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.