Dynamics of unscrewing I'm trying to formulate mathematical model of a nut on screw dynamics. Let's consider nut (with inertia $I_n$ and mass $m_n$) on a screw (with inertia $I_s$) - both rotating with constant angular velocity $\omega_0$ about the vertical axis ($y$). The rotational angular velocity of the nut about the screw frame is equal to $0$. Let's neglect gravity and friction (no meshing/viscous losses). Then screw is immediately stopped (hard-stop). Nut should unscrew with angular velocity $\omega_1$ and linear velocity $v_y$ (along the vertical axis). Considering that screw thread could be represent as inclined plane, the relation between angular and linear velocity is described as $v_y = \dfrac{\omega_1 L}{2 \pi}$, where $L$ is the lead of screw. But I'm wondering how $\omega_1$ is related to $\omega_0$?
 A: Is a cool problem, at first I tought it would be the same but realized the rotation energy was somehow transforming into translational energy. This is my solution.
First we have an initial amount of energy that must be conserved
$$
E_0 = \frac{I_n}{2} w_0^2
$$
clearly for any given moment different from $t=0$ we have
$$
E = \frac{I_n}{2} w_1^2 + \frac{m_n}{2}v_y^2 = \frac{I_n}{2} w_1^2 + \frac{m_n}{2}\left(\frac{\omega_1 L}{2\pi}\right)^2
$$
via conservation of energy we have
$$
\frac{I_n}{2} w_0^2 = \frac{I_n}{2} w_1^2 + \frac{m_n}{2}\left(\frac{\omega_1 L}{2\pi}\right)^2
$$
hence
$$
\frac{\omega_0}{\omega_1} = \pm\sqrt{\frac{m_nL^2+I_n4\pi^2}{4\pi^2I_n}}
$$
we select the one with the positive one because otherwise it would mean the instant tue screw stops, it will start rotating to the oposite direction as before it was stopped and that does not make any sense
$$
\frac{\omega_0}{\omega_1} = \sqrt{\frac{m_nL^2+I_n4\pi^2}{4\pi^2I_n}}
$$
A: If there is no friction the angular and linear velocity of the nut will not immediately change when the screw is stopped.  Over a period of time the nut will rise as it unscrews from the screw (assuming a “right hand screw”). Then there be a loss of kinetic energy matching the gain in gravitational potential energy.
