# 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 $\boldsymbol\Omega'=\boldsymbol\Omega$? Apr 2, 2016 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 $$\vec{v}_A = \vec{v}_B + \vec{\omega} \times (\vec{r}_A - \vec{r}_B)$$ Apr 2, 2016 at 14:41
• @AccidentalFourierTransform But isn't $\vec{\Omega}=\vec{\Omega'}$ what is being proved? Apr 2, 2016 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}