# Setting up the Lagrangian for a pendulum with a circular weight

A pendulum consists of a light inextensible rod $OA$, length $b$, freely pivoted at $O$, attached at $A$ to the rim of a uniform disc of radius $a$ and mass $m$. The disc is free to rotate about $A$. The system moves in a vertical plane containing $O$.

Can you help me to set Lagrangian?

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

I am not sure if this is correct for point $G$ ($G$ is center of mass of a dics): $$\vec r_G=(b\sin\theta+a\sin\phi)\vec i+(b\cos\theta+a\cos\phi)\vec j$$

This is all I've done for now. I would appreciate some help and hints for position vector of the point $G$ and forces acting here.

• Where do $\vec i$ and $\vec j$ point? One has to guess for now... – Ruslan Aug 22 '13 at 13:59
• $\vec i$ horizontaly to the right, $\vec j$ vertically downwards. – gov Aug 22 '13 at 14:02
• $\vec r_G$ looks correct since it's $\vec{OA}+\vec{AG}$. As for forces... is it in uniform gravitation field or completely free? – Ruslan Aug 22 '13 at 14:06
• There is only gravitational force. I am only not sure how to use the fact that a disc rotates about $A$. Do I have to use rotating frame of reference? But no angular velocity is mentioned here. – gov Aug 22 '13 at 14:11
• You already have $\phi$, which accounts for rotation. This is just another coordinate in Lagrangian. – Ruslan Aug 22 '13 at 14:42