# 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.