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

[Orodruin]

The rotation about the CoM is the (instantaneous) motion in the CoM instantaneous rest frame. The separation into CoM angular momentum and angular momentum due to the CoM motion is simply a convenient way of doing the bookkeeping of the angular momentum as the terms separate.

Consider a system of rigidly connected masses $m_i$ with positions $\vec x_i$ and velocities $\vec v_i$ relative to a fixed point of the rotation with angular velocity $\vec \omega$. By assumption of the rigidity:
$$
\vec v_i = \vec\omega \times \vec x_i
$$
and so the total angular momentum relative to the fixed point is
$$
\vec L = \sum_i m_i \vec x_i \times (\vec \omega \times \vec x_i)
= \sum_i m_i [r_i^2 \vec \omega - (\vec\omega\cdot \vec x_i)\vec x_i]
$$
where $r_i$ is the magnitude of $\vec x_i$. This is the usual expression involving what is essentially the definition of the moment of inertia relative to the fixed point.

We can now rewrite this by introducing a reference displacement $\vec x_0$ and the instantaneous rigid velocity $\vec v_0 = \vec\omega \times \vec x_0$ of that point. We introduce $\vec x_i = \vec y_i + \vec x_0$ and $\vec v_i = \vec u_i + \vec v_0$, where $\vec y_i$ and $\vec u_i$ are the position and velocity of mass $i$ relative to the reference displacement and velocity.

The angular momentum about the fixed point now takes the form
$$
\vec L = \sum_i m_i (\vec y_i + \vec x_0)\times (\vec \omega \times \vec y_i + \vec v_0)
= \sum_i m_i[R_i^2 \vec \omega - (\vec y_i\cdot \vec\omega)\vec y_i]
+ M \vec x_0 \times \vec v_0
+ \sum_i m_i [\vec x_0 \times (\vec\omega \times \vec y_i) + \vec y_i \times \vec v_0]
$$
with $R_i$ the magnitude of $\vec y_i$ and $M = \sum_i m_i$ (ie, the total mass). The first term here is now the angular momentum of the system relative to the reference displacement in the instantaneous rest frame of the reference displacement. The second term is the angular momentum of a single particle of mass $M$ at the reference displacement relative to the original fixed point. The last term is quite messy.

However, if you take the CoM as the reference displacement, then $\sum_i m_i \vec y_i = 0$ by definition and the last two terms vanish. Voilà, you have reexpressed the total angular momentum relative to the fixed point as the angular momentum relative to the CoM (in the CoM instantaneous rest frame) plus the angular momentum of a point mass at the CoM moving at the CoM velocity.

The advantage over the first expression is that the moment of inertia relative to the CoM is often much easier to express in terms of standard moments of inertia that have already been computed. Many times simplifying computation.

Iirc, this is also covered in some detail in chapter 10 of my book.

 

[sophiecentaur]

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

Why can't the wheels be perpendicular to the surface with an axle connecting their centers

They would have equal radii?? It may be a good idea just to look at the 'static' geometry first; you may have given yourself a problem with rotational matters by not looking at the question better (e.g. the figid fixing on the rod dictates what will happen.)

Will a cone rotate on a frictionless surface?

What would make it turn? (Or, rotate about what axis?)
@Orodruin has pointed you in the 'clever clogs' direction (post #32) which would sort out your questions probably in one go, I think.

 

[Aurelius120]

They would have equal radii?? It may be a good idea just to look at the 'static' geometry first; you may have given yourself a problem with rotational matters by not looking at the question better (e.g. the figid fixing on the rod dictates what will happen.)

In the 3rd post I explained the arrangement more.
I meant keep the discs perpendicular to the surface and connect their centers via a rod which is not horizontal.
@Lnewqban described that motion in post 4

What would make it turn? (Or, rotate about what axis?)

Isn't static friction not necessary for rolling? I remember reading (on stackexchange maybe) that when a translating object is placed on a surface kinetic friction causes it to roll after which friction doesn't have a role in the rotation or translation.

@Orodruin has pointed you in the 'clever clogs' direction (post #32) which would sort out your questions probably in one go, I think.

Except unfortunately I haven't been taught what a tensor is. All I know is they are half scalar and half vector or something in between

 

[sophiecentaur]

I meant keep the discs perpendicular to the surface and connect their centers via a rod which is not horizontal.

A sort of ball joint with splines at the centre of each disc? I think that would imply the need for a constant velocity joint, as in car steering arrangements.

when a translating object is placed on a surface kinetic friction

After a transitional period, at startup, there may be lateral slip which will involve energy loss ( I think). That's quite a step-up in difficulty imo. The distinction between static and kinetic friction

 

[Aurelius120]

The rotation about the CoM is the (instantaneous) motion in the CoM instantaneous rest frame. The separation into CoM angular momentum and angular momentum due to the CoM motion is simply a convenient way of doing the bookkeeping of the angular momentum as the terms separate.

Consider a system of rigidly connected masses $m_i$ with positions $\vec x_i$ and velocities $\vec v_i$ relative to a fixed point of the rotation with angular velocity $\vec \omega$. By assumption of the rigidity:
$$
\vec v_i = \vec\omega \times \vec x_i
$$
and so the total angular momentum relative to the fixed point is
$$
\vec L = \sum_i m_i \vec x_i \times (\vec \omega \times \vec x_i)
= \sum_i m_i [r_i^2 \vec \omega - (\vec\omega\cdot \vec x_i)\vec x_i]
$$
where $r_i$ is the magnitude of $\vec x_i$. This is the usual expression involving what is essentially the definition of the moment of inertia relative to the fixed point.

We can now rewrite this by introducing a reference displacement $\vec x_0$ and the instantaneous rigid velocity $\vec v_0 = \vec\omega \times \vec x_0$ of that point. We introduce $\vec x_i = \vec y_i + \vec x_0$ and $\vec v_i = \vec u_i + \vec v_0$, where $\vec y_i$ and $\vec u_i$ are the position and velocity of mass $i$ relative to the reference displacement and velocity.

The angular momentum about the fixed point now takes the form
$$
\vec L = \sum_i m_i (\vec y_i + \vec x_0)\times (\vec \omega \times \vec y_i + \vec v_0)
= \sum_i m_i[R_i^2 \vec \omega - (\vec y_i\cdot \vec\omega)\vec y_i]
+ M \vec x_0 \times \vec v_0
+ \sum_i m_i [\vec x_0 \times (\vec\omega \times \vec y_i) + \vec y_i \times \vec v_0]
$$
with $R_i$ the magnitude of $\vec y_i$ and $M = \sum_i m_i$ (ie, the total mass). The first term here is now the angular momentum of the system relative to the reference displacement in the instantaneous rest frame of the reference displacement. The second term is the angular momentum of a single particle of mass $M$ at the reference displacement relative to the original fixed point. The last term is quite messy.

However, if you take the CoM as the reference displacement, then $\sum_i m_i \vec y_i = 0$ by definition and the last two terms vanish. Voilà, you have reexpressed the total angular momentum relative to the fixed point as the angular momentum relative to the CoM (in the CoM instantaneous rest frame) plus the angular momentum of a point mass at the CoM moving at the CoM

And if COM is the fixed point, the angular momentum is the angular momentum in the COM instantaneous rest frame.
So beautiful.

The advantage over the first expression is that the moment of inertia relative to the CoM is often much easier to express in terms of standard moments of inertia that have already been computed. Many times simplifying computation.

Iirc, this is also covered in some detail in chapter 10 of my book.

Sounds like a great book. Is it for after, before or while engineering?

 

[Aurelius120]

@Orodruin
$$\vec L = \sum_i m_i (\vec y_i + \vec x_0)\times (\vec \omega \times \vec y_i + \vec v_0)$$ $$= \sum_i m_i[R_i^2 \vec \omega - (\vec y_i\cdot \vec\omega)\vec y_i]+ M \vec x_0 \times \vec v_0+ \sum_i m_i [\vec x_0 \times (\vec\omega \times \vec y_i) + \vec y_i \times \vec v_0]$$
In the first term, representing angular momentum about instantaneous COM rest frame,
The first part $\sum_i m_iR_i^2 \vec \omega$ is essentially $I\omega$
What happens to the second part? Does it cancel out ?
As stated by @TSny and @haruspex we only used $I\omega$ about two perpendicular axes to compute angular momentum.

 

[Orodruin]

Is it for after, before or while engineering?

It is a mathematical methods textbook. Chapter 10 discusses some applications of the previous chapters - in this case vector algebra/analysis. Would probably be considered late undergrad level.

 

[Orodruin]

@Orodruin
$$\vec L = \sum_i m_i (\vec y_i + \vec x_0)\times (\vec \omega \times \vec y_i + \vec v_0)$$ $$= \sum_i m_i[R_i^2 \vec \omega - (\vec y_i\cdot \vec\omega)\vec y_i]+ M \vec x_0 \times \vec v_0+ \sum_i m_i [\vec x_0 \times (\vec\omega \times \vec y_i) + \vec y_i \times \vec v_0]$$
In the first term, representing angular momentum about instantaneous COM rest frame,
The first part $\sum_i m_iR_i^2 \vec \omega$ is essentially $I\omega$

It is $I\omega$ (relative to the com) up to the second part.

What happens to the second part? Does it cancel out ?

It is the rest of $I\omega$.

Edit: Note that
$$
I_{ab} = \sum_i m_i (R_i^2 \delta_{ab} - y_{i,a} y_{i,b})
$$ by definition.

 

[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.

Order of rotation (as in CW or ACW?) doesn't matter for infinitesimal angles?
Physically it is tough to visualise for sure.

The angular momentum about the fixed point now takes the form
$$\vec L = \sum_i m_i (\vec y_i + \vec x_0)\times (\vec \omega \times \vec y_i + \vec v_0)$$ $$= \sum_i m_i[R_i^2 \vec \omega - (\vec y_i\cdot \vec\omega)\vec y_i]+ M \vec x_0 \times \vec v_0+ \sum_i m_i [\vec x_0 \times (\vec\omega \times \vec y_i) + \vec y_i \times \vec v_0]$$

But mathematically, this formula proves that it is independent of sign of position vector, right?

 

[haruspex]

Order of rotation (as in CW or ACW?) doesn't matter for infinitesimal angles?
Physically it is tough to visualise for sure.

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.

 

[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.