Rotation in a curved surface - relation between $v$ and $\omega$ Consider a fixed curved surface of radius of curvature $R$. A sphere (or cylinder) of radius $r$ is rolling inside it at constant speed. (Note that I mean rolling in the sense of no slipping).
In trying to figure out the relation between $v$ and $\omega$, both of which are constant throughout the motion, I get two different answers from two different approaches.
Firstly, consider the point that lies a distance of $r$ below the center of rotation, in a direction perpendicular to $v$, away from the center of the ramp. Clearly, this point touches the ramp, and has a velocity of $v - r\omega$ (now), which has to be zero (no slip), giving us $v = r\omega$.
Secondly, consider the arc traversed by the center during time $t$. Suppose this arc subtends $\Theta$ at the center of the ramp. So the velocity is $v = (R-r)\Theta/t$. Also the point that touches the ramp travels $R\Theta$. So the angled turned through is $R\Theta / r$, and $\omega = R\Theta / rt$, giving $v = \omega r (R-r) / R$. (Incidentally in the limit of $R \to \infty$, this yields the desired $v = r\omega$).
Surely, one of these is wrong. Which one of these is it and why?
 A: 
So the velocity is $v=(R−r)\Theta/$. Also the point that touches
  the ramp travels $R\Theta$.

Note though that the point itself isn't "moving", so isn't necessarily associated with $\omega$.  

So the angled turned through is $R\Theta/r$

Agreed.

and $\omega=R\Theta/rt$

I don't think so.  Imagine a situation where $r$ is only a tiny bit smaller than $R$.  In such a situation the sphere will rotate very little while the contact point moves through a much larger $\Theta$.  But that formula suggests that near the limit where they are similarly sized, $\omega = \Theta/t$.   
As an example, here the object is at the bottom of the curve.  After a period of time it has rotated to the right.  The contact point and the center of the sphere have rotated through an angle $\Theta = \pi/2$.  But the sphere itself has rotated a much smaller amount.

A: $v=\omega r(R-r)/R$
Your second analysis is indeed correct.
Your first analysis is incorrect when you stated that the point touching the surface moves at $v-r\omega$. This is only true if an object is rolling without slipping on a flat surface.
Consider the distance traveled by the center of mass and the distance traveled by the point on the ground which is touching the ball. If the ball is rolling on a flat surface, these two displacements are the same. The total arc length that a point on the edge of the ball has moved must equal to the latter. Thus, the arc length drawn by a point on the edge must equal to the distance the center of mass has traveled. This is why we say that the speed of the point touching the surface is moving at $v-r\omega$.
However, when the ball is moving inside a concave curvature of radius $R$, the arc drawn by the center of mass is smaller than the arc drawn by the point on the ground touching the ball. If the former has traveled a distance of $d$, the latter has traveled a distance of $dR/(R-r)$. Similarly, the velocity of the latter point must also be off by a factor of $R/(R-r)$. Thus, the velocity of a point on the edge relative to the center of mass must be $vR/(R-r)$.
Thus we have $r\omega = vR/(R-r)$, and $v= \omega r(R-r)/R$.
