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Bead on a Helix, angular velocity
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[QUOTE="Phylosopher, post: 6017519, member: 599850"] [h2]Homework Statement [/h2] [ATTACH=full]227289[/ATTACH] [h2]Homework Equations[/h2] $$\mathcal{L}=T-U$$ $$\omega= \frac{d\phi}{dt}$$ $$I=mr^{2}$$ [h2]The Attempt at a Solution[/h2] My problem is not finding the Lagrangian. But finding the kinetic energy! The translational kinetic energy would obviously be the following: $$K.E t=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2})=\frac{1}{2}m(1+(\frac{R}{\lambda})^{2}) \dot{z}^{2}$$ But as far as I understood, there is rotational kinetic energy as well, namely the following $$K.E r= \frac{1}{2}I \omega^{2} = \frac{1}{2}mR^{2} \frac{\dot{z}^2}{\lambda^2}$$ The helix is just the trajectory of the particle. If you look at the helix from above you see a circle. Furthermore, the angle depend on ##z##, which I believe support my idea. I usually prefer looking at the solutions. Especially when I self study. I looked at two different sources from two different universities, none include rotational kinetic energy! Am I wrong? [URL='http://faculty.tru.ca/rtaylor/phys3200/taylor_7.20.pdf']Source 1[/URL] [URL='https://www.usna.edu/Users/physics//cmorgan/SP333%20Content/Solutions6.pdf']Source 2[/URL] Footnote: Classical Mechanics, By Taylor. Problem 7.20 [/QUOTE]
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Bead on a Helix, angular velocity
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