# Proof of constant angular velocity in rigid body motion

I'm studying rigid body motion on Landau but I'm having troubles to understand this proof of the fact that the angular velocity $\vec{\Omega}$ is constant for a rigid body.

My doubt is about the two last equations in the last two lines of the text. If I use the two I get

$\vec{V'}=\vec{V}+(\vec{\Omega}-\vec{\Omega'})\times \vec{r'} +\vec{\Omega} \times\vec{a}$

How are (31.3) derived from this?

• using $\boldsymbol\Omega'=\boldsymbol\Omega$? – AccidentalFourierTransform Apr 2 '16 at 14:22
• Isn't that the definition of a rigid body? A clump of particles sharing a common angular velocity such that velocities transform as $$\vec{v}_A = \vec{v}_B + \vec{\omega} \times (\vec{r}_A - \vec{r}_B)$$ – John Alexiou Apr 2 '16 at 14:41
• @AccidentalFourierTransform But isn't $\vec{\Omega}=\vec{\Omega'}$ what is being proved? – Sørën Apr 2 '16 at 14:45

You have

$$\boldsymbol{v} = \boldsymbol{V'} + \boldsymbol{\Omega'} \times \boldsymbol{r'}$$

and

$$\boldsymbol{v} = \boldsymbol{V} + \boldsymbol{\Omega} \times (\boldsymbol{r'}+\boldsymbol{a})$$

You collect the $\boldsymbol{r'}$ terms

$$\boldsymbol{v} = \boldsymbol{V'} + \boldsymbol{\Omega'} \times \boldsymbol{r'} = \left( \boldsymbol{V}+ \boldsymbol{\Omega} \times \boldsymbol{a} \right) + \left( \boldsymbol{\Omega} \times \boldsymbol{r'} \right)$$

which is solved uniquely when

\begin{align} \boldsymbol{\Omega'} & = \boldsymbol{\Omega} \\ \boldsymbol{V'} &= \boldsymbol{V}+ \boldsymbol{\Omega} \times \boldsymbol{a} \end{align}