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Angular velocity vector of a point $a$,relative to a stationary point $b$: $$\vec{\omega}=\dfrac{\vec{r}\times\vec{v_{a}}}{|\vec{r}|^2}$$

Where ${\vec{r}}$ is the vector joining $a$ to $b$.

My question is:

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

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, As I present an example: enter image description here

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 formula mentioned in the beginning. The treatment which worked for torque, fails for angular velocity.

Which brings me back to my question:

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

Edit: 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 wont 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.

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  • $\begingroup$ If $\vec r$ goes from a to b, you can choose any axis that start at point a and not parallel to vector r for the angular velocity axis $\endgroup$ – Eli Jun 7 at 13:07
  • $\begingroup$ @Eli my question was for a given axis. I am not asking "what axis should I chose". I am asking for a definition for a given axis. $\endgroup$ – satan 29 Jun 7 at 14:19
  • $\begingroup$ read "my problem with the solution" , This really highlights what i wanted to ask $\endgroup$ – satan 29 Jun 7 at 14:26
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    $\begingroup$ Just to point out that points do not have angular velocity. $\vec{\omega}$ is a property of an entire rigid body and it is considered shared among all points on the body as well as the extended body (rotating frame attached to the body). $\endgroup$ – John Alexiou Jun 7 at 15:34
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    $\begingroup$ Although I can see what you mean: If a rigid body is rotating with some angular velocity , then each point on the rigid body rotates relative to any other point on the rigid body, with the same angular velocity: and this is what is called the angular velocity of the rigid body.This is "intrinsic".(this is what $\vec{\omega}$ in the problem is) $\endgroup$ – satan 29 Jun 7 at 16:32
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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.

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  • $\begingroup$ I had a similar kind of doubt and probably will write a new post on it too soon. I was referring the the book Classical Mechanics, NC Rana and P Joag. They do get an expression for $\omega$ about a point ( say A) in just terms of cross and dot products of radius vector of any 2 points in the rigid body from the point A and go on to prove that $\omega$ for any point in the body is the same about A. But then they state that $\omega$ for any point is even same about any other point. Now this result is what I am trying to prove since they haven't. I couldn't find in in Goldstein. $\endgroup$ – Shashaank Jun 12 at 6:33
  • $\begingroup$ The result doesn't look intuitive to me too. The $\omega$ of any point about the point A should be the same, that is ok. But why should $\omega$ be the same when considered about any other point, say B. I cannot imagine why it should be the same. Can you plz help me or refer a text which discusses these points in detail. Does Landau Lifschitz discuss this in detail. I ask you since I have read few answers of yours on rigid body rotation. I will post a new question precisely asking my doubt. If you could help me or give a reference of a text from where your above proof is it would be great $\endgroup$ – Shashaank Jun 12 at 6:41
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    $\begingroup$ @ja72 That's a nice answer for a rigid body (although, I think some more work is required to argue that $\omega_{i,j}$ is independent of $i,j$). However, I think the OP is concerned about 'angular velocity of a point about an axis' which arguably is a poor choice of a phrase. $\endgroup$ – Vivek Jun 12 at 8:39
  • $\begingroup$ @Shashaank - Are you arguing that Lemma 3 needs more work to show how a single unique $\omega$ satisfies (1)? $\endgroup$ – John Alexiou Jun 12 at 11:48
  • $\begingroup$ @ja72 yeah precisely. I want to prove the uniqueness of $\omega$. I will ask a new question either today or tomorrow giving the "half proof " I have and the difficulty in intuitively accepting the uniqueness of $\omega$. If you wish you might to like to answer it. But yes in short this is what I wish to prove more precisely. $\endgroup$ – Shashaank Jun 12 at 11:54
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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).


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.


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.

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  • $\begingroup$ I had a similar kind of doubt and probably will write a new post on it too soon. I was referring the the book Classical Mechanics, NC Rana and P Joag. They do get an expression for $\omega$ about a point ( say A) in just terms of cross and dot products of radius vector of any 2 points in the rigid body from the point A and go on to prove that $\omega$ for any point in the body is the same about A. But then they state that $\omega$ for any point is even same about any other point. Now this result is what I am trying to prove since they haven't. I couldn't find in in Goldstein. $\endgroup$ – Shashaank Jun 12 at 9:01
  • $\begingroup$ The result doesn't look intuitive to me too. The $\omega$ of any point about the point A should be the same, that is ok. But why should $\omega$ be the same when considered about any other point, say B. I cannot imagine why it should be the same. Can you plz help me or refer a text which discusses these points in detail. Does Landau Lifschitz discuss this in detail. If you could help me or give a reference of a text from where I can find a proof for above is it would be great. $\endgroup$ – Shashaank Jun 12 at 9:03
  • $\begingroup$ @Shashaank It's a standard result. You can check Landau-Lifshitz Mechanics or J. L. Synge Tensor Calculus (which is a rather old fashioned but nice book). You can also try proving it in 3D using the eq 4 of jak72, but with $\omega_{i,j}$, indicating $\omega$ is possibly dependent on $i, j$. Then it's a three liner proof from there to argue that $\omega_{i,j}$ must be independent of $i,j$. $\endgroup$ – Vivek Jun 12 at 9:37
  • $\begingroup$ @Shashaank Goldstein uses Euler's theorem. So he rigorously shows you can write $\vec{v} = \vec{v}_A + \omega_A \times(\vec{r}-\vec{r}_A)$ for a rigid body velocity field wrt point $A$ on the rigid body. You can follow Landau-Lifshitz's argument from here on (i.e. do the same for a point $B$ and then show $\omega_A = \omega_B$). Note you really need a 'non-trivial rigid body' for the argument to go through (i.e. a point particles or linear rigid bodies don't have a well-defined angular velocity). $\endgroup$ – Vivek Jun 12 at 9:45
  • $\begingroup$ Ok let me see the Landau Lifshitz one. Actually Rana and Joag ( if you have read that) use a sort of geometrical proof whereas Goldstein builds on orthogonal transformations. I will check these. Actually, I guess I will ask a separate question as to proving this result. You may have a look at it if you wish to answer it. $\endgroup$ – Shashaank Jun 12 at 10:32
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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 $Q$.


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 $O$.

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  • $\begingroup$ Which textbook could I refer to for that particular formula for the angular velocity you state in the 1st line.... $\endgroup$ – Shashaank Jun 12 at 11:57
  • $\begingroup$ @Shashaank: "Which textbook could I refer to for that particular formula for the angular velocity you state in the 1st line...." -- Having noticed your question only a few hours ago, I report back, for the time being, that I have not found this particular explicit formula in "Goldstein" or "Landau/Lifshitz". For what it's worth, the formula (in some notation) appears in Wikipedia. (And Landau/Lifshitz is mentioning the related "radial velocity component" at least in passing. ...) $\endgroup$ – user12262 Jun 17 at 22:20
  • $\begingroup$ I did try to find this. I am trying to learn concepts in rigig body dynamics like why the angular velocity has to be the same about any axis whether inside the body or outside and how to find the angular velocity if one doesn't know the instantaneous axis of rotation. Can you refer a good book apart from Goldstein. I didn't find it much helpful... If you see my question could you please help me... $\endgroup$ – Shashaank Jul 3 at 5:00
  • $\begingroup$ As a side note this is a specific question of mine physics.stackexchange.com/a/559293/113699 ... If you would like to help me in this would be great or if you can suggest a text or a resource where I can find the answer to it, would be very helpful because I haven't been able to understand it.... $\endgroup$ – Shashaank Jul 3 at 5:06
  • $\begingroup$ @Shashaank: "[...] to find the angular velocity if one doesn't know the instantaneous axis of rotation." -- Seems an interesting problem, depending on what's given or assumed. Would you need to assume that radial and/or axial motion vanishes ? ... "Can you refer a good book apart from Goldstein." -- All my other Mechanics books are in German (including the Landau/Lifshitz ;). Besides: I'm into Kinematics (and Geometry); leaving Dynamics (Variational Calculus) for subsequent, perhaps more focussed study. Besides: PSE beats books, IMHO. "a specific question of mine" -- Only in fall, sorry. $\endgroup$ – user12262 Jul 3 at 21:14

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