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Can you help me to set Lagrangian?

I found that
$$\vec r_A=b\sin\theta\vec i+b\cos\theta\vec j$$
$$\dot{\vec r_A}=b\dot\theta\cos\theta\vec i-b\dot\theta\sin\theta\vec j$$

For point $G$ I've got this ($G$ is center of mass of a diss): $$\vec r_G=(b\sin\theta+a\sin\phi)\vec i+(b\cos\theta+a\cos\phi)\vec j$$
$$\dot{\vec r_G}=(b\dot\theta\cos\theta+a\dot\phi\cos\phi)\vec i-(b\dot\theta\sin\theta+a\dot\phi\sin\phi)\vec j$$

Kinetic energy for point $A$ is
$$T_A=\frac{1}{2}m{r_A}^2$$

$\vec i$ is directed horizontally to the right and $\vec j$ is vertically downwards. Point $O$ is origin of the coordinate system.

How does a kinetic energy for the center of the mass of a disc look like and which forces except gravitational are acting on a systementer image description here?

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closed as off-topic by Emilio Pisanty, user10851, Qmechanic Nov 5 '13 at 21:37

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  • $\begingroup$ It seems to me that this might actually be easier to solve if you work with polar coordinates. $\endgroup$ – Kyle Kanos Aug 22 '13 at 15:44
  • $\begingroup$ I can try. Can you just tell me if kinetic energy of the center of a disc is $T_G=\frac{1}{2}m\dot{r_G}^2$? That's how I should calculate it? And if you can tell me which forces are acting here? $\endgroup$ – gov Aug 22 '13 at 15:49
  • $\begingroup$ I think you will need to incorporate the moment of inertia with the kinetic energy, but I do get the same $\vec{r}_G$ when using Cartesian geometry--which I think might really be the way to solve this, as I didn't note the hinge at $A$ at first. $\endgroup$ – Kyle Kanos Aug 22 '13 at 16:07
  • $\begingroup$ I've found that in the case of a body rotating about a point $Q$ of itself which is anchored, $$T=\frac{1}{2}\vec\omega L_Q$$. In this problem, disc is rotating about $A$. I don't have given angular velocity. Can I write it in terms of $\dot\phi$ or $\phi$ somehow? $\endgroup$ – gov Aug 22 '13 at 16:21
  • $\begingroup$ Yes, $\omega\propto\dot{\phi}$ in this case. $\endgroup$ – Kyle Kanos Aug 22 '13 at 16:28