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Consider a Ring (radius $r$) rolling without slipping with angular speed $\omega$ on the surface. We want to find the radius of curvature of the top most point.

1.The radius of curvature(R): Regardless of the actual path that the particle travels, a particle at every instant can be thought of as tracing a circle, of radius=$R$. The top most point has, at the current instant, Velocity = $2\omega r \hat{i}$, and acceleration =-$\omega^2 R\hat{j}$. Using the radius of curvature formula: $R= v^2/a_{normal}$, we get $R=4r$.Which matches my textbook, and also fact that the radius of curvature of a cycloid, which is the path that any point on the rim of the ring traces, is $4r.$

  1. The Instantaneous axis of rotation(IAR): Each point on a rigid body can be thought of as tracing a circle at an instant, about a point known as the instantaneous center of rotation(c). The axis passing through C is the IAR.

It seems to me from these two statements that the distance of a point from the IAR must be equivalent to its radius of curvature. This is clearly not the case, as in a ring which is rolling without slipping , the IAR is at a distnace $2r$ from the top-most point.

What is the flaw in my interpretation?

Edit: Radius of curvature and Instantaneous Axis of Rotation seems quite similar. However, the top-voted answer has simply tried to explain $R=2r$, using the formula $v^2/a_{normal}$.This answer doesnt explain that why we get two different answers for the seemingly same thing.

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The key difference between Instantaneous Axis Of Rotation(IOAR) and Radius Of Curvature(ROC) is

IOAR: The point about which all points of a rigid body have perpendicular velocities at a particular instant.

ROC: $R= v^2/a_\mathrm{normal}$ where the velocity is perpendicular to the center of curvature which is at a distance $R$ from the particle whose ROC we are calculating.

The confusion (I assume) came from the fact that the particle seems to circle both IOAR and ROC. However for the IOAR we will not be able to satisfy the equation $R= v^2/a_\mathrm{normal}$ where $R$ is the distance from the IOAR. But this equation will give the one and only ROC. However its velocity is perpendicular to both the line joining ROC and IOAR (by definition which also happens to be why you may have confused the two). Therefore the instantaneous axis of rotation cannot be used to calculate radius of curvature.

I would also like to note that while instantaneous axis of rotation depends on the motion at an instant, radius of curvature depends on the trajectory of the individual particle of the rigid body which we are observing.

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    $\begingroup$ Welcome to Physics.SE. When typing equations, you can use standard TeX formatting. I've made an edit for you. $\endgroup$
    – kaylimekay
    Dec 29, 2020 at 4:06

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