How do we formally define the "angular velocity vector" of a point about an axis?

Question

Context:

For instance, the quantity Torque, $\vec{\tau}$ is defined about a point: by the formula $$\vec{\tau}=\vec{r}\times\vec{F}$$ We can use this defintion to define torque about an axis. Let the axis be along the vector $\vec{n}$. If the torque of a force about a point on the axis,=$\vec{\tau_1}$=$\vec{r}\times\vec{F}$, Then, torque of the same force about another point on the axis$=$$\vec{\tau_2}$$=$ $(\vec{r}+\lambda\vec{n})\times\vec{F}$. Clearly, $\vec{\tau_1}.\vec{n}=\vec{\tau_2}.\vec{n}$.

The component of torque about the axis direction, is the same for any point on the axis, and this component is what is defined as the torque about an axis.

I believe, that a similar treatment fails for angular velocity. I present an example:

Two thin circular discs of mass $m$ and $4 m$, having radii of $a$ and $2 a$, respectively, are rigidly fixed by a massless, rigid rod of length $l=\sqrt{24} a$ through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is $\omega$. The angular momentum of the entire assembly about the point ' $O$ ' is $\vec{L}$ (see the figure).

Let $\theta$ be the angle made by the massless rod with the horizontal.The statement(which we had to check):

The Center of mass of the system rotates about the Z axis with an angular speed $\omega/5$.

The solution presented:

The center of mass of the system is at a distance $9l/5$ from point $o$.If angular velocity vector of the center of mass relative to point $o$ is $\vec{\Omega}$,Then $\vec{\Omega}$ will be at an angle $\theta$ from the z-axis. Since the velocity of the center of mass is the disc is $\dfrac{m(a\omega)+4m(2a\omega)}{m+4m}$=$(9/5)a\omega$, it follows that $|\vec{\Omega}|$ =$\dfrac{|\vec{r}\times\vec{v}|}{|\vec{r}|^2}$=$\dfrac{|\vec{r}||\vec{v}|}{|\vec{r}|^2}$=$\dfrac{|\vec{v}|}{|\vec{r}|}$=$a\omega/l$, And thus $\vec{\Omega}.\hat{k}$=$|\vec{\Omega}|\cos\theta=a\omega\cos(\theta)/l=\omega/5$.

My issue with the solution:It seems that they have calculated the z-component of angular velocity of the Center of mass, relative to point $o$. The question asked us to find the "angular velocity vector of the center of mass" about the z axis.

I fail to see how these two are equivalent. The equivalence of these two statements implies that the component about the axis,of the angular velocity vector relative to any point on the axis,is the same.(a concept which worked for torque). This is false, it can be easily shown(using the same procedure as torque).The z component of the Angular velocity vector of the Center of mass are different relative to different point for the Z axis, If we go by the following formula:

$$\vec{\omega}=\dfrac{\vec{r}\times\vec{v_{a}}}{|\vec{r}|^2}$$

Where ${\vec{r}}$ is the vector joining $a$ to $b$. The treatment which worked for torque, fails for angular velocity. Hence, the question.

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Instantaneous axis of rotation and rolling cone motion presents a very similar question. As per its last paragraph, in our example, Velocities of the centers of the discs ** won't** be $a\omega$ and $2a\omega$!! So it seems to me that the presented solution is completely wrong.

The statement in the bold is still un-answered.

Answer 1

How do you formally define angular velocity?

Imagine a rigid body whose center of mass is fixed, and it is free to rotate about it. What are all the allowed motions?

By definition, a rigid body has all the distances between separate particles fixed.

Lemma 1 Image two arbitrary particles on a rigid body with locations $\boldsymbol{r}_i$ and $\boldsymbol{r}_j$ which move over time. The allowed motions are such that the relative velocities must be perpendicular to the separation between the particles $$ (\boldsymbol{v}_i - \boldsymbol{v}_j) \cdot ( \boldsymbol{r}_i - \boldsymbol{r}_j ) = 0 \tag{1}$$

_{Here $\cdot$ is the vector dot product, and boldface letters are vectors.}

Proof

The constant distance (squared) between the points is

$$ d_{ij}^2 = ( \boldsymbol{r}_i - \boldsymbol{r}_j ) \cdot ( \boldsymbol{r}_i - \boldsymbol{r}_j ) \tag{2}$$

Take the time derivative and set it to zero

$$ \frac{\rm d}{{\rm d}t} d_{ij}^2 = 0 \tag{3}$$

Using the product rule

$$ \frac{\rm d}{{\rm d}t} d_{ij}^2 = \frac{\rm d}{{\rm d}t}( \boldsymbol{r}_i - \boldsymbol{r}_j ) \cdot ( \boldsymbol{r}_i - \boldsymbol{r}_j ) + ( \boldsymbol{r}_i - \boldsymbol{r}_j ) \cdot \frac{\rm d}{{\rm d}t}( \boldsymbol{r}_i - \boldsymbol{r}_j ) = 2 ( \boldsymbol{r}_i - \boldsymbol{r}_j ) \cdot \frac{\rm d}{{\rm d}t}( \boldsymbol{r}_i - \boldsymbol{r}_j ) = 0$$

and finally divide by 2 and use $\frac{\rm d}{{\rm d}t} \boldsymbol{r} = \boldsymbol{v}$ to get

$$ ( \boldsymbol{r}_i - \boldsymbol{r}_j ) \cdot ( \boldsymbol{v}_i - \boldsymbol{v}_j ) = 0 \;\checkmark $$

Lemma 2 The only allowed relative motion between two particles is described by a single constant vector $\boldsymbol{\omega}$ which results in velocities perpendicular to it and the separation $$ \boldsymbol{v}_i-\boldsymbol{v}_j = \boldsymbol{\omega} \times ( \boldsymbol{r}_i - \boldsymbol{r}_j ) \tag{4}$$

_{Here $\times$ is the vector cross product.}

Proof

Substitute (4) into (1) to get

$$ ( \boldsymbol{r}_i - \boldsymbol{r}_j) \cdot \boldsymbol{\omega} \times \left((\boldsymbol{r}_i-\boldsymbol{r}_j) \right) \tag{5}$$

Using $\boldsymbol{r}_{ij} = \boldsymbol{r}_i - \boldsymbol{r}_j$ the above is

$$ \boldsymbol{r}_{ij} \cdot \left( \boldsymbol{\omega} \times \boldsymbol{r}_{ij} \right) = 0 \,\checkmark $$

The is a hidden implication here. Since i and j are arbitrary and the above expressions must be true for all pairs of particles, this implies that there is at least one fixed $\boldsymbol{\omega}$ which satisfies (1) since the velocity field seen below in (6) solves (1). This does not exclude the possibility of other varying vectors $\boldsymbol{\omega}_{\rm ij}$ that satisfy (1).

_{In some ways think of $\boldsymbol{\omega}$ as a shortcut to describe the state of motion of the rigid body as the next Lemma shows. But the uniqueness of $\boldsymbol{\omega}$ actually comes from the time derivative on a rotating frame where-by using geometry an expression for the rotation of a vector is developed and then when the time derivative is evaluated the uniqueness of a single rotation axis becomes the uniqueness of $\boldsymbol{\omega}$.}

Lemma 3 The vector $\boldsymbol{\omega}$ describes the direction and magnitude of rotation of the rigid body, which in term is used to find the velocity vector of all points on the body, given the velocity of one point.

Proof If we know the velocity of point j then the velocity of point i is given by (4)

$$ \boldsymbol{v}_i = \boldsymbol{v}_j + \boldsymbol{\omega} \times (\boldsymbol{r}_i - \boldsymbol{r}_j) \,\checkmark $$

Lemma 4 If one point undergoes pure translation, then all other points will retain the velocity component parallel to the rotation axis. The parallel component can be described as a proportion of the rotational velocity.

Proof Again if the motion of point j is known as $\boldsymbol{v}_j = h\,\boldsymbol{\omega}$ where $h$ is a scalar value, then the velocity of all other points are

$$ \boldsymbol{v}_i = h\,\boldsymbol{\omega} + \underbrace{\boldsymbol{\omega} \times (\boldsymbol{r}_i-\boldsymbol{r}_j) }_{\text{always perpendicular to }\boldsymbol{\omega}} \tag{6}$$

Lemma 5 In reverse, given the general velocity vector $\boldsymbol{v}_j$ if a known location $\boldsymbol{r}_j$, one can find at least one location in space $\boldsymbol{r}_i$ whose velocity vector is strictly parallel to the rotation vector. This describes the instant axis of rotation and is found with $$\boldsymbol{r}_i = \boldsymbol{r}_j + \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_j}{ \| \boldsymbol{\omega} \|^2} \tag{7}$$

_{Here $\| \boldsymbol{\omega} \|$ is the rotational speed, and $\|\boldsymbol{\omega}\|^2 = \boldsymbol{\omega} \cdot \boldsymbol{\omega}$.}

Proof Use (7) in (4) to show that only $\boldsymbol{v}_i = h\,\boldsymbol{\omega}$ is allowed

$$\boldsymbol{v}_i - \boldsymbol{v}_j = \boldsymbol{\omega} \times \left( \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_j }{\| \boldsymbol{\omega} \|^2} \right) = \frac{ \boldsymbol{\omega} (\boldsymbol{\omega} \cdot \boldsymbol{v}_j) - \boldsymbol{v}_j ( \boldsymbol{\omega} \cdot \boldsymbol{\omega}) }{\| \boldsymbol{\omega} \|^2} $$

_{Here I am using the vector triple product identity $a \times ( b \times c) = b(a \cdot c) - c (a \cdot b)$.}

$$\boldsymbol{v}_i - \boldsymbol{v}_j = \frac{ \boldsymbol{\omega} (\boldsymbol{\omega} \cdot \boldsymbol{v}_j) }{\| \boldsymbol{\omega} \|^2} - \boldsymbol{v}_j$$

$$ \boldsymbol{v}_i = \left(\frac{ \boldsymbol{\omega} \cdot \boldsymbol{v}_j }{\| \boldsymbol{\omega} \|^2} \right) \boldsymbol{\omega} = h\,\boldsymbol{\omega}\,\checkmark$$

Lemma 6 The parallel scalar (pitch) value is found from the motion of an arbitrary point j and the rotation with $$ h = \frac{ \boldsymbol{\omega} \cdot \boldsymbol{v}_j }{\| \boldsymbol{\omega} \|^2} \tag{8} $$

Proof See proof of previous lemma.

Answer 2

For a point, you do talk about its velocity. By abuse of language, for an axis aligned with $\vec{n}$, I've seen people refer to angular velocity of the point about the axis as $\dot{\theta}\hat{n}$, that is, if you express the coordinates of the point in a cylindrical coordinate system with coordinates $(\rho, \theta, z)$ in which $\hat{n}$ is aligned with the cylindrical axis.

As far as I understand, this is not standard and generally rather ambiguous.

There is a physical way to imagine this, though. Imagine a/an (infinitely extended) rigid body which can perform only screw motion and/or rotation about the given axis, in a manner such that the given moving point is stationary wrt to this rigid body. Then, the "angular velocity of the point defined wrt the axis" is identical to the angular velocity of the aforementioned rigid body.

However, I'd say it's best to avoid talking about angular velocity of a point defined this way, since it depends on the choice of the origin (even in the same reference frame).

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On the other hand, for a rigid body, there is a very precise notion of angular velocity, which doesn't depend on the choice of the origin as long as you stick to the same frame.

To define the angular velocity of a rigid body, you need to know the velocity field $\mathbf{v}(\mathbf{r})$ in a given frame. The rigidity constraint then implies that in a Cartesian coordinate system the velocity field can be decomposed as,

$$\mathbf{v}(\mathbf{r}) = \mathbf{v}_0+\mathbf{\omega}\times(\mathbf{r}-\mathbf{r}_0).$$

It can be shown that $\omega$ is independent of the origin of the coordinate system (eg. cf. Landau-Lifshitz Mechanics).

So, you can see that $\mathbf{\omega}$ is a quantity that comes out of the collective motion of all points on the rigid body. And this is the standard, commonly-accepted definition of angular velocity for a rigid body.

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ABUSE OF LANGUAGE

The question talks about "the angular velocity of a rigid body about the 'axis'" - this is a meaningless phrase if you do not specify the frame of reference. For instance, one could switch to another frame rotating about the axis with some angular velocity $\Omega$. If you do so, the angular velocity of the rigid body about the axis would change, even though the said 'axis' wouldn't acquire any additional velocity in the changed frame of reference.

Long story short: An axis doesn't specify a reference frame. You need two more perpendicular axes (and what they are doing) to complete the story.

I believe the question, as it stands, has plenty of abuse of language and is fairly ambiguous.

Answer 3

Angular velocity vector of a point $A$, relative to a stationary point $B$:

$\vec\omega_B[ \, A \, ] = \frac{(\vec r_A - \vec r_B) \times \vec v_B[ \, A \, ]}{(| \vec r_A - \vec r_B |)^2} $

Right. (And I hope you don't mind my choice of notation.)

Correspondingly, the velocity vector of $A$ relative to (the inertial system containing) $B$ can be decomposed as

$\vec v_B[ \, A \, ] = \vec v_B[ \, A \, ]^{\text{(radial)}} + \vec v_B[ \, A \, ]^{\text{(tangential)}} $,

where

$\vec v_B[ \, A \, ]^{\text{(radial)}} := (\vec r_A - \vec r_B) \frac{\vec v_B[ \, A \, ] \cdot (\vec r_A - \vec r_B)}{(| \vec r_A - \vec r_B |)^2}$,

such that also

$\vec\omega_B[ \, A \, ] = \frac{(\vec r_A - \vec r_B) \times \vec v_B[ \, A \, ]^{\text{(transversal)}}}{(| \vec r_A - \vec r_B |)^2}$.

How do we formally define the "angular velocity vector" of a point about an axis?

For a straight, thin axis with direction $\vec x$ and point $P$ on the axis which is (instantaneously) closest to $A$, i.e. such that

$\vec x \, \cdot \, (\vec r_A - \vec r_P) = 0$,

I'd suggest

$\vec\omega_P[ \, A \, ]^{(\text{direction } \vec x)} := (\vec x) \frac{\vec x \, \cdot \, \vec\omega_P[ \, A \, ]}{(| \vec x |)^2}.$

Accordingly the tangential velocity vector of $A$ relative to $P$ can be further decomposed as

$\vec v_P[ \, A \, ]^{\text{(tangential)}} = \vec v_P[ \, A \, ]^{(\text{tang. along } \vec x)} + \vec v_P[ \, A \, ]^{(\text{tang. across } \vec x)},$

such that

$\vec\omega_P[ \, A \, ]^{(\text{direction } \vec x)} = \frac{(\vec r_A - \vec r_P) \times \vec v_P[ \, A \, ]^{(\text{trans. across } \vec x)}}{(| \vec r_A - \vec r_P |)^2}$.

While the component $ \vec v_P[ \, A \, ]^{(\text{tang. along } \vec x)} = 0$, point $P$, being the axis point closest to $A$, remains fixed. (Which may be convenient in certain calculations.)

In reference to some other point $Q$, which also belongs to the axis under consideration, and which may be conveniently fixed, while point $P$ is defined only instantaneously and may be changing due to non-zero $\vec v_P[ \, A \, ]^{(\text{tang. along } \vec x)} $

$ \vec r_A - \vec r_P = (\vec r_A - \vec r_Q) - (\vec r_P - \vec r_Q) = (\vec r_A - \vec r_Q) - (\vec x) \frac{(\vec r_A - \vec r_Q) \, \cdot \, \vec x}{(| \vec x |)^2},$

and the angular velocity vectors $\vec\omega_P[ \, A \, ]$ as well as $\vec\omega_P[ \, A \, ]^{(\text{direction } \vec x)}$ may be accordingly expressed in refrence to $Q$.

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p.s.

As I present an example: [...] $|\vec \Omega| = \omega / 5$.

I find this rather obvious just by considering geometry and kinematics:

The "rolling radius around point $O$" of the (instantaneous) contact point of the small disk with the table surface is $R_{sd} = \sqrt{ \ell^2 + a^2 } = \sqrt{ (\sqrt{24}~a)^2 + a^2 } = 5~a$, which is obviously $5$ times the radius of the small disk.

Likewise is the "rolling radius around point $O$" of the (instantaneous) contact point of the large disk with the table surface $R_{ld} = \sqrt{ (2~\ell)^2 + (2~a)^2 } = 10~a$, i.e. $5$ times the radius of the large disk.

Consequently it takes five full rotations of the two-disk cone, around its axis, in order to complete one full round rolling on the table around point $O$.