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What is relative angular velocity of one point, say A, with respect to another point, say B? Both the points lie on the same rigid body which is rotating with constant angular velocity ω about a fixed axis.

Edit:

Here is the figure enter image description here

The above body is rigid. For simplicity consider the rod joining A and B to be massless. So is the relative angular velocity of A with respect to B be zero? And if this is the case then how my question is different from Relative angular velocity

I think i'm missing something.

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  • $\begingroup$ I will return with a proper answer. $\endgroup$
    – Tony Stark
    Commented Aug 30, 2020 at 9:41
  • $\begingroup$ They are at rest wrt each other thus answer is $0$ $\endgroup$
    – Protein
    Commented Aug 30, 2020 at 14:26
  • $\begingroup$ This question should be helpful. $\endgroup$
    – Protein
    Commented Aug 30, 2020 at 14:29
  • $\begingroup$ @Eli's answer below gives a mathematical proof that the relative angular velocity is zero. From an intuitive point of view, imagine sitting at point B and looking at point A. As the rigid body rotates, A remains fixed your field of view, so no angular velocity with respect to you. $\endgroup$ Commented Aug 30, 2020 at 14:30
  • $\begingroup$ @dark_prince Even I feel the same as others and the answer should be zero. $\endgroup$
    – Tony Stark
    Commented Aug 30, 2020 at 14:37

2 Answers 2

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I tried above to give an intuitive explanation of why the answer was zero. I will try to do the same for the new answer, ω (I'm flexible !). So again, imagine you are sitting at point B and looking at point A. As the rigid body rotates, A remains fixed your field of view, which led me before to say the relative angular velocity was zero. But as the object rotates, A faces different directions in the environment of the object so it appears to rotate once for each rotation of the object (as our moon rotates once a month despite always showing the same face to us). Make sense?

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  • $\begingroup$ Yes, perfectly! So that's what happening even if B's head remain fixed A will appear to move. $\endgroup$ Commented Aug 31, 2020 at 23:25
  • $\begingroup$ Another thing to note: relative velocity should not be zero to have a non zero relative angular velocity. Imagine B to be situated below A at the same distance from axis; here relative angular velocity will be zero because relative velocity is zero. Your analogy fits here if we look from B, A will face the same direction i.e. upwards. B will always look upwards wrt to the environment $\endgroup$ Commented Sep 1, 2020 at 0:00
  • $\begingroup$ I'm not following what you mean by below A, but it would seem no matter where B is an observer on B would notice that a spot on A faced in succession 360 degrees around the environment of the object thus again appearing to rotate once for each rotation of the object. $\endgroup$ Commented Sep 1, 2020 at 0:49
  • $\begingroup$ Assume a cylinder, A is on top face and B is on bottom face, both at the boundary(distance from the axis is same). Now B will look at A in upward direction and wrt to the environment B's view will not change. So B will notice the angular velocity of A is zero. (here relative velocity of A wrt B is zero and so is relative angular velocity) $\endgroup$ Commented Sep 1, 2020 at 1:07
  • $\begingroup$ The axis of rotation of the cylinder is through the center of the faces? If A had a mark on it, it seems to me that B would still see that mark in turn face 360 degrees around the environment of the object for each rotation of the object. So A would still have a relative angular velocity of ω. But @Eli's proof assumes they lie in a plane perpendicular to the axis of rotation ?? $\endgroup$ Commented Sep 1, 2020 at 2:35
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enter image description here

The relative angular velocity$~\vec{\omega}_{r}~$ can obtain from this equation:

$$\vec{\omega}_{r}=\frac{\vec{R}_{AB}\times \vec{V}_{AB} }{\vec{R}_{AB}\cdot \vec{R}_{AB}}\tag 1$$

with :

$$\vec{R}_{AB}=\vec{R}_{B}-\vec{R}_{A}$$ $$\vec{V}_{AB}=\vec{V}_{B}-\vec{V}_{A}$$

equation (1)

$$\vec{\omega}_{r}=\frac{\left(\vec{R}_{B}-\vec{R}_{A}\right)\times \left(\vec{V}_{B}-\vec{V}_{A}\right) }{\vec{R}_{AB}\cdot \vec{R}_{AB}}\tag 2$$

with $~\vec{V}_A=\vec{\omega}\times \vec{R}_A~$ and $~\vec{V}_B=\vec{\omega}\times \vec{R}_B~$

equation (2)

$$\vec{\omega}_{r}=\frac{\vec{R}_{AB}\times (\vec{\omega}\times \vec{R}_{AB})}{\vec{R}_{AB}\cdot \vec{R}_{AB}}=\frac{(\vec{R}_{AB}\cdot \vec{R}_{AB})\vec{\omega} - ( \vec{R}_{AB}\cdot \vec{\omega})\vec{R}_{AB}}{\vec{R}_{AB}\cdot \vec{R}_{AB}}\tag 3$$

Now if A and B lie in the plane perpendicular to ω then $$\vec{R}_{AB}\cdot \vec{\omega} = \vec{0}$$

equation (3) becomes:

$$\vec{\omega}_{r} = \frac{(\vec{R}_{AB}\cdot \vec{R}_{AB})\vec{\omega}}{\vec{R}_{AB}\cdot \vec{R}_{AB}} = \vec{\omega}$$

thus the relative angular velocity is ω.

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  • $\begingroup$ Please see edited question. $\endgroup$ Commented Aug 31, 2020 at 12:11
  • $\begingroup$ @dark_prince $\overrightarrow{a}\times \overrightarrow{b}\times \overrightarrow{a}=-\overrightarrow{b}\times \overrightarrow{a}\times \overrightarrow{a}=\overrightarrow{0}$ $\endgroup$
    – Eli
    Commented Aug 31, 2020 at 13:28
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    $\begingroup$ NO!! wikimedia.org/api/rest_v1/media/math/render/svg/… $\endgroup$ Commented Aug 31, 2020 at 13:39
  • $\begingroup$ You can't do that. Search vector triple product $\endgroup$ Commented Aug 31, 2020 at 13:40
  • $\begingroup$ O.k I agree with you . I will correct my results $\endgroup$
    – Eli
    Commented Aug 31, 2020 at 14:07

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