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Let me explain my doubt with help of an example consider a disc of radius $2R$ rotating with contant angular velocity $\omega$ locate two particles $A$ at a distance $R$ from centre and $B$ at a distance $2R$ from the centre, we know that $\omega_A=\omega$ and $\omega_B=\omega$ ,Also $v_A=R\omega$ and $v_B=2R\omega$ , now $\omega_{B/A}=\dfrac{v_{B/A}}{r_{B/A}}=\omega$ but if I use the definition $\omega=\dfrac{d{\theta}}{dt}$ we get a contradicting result, $\omega_{B/A}=\dfrac{d{\theta}}{dt}=0$ since $\theta$ is the angle between the line joining them which does not change with time. My doubts

  1. Is $\omega=\dfrac{d{\theta}}{dt}$ always valid , even for rotating frames, If yes how do I derive ralative angular velocity between the two particles only using this definition?
  2. Am I making any conceptual error?

EDIT 1 : For a rigid body we can say each point turns relative to other point of the rigid body with same angular velocity at a given instant. So $\omega_{B/A}$ is indeed equal to $\omega$.Then why the contradiction occurs?

EDIT 2: https://www.brown.edu/Departments/Engineering/Courses/En4/notes_old/RigidKinematics/rigkin.htm Please check this link and see under 5.1.3 section where they derive equations governing general plane motion , there also they have used $\omega _{B/A}= \omega$ of the rigid body where A and B are any two arbitrary points.

EDIT 3: See that the equation $v_{A}=v_{B}+\omega_{B/A}×r_{A/B}$ where $\omega_{B/A}=\omega$ furthermore if we transform the equation we can se $v_{B}=v_{A}+\omega×r_{B/A}$ hence cocluding my point that $\omega_{B/A}=\omega_{A/B}=\omega$

EDIT 4: you can also see what I am trying to say here https://en.m.wikipedia.org/wiki/Angular_velocity under Rigid Body Consideration , Consistency.

EDIT 5: Here are some more links to support $\omega_{B/A}=\omega$

  1. Relative angular velocity of point with respect to another point
  2. Angular velocity about an arbitrary point

"Consider a rigid body rotating with angular velocity $\omega$ . Now, we know that this $\omega$ is an intrinsic property for the rigid body, in the sense that: Each point on the rigid body rotates with $\omega$ relative to any other point on the rigid body."

  1. Relative angular velocity

  2. "An important feature of spin angular momentum is that it is independent of the coordinate system. In this sense it is intrinsic to the body; no change in coordinate system can eliminate spin, whereas orbital angular momentum disappears if the origin is chosen to lie along the line of motion." , Source: An introduction to mechanics Textbook by Daniel Kleppner and Robert J. Kolenkow.

Akshaj Bansal
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  • yes you are right –  Jun 06 '21 at 18:43
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    I think the problem is with your definition of relative angular velocity, it is $\omega_B-\omega_A$, not $(v_B-v_A)/(r_B-r_A)$ –  Jun 06 '21 at 18:47

2 Answers2

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The mistake here is in the fundamental definition of relative angular velocity.

The relative angular velocity of B with respect to A is defined as : $$\omega_{BA}=\omega_B-\omega_A$$ Since $v=r\omega$ we can say $$\omega_{BA}=\frac{v_B}{r_b}-\frac{v_A}{r_A}$$ which is very different from $\frac{v_{BA}}{r_{BA}}$ mathematically.

On evaluating $\omega_{BA}=\frac{v_B}{r_b}-\frac{v_A}{r_A}$ we get the answer zero which is consistent with the results given by $\omega = \frac{d\theta}{dt}$. Thus the definition $$\omega = \frac{d\theta}{dt}$$ is absolutely correct.

EDIT

In the link attached it says: " We will examine the motion of this body in both, the fixed reference O shown, as well as relative to a non-rotating reference attached to point B." This means that the rotation of the body is about the point B. So $\omega_{B}$ is practically zero while $\omega_{A}$ is equal to $\omega$.Then $$\omega_{AB}=\omega_{A}-\omega_{B}$$$$\omega_{AB}=\omega-0= \omega $$

But if you considered the scenario I have introduced in the diagram where the rotation neither about point A or B but about the centre, the results are consistent with my explanations.

enter image description here

For another intuitive understanding think of how the Earth rotates about its axis but to us everything appears to be at rest at all times.

Möbius
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  • But for a rigid body we can say each point turns relative to other point of the rigid body with same angular velocity at a given instant – Akshaj Bansal Jun 07 '21 at 04:41
  • No. A rigid body is defined in such a way that the distance between any two of its points have same distance at any time. Since the distance (relative position) between any two points is fixed, both relative translational velocity and relative angular velocity are zero. – Генивалдо Jun 07 '21 at 06:19
  • Why relative angular velocity needs to be zero the distance between all the particles is fixed even when they are rotating with same $\omega$ wrt each other. – Akshaj Bansal Jun 07 '21 at 08:07
  • @Möbius please look at the edits I made. – Akshaj Bansal Jun 07 '21 at 12:47
  • @Möbius let us place an observer at A facing towards north now after time t it will face towards north only and to observe B it has to rotate its head constantly at angular velocity $\omega$ to observe B at all point of times. And please check my Edit 4 as well – Akshaj Bansal Jun 07 '21 at 14:07
  • After some time if observer is still facing north that means he has zero angular velocity.But B has moved angle theta in this time .Then angle between A and B is theta which has been covered in t time.Then with respect to A angular velocity of B is theta/t equal to omega.This does not agree with my explanation because in my example observer at A rotates with A so he too has angular velocity omega has relative angular velocity is zero.But here observer at A is not rotating with A so his angular velocity is zero making relative angular velocity omega – Möbius Jun 07 '21 at 15:51
  • Let me know what you think – Möbius Jun 07 '21 at 16:25
  • Yes that's why i say $\omega_{B/A}$ is $\omega$ when we would observe B from A our observer should be at rest wrt A in that case my result follows whereas in you example you make observer also rotate wrt A which in turn doesnt imply that you are observing B wrt A. – Akshaj Bansal Jun 07 '21 at 16:27
  • Your mistake is in the understanding of frame of reference .The frame of reference fixed to A from which we observe B is actually rotating with A just like you are rotating with the earth and everything you observe on the surface of the planet is also rotating .But if you want the frame of reference to not rotate you have to specify that like it was specified in the link by saying "a non rotating frame of reference at A" with special emphasis on non rotating.If you don't say it that by default means your frame of reference is rotating with A. – Möbius Jun 07 '21 at 16:45
  • Then what about all the links that I added in EDIT 5 say the same thing. – Akshaj Bansal Jun 07 '21 at 17:06
  • @Möbius leave my interpretation just check out edit 4 and 5 please – Akshaj Bansal Jun 07 '21 at 21:21
  • Let us continue this discussion in chat. – Möbius Jun 08 '21 at 10:49
  • Tell me, if every point on the surface of a rotating body has omega angular velocity with respect to any other point on the surface of that body then how come the earth rotates about its axis but we see all the trees and buildings stationary ? – Möbius Jun 08 '21 at 10:59
  • The statement holds only for non-rotating reference frames but B is a rotating reference frame. Besides ours is a case of orbital angular momentum not spin angular momentum coz although the disk is spinning points A and B are orbiting around the centre. – Möbius Jun 08 '21 at 18:56
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θ is measured in the fixed reference system, even if measured between the two points A and B it changes when the disc rotates: it is not zero.

however, wrt the rotating reference system of the disc itself the angular velocity of the points of the disc is zero because the disc is rigid... if the disc was "soft" or a fluid, the things would be different and may follow a vortex with non uniform angular velocities.

mangiumicro
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