Question concerning rigid bodies

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The discussion centers on the relationship between the velocities of particles in a rigid body as described by Landau's Mechanics. It explains how the velocity of a particle can be expressed in terms of the body's angular velocity and the position vector from a chosen origin. By substituting a new origin, the relationship between the velocities remains consistent, demonstrating that if the original velocity and angular velocity are perpendicular, the new velocities will also be perpendicular. The mathematical proof involves dot products and properties of cross products, confirming the orthogonality condition. The conclusion affirms the correctness of the reasoning presented.
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The following comes from Landau's Mechanics, pages 97 - 98.

For a particle in a rigid body, v = V + W x r -- (1)

where for some origin O of the moving body measured in the "fixed" system of
co-ordinates, v = particle's velocity in body in the "fixed" system, V = velocity of O in "fixed" system , W is the body's angular velocity in *fixed system", x is a cross product and r the particle's radial vector within body measured from O.

For another origin, O' distance a from O, r = r' + a, and substituting in
(1) gives:

v = V + W x a + W x r'. The definition of V' and W' shows that v = V' + W' x
r' and so

V' = V + W x a, W' = W -- (2)

He then says that the first part of (2) shows that if V and W are
perpendicular for a choice of origin O, then V' and W' are also
perpendicular for O'. Why?

Thanks in advance.
 
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If you dot (*) equation (2a) with W, you get

W*V' = W*(V + W x a)=W*V +W*(Wxa) = W*V + 0 , since W is orthogonal to any vector crossed with W.

So, W*V' = W*V.

Since W'=W, by equation (2b), the left hand side is rewritten so that
W'*V' = W*V

So, if W*V=0, then W'*V'=0.

I think this is correct.
 
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