Why is the resultant velocity of a particle inside a body undergoing rolling without slipping always perpendicular to the line segment connecting it and the instantaneous axis of rotation?
$P_2V_2$ is the resultant velocity.
$P_2V_2=\omega R^2+V_{CM}$
$P_0$ is the instantaneous axis of rotation.
Rephrased question: why is $P_0P_2$ always perpendicular to $P_2V_2$ for all $P_2$ situated inside the circle?
If ${\bf v}$ is the velocity of some material point belonging to the disk, and ${\bf v}_0$ is the velocity of the point of contact, then, since they are material points belonging to the same rigid body, they obey
${\bf v} - {\bf v}_0 = \boldsymbol{\omega} \times ({\bf r} - {\bf r}_0)$,
where ${\bf r}$ and ${\bf r}_0$ are position vectors to the respective points and $\boldsymbol{\omega}$ is the angular velocity of the rigid body.
The no slip/penetration condition provides ${\bf v}_0 = {\bf 0}$. Therefore, the velocity of a material point is given by
${\bf v} = \boldsymbol{\omega} \times ({\bf r} - {\bf r}_0)$.
Therefore, ${\bf v}$ is perpendicular to ${\bf r} - {\bf r}_0$ for all ${\bf r}$ belonging to the rigid body.
