# Does rigid body rotation always add a new independent variable?

I want to talk about the constrain added by introducing rotation of a rigid body to a simple case:

An homogeneous ring at rest is dropped from height $$H$$ of an declined surface without any kind of friction. To find it's translational velocity once it hits the horizontal floor.

Conservation of energy gives: $$2g(H-R) = v^2 + R^2\omega ^2$$

I want to know if $$\omega$$ and $$v$$ are independent variables, usually correlated with the additional assumption of rotation without slipping, (and thus the problem is unsolvable without any additional assumptions ) or if there is no need for more information in order for the problem to be solved and rotation without slipping is a natural consequence of this configuration and therefore $$\omega$$ and $$v$$ are not independent in first place.

• Without friction the ring won’t start rotating. – G. Smith Sep 2 at 22:17
• the own weight creates a momentum at the base doesn't it? – mranon Sep 2 at 23:19