# Should rotation of a rigid body confined to a sphere couse it to divert from a big circle?

Consider an axially symmetric body constrained to a unit sphere in such a way that the symmetry axis of the body is always normal to the sphere.

Edit: I guess what I really wanted to ask (although it may not look like it) is whether the following reasoning is correct. In the scenery described at the beginning of my post, consider an initially stationary body spinning around its symmetry axis. If the body gains a velocity $$\vec{v}$$ tangent to the sphere, will it move along a big circle (geodesic for a point mass) or will it divert from it? The linear movement will tilt the angular momentum (because the symmetry axis must stay aligned with the normal to the sphere) and this change should produce a torque $$\frac{dL}{dt} = \tau = \vec{n} \times F$$ where $$F$$ is perpendicular to both $$n$$ and $$v$$. Therefore, $$F$$ should chang the direction of $$\vec{v}$$ resulting in diverting from the big circle. Does it make sense?

Old question: What is the Lagrangian of such system? Does it make sense to write it in a way similar to the below? $$L = \frac12 I_{\perp}\|\vec{n} \times \vec{v}\|^2 + \frac12 I_{ax} \epsilon^2$$ where $$\vec{v}$$ is the linear velocity, $$\vec{n}$$ is the normal vector to the sphere, $$I_{ax}$$ is the moment of inertia for spinning around the symmetry axis, $$I_{\perp}$$ is the moment of inertia for revolving around the sphere (in any direction as the body is axially symmetric) and $$\epsilon$$ is the angular velocity of the rotation around the symmetry axis. My uncertainty is with $$\epsilon$$. What should it be? How should be its corresponding coordinate defined? It should describe rotations in the tangent plane, so is the coordinate just the angle between a vector in the plane (corresponding to the direction the body is "facing") and the vector (for example) facing north? Can it be somehow expressed globally?

$$L=\frac{1}{2}mR^2(\dot{\theta}^2+\sin^2\theta ~\dot{\phi}^2) + \frac{1}{2}I\omega^2$$