# Question concerning rigid bodies

1. Aug 20, 2004

### 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.

2. Aug 20, 2004

### 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.

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