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If all the points in a rigid body have same angular velocity (say, $\vec{\omega}$), then why is angular velocity of point $A$ in the body w.r.t. point $B$ in the body is still $\vec{\omega}$ and not $\vec{\omega}-\vec{\omega}=0$?

In satellite problems, the time after which the two satellites will meet again is given by $\frac{2\pi}{\text{relative angular velocity of satellites}}=\frac{2\pi}{{\omega}_2-{\omega}_1}$ (assuming both the satellites are moving clockwise). Using the same logic to find angular velocity of point $A$ in a rigid body w.r.t. another point $B$ gives $\omega-\omega=0$ but it actually is $\omega$ and I can visualize it no problem, but I can't work it out using equations. What am I missing?

kbakshi314
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3 Answers3

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In the context of rigid body mechanics the "phrase angular velocity of point A in the body w.r.t point B in the body" is meaningless because points do not have angular velocities, but bodies (and their frames) do.

Points do have translational velocity and you can express their relative velocities as observed by an inertial reference frame.

Two bodies can have two different angular velocities and this is how they are used in the contact of satellites. In fact, in the transit equations, it is not the angular velocity of the satellite in question, but the orbital velocity expressed in terms of $\vec{\omega}$. The satellite might be rotating or pointing to a fixed point in space and still be orbiting the earth.

So the relative orbital angular velocity of two satellites is what we are talking about here. not the relative angular velocities of two points. Besides two satellites are not a rigid body and therefore do not share angular velocity.

John Alexiou
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  • "Besides two satellites are not a rigid body and therefore do not share angular velocity." So let's consider two points on a rotating rigid body, at different distances from the axis of rotation. The tangential velocity of the point further from the axis is larger than the tangential velocity of the point nearer to the axis. There is thus a relative velocity between the two points. How is that consistent with the fact that the body is rigid? – Not_Einstein Mar 20 '21 at 15:01
  • @Not_Einstein The key-words are translational (linear) velocity vs. rotational (angular) velocity. My second paragraph addresses that. $\vec{v}$ are different, but $\vec{\omega}$ the same. The body is rigid because rotation preserves all lengths between particles. The general motion (called Chasle's Theorem) of a rigid body is a rotation about an arbitrary axis coupled with a translation along this axis. – John Alexiou Mar 20 '21 at 15:13
  • I'm trying to picture what an observer would see who is fixed at the point nearer the axis mentioned in my comment and looking at the point further from the axis i.e. a reference frame with origin at the nearer point. The distance to that further point would not change but the background beyond that point would be moving so it would look like that further point is moving at right angles to the line connecting the two points. That is, the further point would appear to have a velocity relative to the nearer point. Make sense? – Not_Einstein Mar 20 '21 at 19:01
  • @Not_Einstein - yes, but that is not angular velocity. What you can observe is translational velocity. Unless the observes is co-rotating then you wont see anything move. – John Alexiou Mar 20 '21 at 19:29
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Angular velocity must be measured with respect to a point. For rigid bodies, the angular velocity of point A relative to any point B within the body must be the same as relative to any other point C in that same body. That keeps the relative coordinates from changing (it's a rigid body).

You misapply the idea of relative angular velocity by choosing another point in the body. Both of those are already angular velocities relative to any point in the body.

Aside: also, the translational velocity of point A relative to point B must be zero within a rigid body using body-centered coordinates.

Bill N
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  • It is improper to answer new questions in comments. You can always post another question so that everyone will get a notice about it. – Bill N Mar 18 '21 at 23:45
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The distance $d$ between A and B must be constant, otherwise it is not a rigid body. So the only allowable movement is transversal to the line joining them. The ratio $\frac{v_t}{d}$ is called angular velocity $\omega$.

If $\omega = 0$ there is no allowable movement between A and B. In this case, for any other point C of the body, $V_c$ must also be zero, otherwise or the distance AC or BC (or both) would change what is not possible for a rigid body.

So, $\omega = 0$ for any point w.r.t. another one, means $\omega = 0$ for all other points.

Claudio Saspinski
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