# How do you prove that an instantaneous center of rotation exists?

Fix a reference frame $R$ with axes $x,y,z$ and origin $O$ and suppose that the solid body $B$ is moving with a planar motion, let's say, parallel to the plane $x,y$. If $O'$ is a fixed point of $B$, but generally moving in $R$, the velocity of a point $P \in B$ in $R$ satisfies $$\vec{v}_P(t) = \vec{v}_{O'}(t) + \vec{\omega}(t)\times \vec{O'P}(t)\:.$$ Since the motion is planar $\vec{\omega}(t) = \omega(t) {\bf e}_z$ and $\vec{v}_{O'}(t) =v_{O'x}(t){\bf e}_x + v_{O'y}(t){\bf e}_y $.

The question now is if it is possible to fix $P(t)$, i.e., a point on $B$ at a given instant $t$, such that $$\vec{v}_{O'}(t) + \vec{\omega}(t)\times \vec{O'P}(t)=0\:.$$ Here $\vec{\omega}(t)$ and $\vec{v}_{O'}(t)$ are given. So the equation for the unknown $\vec{X}= \vec{O'P}(t)$ is $$\vec{\omega}(t) \times \vec{X}= -\vec{v}_{O'}(t)\:.$$ If $\vec{X}= X{\bf e}_x+ Y{\bf e}_y + Z{\bf e}_z$, the above equation reads $$\omega(t) X = v_{O'y}(t)\:, \quad \omega(t) Y = -v_{O'x}(t)$$ so that $$X = v_{O'y}(t)/\omega(t) \:, \quad Y = -v_{O'x}(t)/\omega(t)\:, \quad Z=0$$ defines a point on $B$ (more precisely a point in the rest space of $B$ but not necessarily coinciding with a material point of $B$) which has zero velocity at time $t$ in $R$.

I stress that if the motion is not planar, the statement is generally false. This is evident when trying to solve the system of equations above in the general case.