A sphere released from rest from the top of a hemisphere

[RoyalCat]

EDIT:
Problem solved, the mistake was in the statement that the point of contact traces out an arc of length $$\theta r$$ on the hemisphere. This statement neglects the change in orientation of the sphere relative to the hemisphere.

An object with cylindrical symmetry (Full/hollow sphere, cylinder, ring, point mass, etc.) of radius r is released from rest, from the top of a hemisphere of radius R, and rolls without slipping.

At what angle, $$\phi_c$$ will it leave the hemisphere?

The approach is fairly straightforward, an energy equation relating the energies at the top of the hemisphere and at the point of release (Kinetic energy of the CM, kinetic energy of rotation about the CM, gravitational potential energy), and an equation of forces in the radial direction at the point of disengagement (Normal force goes to 0, gravity provides centripetal acceleration). Putting all these together with the geometrical constraint of rolling without slipping allows us to solve for the critical angle.

The problem here lies in the geometry and in relating the velocity of the center of mass to the angular velocity about the center of mass.

On the one hand, since we have rolling without slipping (Albeit along a curved surface), following the naive approach, one can make the case that $$V_{cm} = \omega r$$ since that would ensure 0 relative velocity between the point of contact and the surface of the hemisphere.

On the other hand, taking the geometric approach: (See attachment)
As the object rotates through an angle $$\theta$$, the point of contact traces out a path of length $$\theta r$$
This path is a circular arc of angle $$\phi$$ and of radius $$R$$
The path the center of mass takes, is marked $$s$$
It subtends the same angle $$\phi$$ but is of radius $$R+r$$

Equating the two expressions for $$\phi$$ one finds:
$$\frac{s}{R+r} = \frac{\theta r}{R}$$

Solving for $$s$$ and differentiating with respect to time, noting that $$\dot s = V_{cm}$$ one finds:
$$V_{cm} = \omega r \frac{R+r}{R}$$This is all making me feel quite uneasy, as it is hard to reconcile the two. Taking the geometric approach leads to a contradiction, since it gives the point of contact velocity relative to the surface of the hemisphere, but the naive approach doesn't square with the geometry (Pun not intended).

Any guidance would be greatly appreciated. :)

P.S.
The final answers don't differ by much.
If we assign $$I=kmr^2$$
The two answers are:
Naive approach: $$\cos{\phi_c} = \frac{2}{3+k}$$
Geometric approach: $$\cos{\phi_c} = \frac{2}{3+k(\frac{R}{R+r})^2}$$

 

[AvroArrow]

The difference is negligible, and should be even smaller for full spheres, since k goes to 0.5. But the contradiction still bugs me.