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 $\omega_0$?
2 Answers
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
$$ \frac{\omega_0}{\omega_1} = \sqrt{\frac{m_nL^2+I_n4\pi^2}{4\pi^2I_n}} $$
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$\begingroup$ Thank you for your answer. This reasoning seems correct :) $\endgroup$ Commented Jul 18, 2020 at 22:56
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.
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$\begingroup$ Thank you for your answer. Yes, I made an assumption that screw helix type is "right-hand", but I also assumed that there is no gravitational force affecting the system. Despite no losses of kinetic energy I think the part of rotational kinetic energy will transform into translational kinetic energy. As @mytorojas described below. Is this reasoning correct? The next step in my model is to replace "hard-stop" by torque $\tau(t)$ that comes from stopping screw. I will try to develop mathematical model based on Euler–Lagrange method. But first I want to be sure about the energy equations. $\endgroup$ Commented Jul 18, 2020 at 22:55
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$\begingroup$ Sorry,I got it wrong. The answer from Mytorojas is probably what you are looking for. $\endgroup$ Commented Jul 19, 2020 at 19:48