear velocities will be in the same direction (with different magnitudes). Also, their angular velocities will be in the same direction (counter-clockwise as seen from above).

For the inside axis, the balls have the same counter-clockwise angular velocity as seen from above. If one ball makes 5 revolutions per second, so does the other ball. The linear velocities are now opposite in direction and have different magnitudes.

 

[Aurelius120]

View attachment 347427
Suppose the dumbbell rotates about an "outside" axis. Both balls will have the same angular speed $\omega$. Their linear velocities will be in the same direction (with different magnitudes). Also, their angular velocities will be in the same direction (counter-clockwise as seen from above).

For the inside axis, the balls have the same counter-clockwise angular velocity as seen from above. If one ball makes 5 revolutions per second, so does the other ball. The linear velocities are now opposite in direction and have different magnitudes.
View attachment 347429

But doesn't that change the system completely?
Applying that to wheels,
The wheels will stop rotating about the z-axis as cone would have and rather rotate about each other. But that is not the motion described in the question?
Aren't we supposed to analyse the system's motion as it is but about a different axis?

 

[TSny]

Let $\vec {\omega}^{net}$ be the total angular momentum velocity vector of the assembly at some instant of time. This is the vector sum of the "spin" angular velocity $\omega$ along the axle and the "precession" angular velocity $\omega/5$ along the z-axis.

An axis along the axle of the assembly is a "principal" axis of the assembly. Thus, the total angular momentum vector of the system will have a component along the axle given by $L_a = I_a \omega^{net}_a$ where $I_a$ is the moment of inertia of the assembly about the axle and $\omega^{net}_a$ is the component of $\vec {\omega}^{net}$ along the axle. ("$a$" is for "axle".)

Consider another axis that (i) passes through the center of mass of the assembly, (ii) is perpendicular to the axle, and (iii) lies in the plane containing the axle and the z-axis. This axis is also a principal axis. The total angular momentum vector will have a component along this axis given by $L_p = I_p \omega^{net}_p$ where "$p$" indicates components along this perpendicular axis.

The magnitude of the total angular momentum about the center of mass is $L_{cm} = \sqrt{L_a^2 + L_p^2}$.

You already know $I_a$.

You need to calculate $I_p$. This is similar to the dumbbell.

 

[haruspex]

I was going like this
Since linear velocity is same irrespective of axis
I equated the product of angular velocity and perpendicular distance from axis in each case
As in
$$\omega_z × r_z= \omega_{COM}×r_{COM}$$

But the angular momenta of the COMs are inadequate. They are not point masses.
As I wrote, you need to take these steps:

1. Resolve the precession about the z axis into a rotation along the axle of the 'cone' (pair of discs) and a rotation about an axis which is normal to that and lies in the zx plane.
2. For the component along the axle of the cone, you can simply add it to the given $\omega$ (but the sign will be opposite, I believe). From that you can calculate that component of the angular momentum.
3. For the other component of the precession, the axis is parallel to the discs. Find how far each disc is from their common mass centre, compute the two angular momentum components using the parallel axis theorem and add them.
4. Use Pythagoras to combine the two orthogonal angular momentum components.

@TSny has said the same in algebra. Having it expressed in two ways should help.

 

[Aurelius120]

@TSny has said the same in algebra. Having it expressed in two ways should help.

Yes thank you.
What was bugging me was $\vec L=m\vec r\times \vec v$, gave one positive and one negative value for the wheels because the direction of $\vec r$ was opposite.
But

They are not point masses

So this formula failed.

I was considering the system to be still rotating about z-axis while trying to compute it's angular velocity about COM
Unlike this:

You need to calculate Ip. This is similar to the dumbbell.

 

[Aurelius120]

Not to sound disrespectful but
I thought I got it but I didn't.
So forgive my bs here and in 44
@haruspex @TSny
I understand the method you are using and how it is working.

But I don't understand why it is working.
As in why are you considering it to be rotating around the axis through COM and perpendicular to axle? Isn't it actually rotating about z-axis making both their linear velocities in same direction and thus both having different angular velocity about the required COM axis? (which also seems impossible as they are rigidly connected)

 

[haruspex]

why are you considering it to be rotating around the axis through COM and perpendicular to axle?

It is a valid way of decomposing a rotational velocity vector.
I admit it is less obvious than with linear motion. Displacements can be easily shown to add vectorially, and it therefore works for linear velocities and accelerations by differentiation wrt time. But it does not work for rotations through non-infinitesimal angles: the order of two such rotations matters. But again, it does work for infinitesimal angles, so it also works for angular velocities and accelerations.

 

[Aurelius120]

It is a valid way of decomposing a rotational velocity vector.
I admit it is less obvious than with linear motion. Displacements can be easily shown to add vectorially, and it therefore works for linear velocities and accelerations by differentiation wrt time. But it does not work for rotations through non-infinitesimal angles: the order of two such rotations matters. But again, it does work for infinitesimal angles, so it also works for angular velocities and accelerations.

Thanks
Just as a way to visualise:Is the rotation about COM axis what would happen if someone suddenly put a rod at the COM causing it to come to rest immediately while the wheels because of their inertia rotate about that axis?