For a rigid body undergoing planar motion I can associate to it the concept of Instantaneous Axis of rotation (IAOR), the point from which I can see the body undergoing pure rotation. Consider any point on the rigid body.
Here is my concern. Why is the radius of curvature of the point's trajectory and the it's distance from the IAOR not the same? Take for example a rolling disc on a smooth horizontal surface.
The radius of curvature of the top most point is $4r$ but the distance of that point from the axis of rotation is $2r$ (Here the IAOR is the bottom most point of the disc). What I think is that it's distance of the point from IAOR should be the same as the radius of curvature. Where am I wrong?
I calculated the radius of curvature of the topmost point. It's velocity w.r.t. ground is $2v$, so according to the equation of kinematics centripetal acceleration of a particle whose trajectory is known beforehand would be $$R_c=\frac{(2V_0)^{2}}{a_c}.$$ The centripetal acceleration would remain the same $=\omega^{2}r$ or $\frac{V_0 ^2}{R}$. Plugging in the values we get the radius of curvature $=4R$ ( where $R$ is the radius of the rolling body).