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- Thread starter Silviu
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- #1

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- #2

tnich

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$$z=f(r)$$

I think your problem is that f represents a distance in the first of these equations and an angle in the second.$$f=\phi-\omega t$$

- #3

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Shouldn't ##\lambda## take care of this automatically? Like shouldn't I get the right units, just by applying the formula? The Lagrangian I obtained is the same as the one they obtained so I assume it is correct. Edit: the ##f## that appears in ##z=f(r)## is not the same as the ##f=\phi - \omega t##, in case I wasn't clear there. In the second case I just gave it a general name, to reflect the way I applied the formula.I think your problem is that f represents a distance in the first of these equations and an angle in the second.

- #4

tnich

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##\lambda## has units consistent with your original equation, so the problem is probably somewhere else.Shouldn't ##\lambda## take care of this automatically? Like shouldn't I get the right units, just by applying the formula? The Lagrangian I obtained is the same as the one they obtained so I assume it is correct. Edit: the ##f## that appears in ##z=f(r)## is not the same as the ##f=\phi - \omega t##, in case I wasn't clear there. In the second case I just gave it a general name, to reflect the way I applied the formula.

- #5

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Yeah, but I am not sure what I am doing wrong...##\lambda## has units consistent with your original equation, so the problem is probably somewhere else.

- #6

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Of course, you book is correct too. To get the force, you consider the piece with the Lagrange parameter as a potential for the constraint forces (in this case time dependent, but that doesn't matter). Then the force is given as the gradient of this "potential". Now you have to define the gradient in terms of your generalized coordinates. This is fortunately done in any textbook, because you have cylinder coordinates, and the gradient is

$$\vec{\nabla} \Phi(r,\varphi,z)=\vec{e}_r \partial_r \Phi + \frac{\vec{e}_{\varphi}}{r} \partial_{\varphi} \Phi+\vec{e}_z \partial_z \Phi.$$

Note that you don't need to take into account the derivatives of ##\lambda##, because the constraint equation makes this contribution obviously vanish.

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