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Lagrangean and non inertial frame
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[QUOTE="LCSphysicist, post: 6620971, member: 675151"] [B]Homework Statement:[/B] Smooth rod OA of length l rotates around a point O in a horizontal plane with a constant angular velocity $\delta$. The bead is fixed on the rod at a distance a from the point O. The bead is released, and after a while, it is slipping off the rod. Find its velocity at the moment when the bead is slipping? [B]Relevant Equations:[/B] . I have tried to solve this problem using the lagrangean approach: $$L = T - V = m((\dot r)^2 + (r \dot \theta)^2)/2 - 0 = m((\dot r)^2 + (r \delta)^2)/2 - 0 $$ The problem is that the answer i got is the right answer at the smooth rod referencial, that is, at the non inertial frame. Now we can easily translate my answer of v to the inertial frame, but that is not the point of my question. The point is, when did i have assumed i was at the rod frame? That is, i have just written the kinect energy in polar coordinates, so instead of $$\sum m x_i x^{i} / 2$$ i have written it in polar language. All of sudden, am i now in a non inertial frame? Do polar coordinates are by definition non inertial? Should have a potential here? I can't see where does it come from. In fact, another point i can't understand is, in a inertial frame there is not a potential energy, and in fact i have used (V=0). But yet, my answer is given in a non inertial frame... [/QUOTE]
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Lagrangean and non inertial frame
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