# Converting angular velocity to linear velocity through friction

A very basic question here; it's related to this one, but not quite the same.

If a rotating rigid body (a sphere for the sake of discussion) with mass $m$, radius $r$ and inertial tensor $I$ has initial linear velocity $v_0=0$ and non-zero, initial angular velocity $\omega_0$, what are the idealized final angular and linear velocities if the body finds itself instantaneously on a surface with infinite friction and mass, assuming $\omega_0$ is parallel to the surface (and the surface doesn't absorb any of the energy, nothing is lost to heat)?

The body should end up rolling along the surface at some constant velocity. So I think we know that $v_f = r\times \omega_f$ with $r$ anti-parallel to the surface normal. Clearly the inertial tensor must come into play since if all of the mass is concentrated on the outside of the body, it has a lot more angular momentum than if the mass is mostly at the center of the body. I tried using conservation of momentum here, but the units don't match up between angular momentum and linear momentum.

Angular momentum is not conserved since there is an external force applied to the body - the friction.

According to the condition there is no energy loss, so you can use energy conservation law. Since the friction is infinite the body will instantly start rolling without any sliding. The initial rotational energy $$E_0 = \frac{I \omega_0^2}{2}$$ will be distributed between translation movement with velocity $v_1$
and rotation with new angular velocity $\omega_1$: $$\frac{I \omega_0^2}{2} = \frac{I \omega_1^2}{2} + \frac{m v_1^2}{2}$$

The velocities $v_1$ and $\omega_1$ should fit the condition of rolling without sliding: $$\omega r = v$$

• This is not true, the energy is not conserved either, because the friction is doing impulsive work. The angular momentum at the contact point is conserved. – Ron Maimon Dec 24 '11 at 14:29
• @RonMaimon, yes the translational momentum of the body changes and momentum of the surface changes too. So some part of the kinetic energy is transfered to the surface but its mass is implicitly assumed to be huge and this energy loss is negligible. – Maksim Zholudev Dec 24 '11 at 15:17
• Conservation of energy seems right. This should work for what I need (although I'll need it in 3d; so $\omega r$ needs to be $\omega \times r$ which means that I've got to do a little bit of playing to solve for $v_1$). Thanks. – JCooper Dec 24 '11 at 16:34