Pitch in Screw Motion of Rigid Body

Question

Does anyone know how to prove these statements?

Answer 1

So you have the velocity vector $\boldsymbol{v}$ at some arbitrary point which is part of a rotating body with rotation $\boldsymbol{\omega}$.

And you compose this vector by $$ \boldsymbol{v} = \boldsymbol{q} \times \boldsymbol{\omega} + h\, \boldsymbol{\omega} \tag{1} $$

To find $h$ and $\boldsymbol{q}$, use the dot product first

$$ \require{cancel} \begin{aligned} \boldsymbol{\omega} \cdot \boldsymbol{v} & = \cancel{\boldsymbol{\omega} \cdot ( \boldsymbol{q} \times \boldsymbol{\omega})}+ \boldsymbol{\omega} \cdot h \boldsymbol{\omega} \\ h & = \frac{ \boldsymbol{\omega} \cdot \boldsymbol{v} }{\boldsymbol{\omega}\cdot \boldsymbol{\omega}} \end{aligned} \tag{2}$$

and then the vector triple product identity $a\times(b\times c) = b(a\cdot c) - c(a \cdot b)$

$$ \require{cancel} \begin{aligned} \boldsymbol{\omega} \times \boldsymbol{v} & = \boldsymbol{\omega} \times ( \boldsymbol{q} \times \boldsymbol{\omega})+ \cancel{ \boldsymbol{\omega} \times h \boldsymbol{\omega}} \\ & = \boldsymbol{q}(\boldsymbol{\omega}\cdot\boldsymbol{\omega}) - \cancel{\boldsymbol{\omega} ( \boldsymbol{\omega}\cdot \boldsymbol{q})} \\ \boldsymbol{q} & = \frac{ \boldsymbol{\omega}\times\boldsymbol{v}}{\boldsymbol{\omega}\cdot\boldsymbol{\omega}} \end{aligned} \tag{3}$$

the above requires that $\boldsymbol{q}$ is perpendicular to $\boldsymbol{\omega}$. As a result equation (3) retuns the point on the rotation axis closest to the point where $\boldsymbol{v}$ is measured.