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I'm trying to understand the concept of angular velocity. I read this paragraph on Wikipedia, which asserts that if a point $p$ has angular velocity $u$ within a coordinate frame $F_1$ which itself has angular velocity $v$ within some other frame $F_2$, then the point has angular velocity $u+v$ with respect to the second frame.

I want to set aside the specific formula $u+v$ and just focus on the abstract form $f(u, v)$ - in other words, just the fact that the angular velocity of $p$ with respect to $F_2$ can be deduced from the angular velocities of $p$ with respect to $F_1$ and of $F_1$ with respect to $F_2$. That seems like it ought to be impossible to me. In the diagram below, $C$ has the same angular velocity $u$ with respect to $B$ in both figures and $B$ has the same angular velocity $v$ with respect to $A$ in both figures, yet in the first figure $C$ has angular velocity $u+v$ with respect to $A$, whereas in the second figure $C$ has angular velocity $u-v$, zero if we set $u=v$.

enter image description here

What am I failing to understand?

Qmechanic
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Jack M
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  • Just note that angular velocity applies to all points on a rigid body and that a single point does not have angular velocity. Also, note that learning from Wikipedia is tricky because it is very inconsistent and sometimes erroneous. – John Alexiou Apr 29 '20 at 12:26
  • Can you please explain what the figures represent. There are circles with points and arrows which appear tangential but could indicate out of plane. Where is $u$ and $v$ in the figures? – John Alexiou Apr 29 '20 at 12:28
  • @ja72 Everything is happening in a flat plane. $C$ is rotating around $B$, $B$ is rotating around $A$. The arrows attached to $B$ and $C$ represent their velocity vectors and are $u$ and $v$ respectively. – Jack M Apr 29 '20 at 13:06
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    The vectors $u$ and $v$ are described as rotational velocity in the question, but drawn as translational velocities in the diagram. Please clarify. Your question is still not clear to me. I want to help, but I don't know what exactly are you asking. – John Alexiou Apr 29 '20 at 14:38
  • @ja72 "Also, note that learning from Wikipedia is tricky because it is very inconsistent and sometimes erroneous." Wiki does contains some errors but compared to other encyclopedic resources it does quite well. – Gert Apr 29 '20 at 14:41
  • @ja72 The angular velocity of a point $A$ relative to a point $B$ is determined by the axis of rotation and the radians per second, right? In this case everything is happening in a plane, so we only have to worry about radians per second. – Jack M Apr 29 '20 at 14:54
  • @ja72 My understanding of that Wikipedia article is that if we know the angular velocity of $A$ about $B$ and of $B$ about $C$, then we can determine the angular velocity of $A$ about $C$. In figures 1 and 2, the angular velocity of $B$ about $A$ and of $C$ about $B$ are both the same (in each case they are rotating clockwise with some given speed), and yet sure the angular velocity of $C$ about $A$ is different in each figure. – Jack M Apr 29 '20 at 14:54
  • Per my first comment. There no such thing as angular velocity of point about another point. Angular velocity is a property of the entire reference frame or body, and where the center of rotation is does not affects its value. Also read this wikipedia section I authored to show the relationship between absolute and relative centers of rotation. – John Alexiou Apr 29 '20 at 16:13
  • @ja72 I'm reading Physics For Mathematicians by Spivak in which he defines "the angular velocity of a particle" to be the cross-product $c\times v$ where $c$ is the position of the particle and $v$ its velocity. Is this concept usually known by a different name? – Jack M Apr 29 '20 at 16:43
  • By definition a single particle has no spatial dimensions (it's a point of zero size) and hence rotation of a point is nonsensical. Rotation arises when a collection of particles (the body) move together such that their distances remain fixed. See Chasles theorem. Rotational velocity is the fixed vector $\vec{\omega}$ that is used to describe the individual velocity of each particle in a rigid body as $$\vec{v} = \vec{\omega} \times \vec{r}$$ – John Alexiou Apr 29 '20 at 16:54
  • @ja72 I always thought angular velocity was just a useful convention for representing a uniform circular motion. You take a vector parallel to the axis of rotation, and set its magnitude equal to the radians per second of the rotation. That's always how I understood it at school. Does that concept not exist at all? – Jack M Apr 29 '20 at 17:10
  • You are taking about representing orbital motion with a rotation. That only works when the radial distance is constant, and hence the frame is rotating, and the one particle is riding along with the frame. We're still talking about the rotational velocity vector as defined above. – John Alexiou Apr 29 '20 at 18:30
  • Let us continue this discussion in chat. – John Alexiou Apr 29 '20 at 18:54

1 Answers1

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You are asking about the algebra of angular velocities, especially as it relates to the area of kinematics.

Consider the tea-cup ride at Disney. You have a platter (orange) rotating about A with angular speed $\Omega$. At some radius $R$ away form the center a 2nd object (green) is pinned at point B with relative speed $\dot \theta$. This is to denote that $\theta$ is an orientation angle. On this object there are two attached points, C and D, that at some instance in time are located as shown below:

teacups

  • What is the rotational velocity of A?

    Every particle attached to the platter moves with translational velocity except the center of rotation. Nevertheless we can state that the body rotates with $\Omega$ and the translational speed of each particles is given by $$ v= R\, \Omega $$ where $R$ is the radial distance from the center of rotation. You can say that B rotates about A, but in reality the entire body is rotating about A. A more accurate description would be that B orbits A.

  • What is the rotational velocity of B?

    This depends on which body is B belong to. The bottom of the pin connecting the two bodies rotates with $\Omega$ and the top of the pin with $$\omega = \Omega + \dot\theta$$ since $\dot \theta$ is the relative motion between the two bodies.

  • What is the translational velocity of points A, B, C and D.

    • $v_A = 0$

      since it is on a fixed point

    • $v_B = R\,\Omega$

      as we saw before

    • $v_C = v_B + \omega \, r = \Omega \,R + (\Omega + \dot \theta) \, r = \Omega ( R+r) + \dot \theta \, r $

      To be interpreted as the velocity of the platter at C plus the relative velocity of the second body.

    • $v_D = v_B - \omega \, r = \Omega \,R - (\Omega + \dot \theta) \, r = \Omega ( R-r) - \dot \theta \, r $

      To be interpreted as the velocity of the platter at C minus the relative velocity of the second body.

As you can see, relative translational velocities switch signs between C and D. This is because the position of the point is either radially out or radially in from the reference point B.

On the contrary rotational velocity of the second body is only added the motion of the platter, because the location of the pin at B bears no significance in the rotational kinematics.

John Alexiou
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  • Thanks very much. I don't understand the equation $\omega=\Omega+\dot\theta$. What exactly is $\omega$? I don't understand how it can be the rotational velocity of the top of the pin, which is surely just $\Omega$. – Jack M Apr 29 '20 at 21:56
  • Ok we are getting somewhere now. $\omega$ is the rotational velocity of the green body. That is if you want to find the translational velocity of C relative to B you need to do $$ v_C - v_B = \omega , r $$ Rotational velocities add vectorially, and in this case you add the green body rot. velocity to the relative rot. velocity to get the rot. velocity of the green body. Or otherwise the relative rot. velocity is $$ \dot \theta = \omega - \Omega$$ – John Alexiou Apr 29 '20 at 22:55
  • So just to make sure I understand,are $\omega$, $\Omega$ and $\dot\theta$ all vectors? And the expression $\omega r$ in your previous comment is to be interpreted as a cross product, where $r=C-B$? – Jack M May 01 '20 at 11:36
  • Yes, The true 3D kinematics of a pin joint where $B^\star$ is the bottom of the joint (riding on orange body) is $$ \vec{\omega} = \vec{\Omega} + \hat{z} \dot \theta $$ and $$\vec{v}B = \vec{v}{B^\star}$$ – John Alexiou May 01 '20 at 12:49
  • But then surely the correct equation should be $v_C-v_B=\dot\theta\hat z\vec r$, no? Isn't the point $C$ orbiting the point $B$ at $\dot\theta$ radians per second? – Jack M May 01 '20 at 13:02
  • The rigid transformation of velocities is $$ \vec{v}{B^\star} = \vec{v}_A + \Omega \hat{z} \times \vec{R} $$ and $$ \vec{v}{C} = \vec{v}_B + \omega \hat{z} \times \vec{r}$$ – John Alexiou May 01 '20 at 22:05
  • If I don't understand after this I'll stop pestering you and seek out a book with a chapter on angular velocity - but could you please provide a similar equation to those two, but with $\dot\theta$ playing the role of $\Omega$/$\omega$? Because I'm clearly wildly misunderstanding what $\dot\theta$ represents physically, since I would have it in place of $\omega$ in your second equation. In any case, thanks for all your time. – Jack M May 01 '20 at 22:38
  • $\theta$ is the orientation angle of the green part relative to the orange part. Draw a line across both of them at some point, and then later measure the angle $\theta$ that the two lines segments make. $\dot \theta$ is the speed at which this angle changes. If both objects have the same angular velocity, then $\dot \theta = 0$. – John Alexiou May 01 '20 at 23:36