1
$\begingroup$

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.

enter image description here

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?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ using $\boldsymbol\Omega'=\boldsymbol\Omega$? $\endgroup$
    – AccidentalFourierTransform
    Apr 2, 2016 at 14:22
  • $\begingroup$ Isn't that the definition of a rigid body? A clump of particles sharing a common angular velocity such that velocities transform as $$\vec{v}_A = \vec{v}_B + \vec{\omega} \times (\vec{r}_A - \vec{r}_B)$$ $\endgroup$
    – John Alexiou
    Apr 2, 2016 at 14:41
  • $\begingroup$ @AccidentalFourierTransform But isn't $\vec{\Omega}=\vec{\Omega'}$ what is being proved? $\endgroup$
    – Sørën
    Apr 2, 2016 at 14:45

1 Answer 1

Reset to default
2
$\begingroup$

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

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

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.