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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.

dark_prince
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2 Answers2

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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?

Not_Einstein
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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.

Eli
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