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Consider a rigid body rotating with angular velocity $\vec{\omega}$. Now, we know that this $\vec{\omega}$ is an intrinsic property for the rigid body, in the sense that:

Each point on the rigid body rotates with $\vec{\omega}$ relative to any other point on the rigid body.

Now consider a point A outside the rigid body (i.e not on it). Is the angular velocity vector relative to A same for all points? i.e

Does each point rotate with say, $\vec{\Omega}$ relative to A?

I am not able to prove/disprove this rigorously. The application of certain formulae like $\vec{\tau}=d\vec{L}/dt$ and $\vec{\tau}=I\vec{\alpha}$ depend upon the validity of this statement , since while taking the sum over discrete particles,(we assume the rigid body to be a collection of discrete particles) we take $\vec{\omega}$ outside the sum (since we assume they are the same for all particles).

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You are talking about the concept of the "extended rigid body" which encompasses every point in space as co-rotating with the body regardless if it is physically located on the body, or not.

The conceptual jump here is as follows:

You are used to thinking of particles riding on a rigid body an tracking their velocity as they move around. But if you have a magical magnifying glass and wherever you place in space you measure the velocity of whatever particle happens to be passing underneath it then you can still use $$\vec{v} = \vec{\omega} \times \vec{r}$$ but $\vec{r}$ is the location of the magnifying glass and not any particular particle on the body. Thus you can freely move the magnifying glass anywhere in space, and imagine what would the velocity of a particle would be if it was under it, even if the body does not extent to this location.

Thus the concept of the extended body is needed where $\vec{v} = \vec{\omega} \times \vec{r}$ becomes a field equation that applies to all points in space. This is also called a rotating frame, and crucial to dynamics is to be able to take derivatives of vectors that ride on rotating frames (regardless if they reside on a physical body or on the extended body).


A little more rigorously you can show that two points in space that are riding on the same rotating extended frame and keeping their distance fixed must obey the kinematics of $$\vec{v}_A = \vec{v}_B + (\vec{r}_B-\vec{r}_A) \times \vec{\omega} \tag{1}$$ regardless of where the points are. Thus if all points follow the same kinematics, they must share the same $\vec{\omega}$, or else the above equation would not be valid at all points.

The length between two points is $\ell = \sqrt{ (\vec{r}_A-\vec{r}_B) \cdot (\vec{r}_A-\vec{r}_B) }$ or

$$ (\vec{r}_A-\vec{r}_B) \cdot (\vec{r}_A-\vec{r}_B) = \ell^2 = \text{(const.)} $$

of which you take the time derivative

$$ 2(\vec{v}_A-\vec{v}_B) \cdot (\vec{r}_A-\vec{r}_B) = 0 $$

now use (1) above to get

$$ 2 ( \vec{v}_B + (\vec{r}_B-\vec{r}_A) \times \vec{\omega} - \vec{v}_B) \cdot (\vec{r}_A-\vec{r}_B) = 2 ( (\vec{r}_B-\vec{r}_A) \times \vec{\omega} ) \cdot (\vec{r}_A-\vec{r}_B) \equiv 0 $$

and prove (1)

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No, Consider a particle(say A) at rest(with respect to earth) at a distance X (outside the rigid body) from the axis about which the rigid body is rotating. Now consider another particle (B) on the rigid body at a distance R from the axis of rotation. I will do all calculations with respect to earth.

Velocity of B = $\omega$R,

Velocity of A = 0,

Velocity of A with respect to B = 0 - $\omega$R = -$\omega$R.

Now, from the def. of angular velocity,

angular velocity of A with respect to B = -$\omega$R/(X-R)

Which is clearly dependent on X hence it will be not same.

Hope this helps

Suggestions/corrections[if any ;) ] are welcomed.

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Does each point rotate with say, Ω⃗ relative to A?

No.

Suppose A is comoving with the COM of the body. It is a static point in the frame of the COM. Take a straight line from A passing by the COM of the rigid body.

There are 2 points where the line cross the surface of the body, one closer and one farer.

If the body has an instantaneous angular velocity vetor $\omega$ with respect to the COM, the velocities measured from A are the same as measured from COM because they are in the same frame:

$v_c = \omega (r_{COM} - r_c)sin(\theta)$
$v_f = -\omega (r_f - r_{COM})sin(\theta)$

$\theta$ is the angle between the line and the instantaneous axis of rotation of the body. The velocities have opposite signs because they are at opposed sides of the rotation axis

For the 2 points have the same angular velocity with respect to A, the ratios of linear velocity and radius must be the same:

$$\frac{\omega (r_{COM} - r_c)sin(\theta)}{r_c} = -\frac{\omega (r_f - r_{COM})sin(\theta)}{r_f}$$

Unless $sin(\theta) = 0$, what means that A is colinear with the rotation axis of the body, the above expression requires: $r_f = r_c \implies$ the rigid body reduces to a point mass.

Another possibility is if A rotates with the body. In this case it belongs to it, even while separated from some distance.

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Yes. For intuitive picture, consider this-extend the rigid body from a point on it to the point A by a thin rigid rod (such that the point A just inside the rod). Consider the body and the rod to be a single rigid body. Now use your first statement in bold.

This is true when the instantaneous angular velocity (which is the ratio $d\Omega/dt$) of the body is to be looked at and not the relative position of point A w.r.t. the body or its rotation axis when looked in an inertial space of coordinates.

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  • $\begingroup$ I am not sure this is correct. The first statement in bold is based on the fact that the distances between points on a rigid body are fixed. This constraint obviously does not hold for points outside the rigid body, and so I believe your method of "extending" is not correct, since for the "extended portion", the fixed-distance-constraint does not hold. $\endgroup$
    – satan 29
    Commented Dec 22, 2020 at 14:19
  • $\begingroup$ @satan29 I meant 'extend' it with a 'rigid' rod. In other words, take a rigid rod, fix its one end on any point of the body so that the other end just encompass the outside point of concern. (Infact, it will do the trick as long as the point lies anywhere inside the rod and not just near the end) $\endgroup$ Commented Dec 22, 2020 at 14:24

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