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A particle of mass m moves frictionlessly under the influence of gravity on a curve defined by:

$x=a(\phi+\sin\phi)$ and $y=a(1-\cos\phi)$.

a) Set up the terms for the kinetic and potential energy.

b)Use a suitable generalized variable $q=f(\phi)$ to turn both terms into a purely quadratic form.

c)Set up the Lagrange-function $L=L(q,\dot{q})$ and derive the equations of motion from that.

We just started talking about generalized variables and Lagrange functions in class and to be honest I don't really understand the concept yet.

I mean I don't even know how to derive the kinetic and potential energy.

My only ideas for the kinetic energy were to get $\dot{x}$ and $\dot{y}$ and then going with $\frac{1}{2}m(\dot{x}^2+\dot{y}^2)$ where as $\dot{x}=a(\dot{\phi}+\dot{\phi}\sin{\phi})$ and $\dot{y}=-a\dot{\phi}\cos{\phi}$, right?

Simplification on KE: $$1/2m(a^2(\dot{\phi}^2+2\dot{\phi}^2\sin(\phi)^2+\dot{\phi}^2\sin(\phi)^2)+a^2\dot{\phi}^2\cos{\phi}^2)=1/2ma^2\dot{\phi}^2\cos(\phi)$$

Following Divergent Queries' comment the potential energy is then just $V=mga(1-\cos(\phi))$? Anyway, I hope someone with more knowledge with this than me could help me.

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Nicole Parson
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  • How is $z$ defined, or is that a typo? A natural choice may be to define the potential energy as $mgy$ (assuming that gravity acts along the $y$-axis in the negative direction), so it would be $0$ when $y=0$. – hexaflexagonal Jul 08 '15 at 20:41
  • Oh, you are right. My mistake, I will edit that. I was thinking of the $\mathbb{R}^3$ coordinate system. – Nicole Parson Jul 08 '15 at 20:45
  • You seem to be doing pretty well with the kinetic energy so far. Try substituting for $\dot x$ and $\dot y$ in your KE formula (so that the energy is expressed as a function of $\phi$ and $\dot\phi$) and try to simplify the formula as much as you can. – David K Jul 08 '15 at 20:48
  • @DavidK Okay, as long as I didn't do some arithmetic mistakes the simplification I did in the post should be correct, right? – Nicole Parson Jul 08 '15 at 20:57
  • Are you sure of $x=a(\phi+\sin\phi)$ instead of $x=a(\phi-\sin\phi)$?. With the signe minus you have the parametrics of a cycloid which seems appropriate for this problem of physics. – Piquito Jul 08 '15 at 21:00
  • I'm not sure. At least that's what it says in my textbook. It could be a typo of the book but I think that's unlikely. – Nicole Parson Jul 08 '15 at 21:06
  • By the way, when you crosspost the exact same question to multiple SE sites, you should put links in each of the questions pointing to the copies of the question on the other site(s), like this one: http://physics.stackexchange.com/q/193175 (and the physics question should link back to this one, but I can't do that for you at this time). – David K Jul 09 '15 at 03:31

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Your formula for kinetic energy, $E_k = \frac12 m\left(\dot{x}^2+\dot{y}^2\right)$, looks good. Expressing $\dot x$ and $\dot y$ in terms of $\phi$ is a good idea, but note that $\frac{d}{dt}\sin u = \dot u \cos u$, not $\dot u \sin u$, and $\frac{d}{dt}\cos u = -\dot u \sin u$, not $\dot u \cos u.$

You also have an error in your simplification of the kinetic energy (or what would be the kinetic energy if the formulas for $\dot x$ and $\dot y$ were what you wrote).

I find that $\dot{x} = a (\dot\phi + \dot\phi \cos\phi) = a \dot\phi(1 + \cos\phi)$ (factoring out the $\dot\phi$ since it seemed to make the formulas just a little easier to work with) and $\dot{y} = a \dot\phi \sin\phi.$ The kinetic energy then simplifies as shown below. (But you might want to try it yourself, again, with the correct formulas for $\dot x$ and $\dot y$ before you read further.)

$$\begin{eqnarray} E_k &=& \frac12 m\left(a^2 \dot\phi^2 (1 + \cos\phi)^2 + a^2 \dot\phi^2 \sin^2\phi\right) \\ &=& \frac12 m a^2 \dot\phi^2 \left((1 + \cos\phi)^2 + \sin^2\phi\right) \\ &=& \frac12 m a^2 \dot\phi^2 (1 + 2\cos\phi + \cos^2\phi + \sin^2\phi) \\ &=& \frac12 m a^2 \dot\phi^2 (2 + 2\cos\phi) \\ &=& m a^2 \dot\phi^2 (1 + \cos\phi). \\ \end{eqnarray}$$

David K
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