Question concerning rigid bodies

[jmc8197]

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.

 

[robphy]

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.

 

[GSXR750]

The reason why V' and W' are also perpendicular for origin O' is because of the definition of the cross product. The cross product of two vectors is always perpendicular to both of the vectors. In this case, V' and W' are the cross product of V and W with a, which means they are perpendicular to both V and W. And since a is also perpendicular to V and W, V' and W' are perpendicular to a as well. Therefore, V' and W' are perpendicular for origin O' as well.