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Consider a pendulum consisting of two parts: a uniform rod of massm, lengthl, negligible thickness and with one end fixed; and a uniform disk of mass [tex]\mu[/tex] and radius [tex]\rho[/tex].

The rod is moving in a plane, and the disk is attached at a pointPon its boundary to the non-fixed end of the rod, in such a way that it can freely rotate aboutPin the plane in which the rod is moving.

Obtain the Lagrangian and the equations of motion.

I suppose I should divide this into two parts; one for the rod and one for the disk.

For the rod I get

[tex]T_1 = \frac{1}{2} I_1 \dot{\theta}_1^2 + \frac{1}{2} m(\dot{x}_1^2 + \dot{y}_1^2)[/tex]

,where the index 1 is the rod, and

[tex]V_1 = mgh = mg \frac{l}{2} (1 - \cos \theta_1)[/tex].

The moment of inertia for the rod is

[tex]I_1 = \frac{1}{3} m l^2[/tex].

So far so good, I think.. But how should I do with the disc? Should I treat this the same way and just use a superposition of the two lagrangians? And how do I get the moment of inertia?

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# Rigid body motion

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