Lagrangian of System w/ 2 M masses & Rigid Rod - Physics Homework

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SUMMARY

The discussion focuses on deriving the Lagrangian for a system involving two masses, m1 and m2, connected by a rigid rod and suspended by strings of lengths L1 and L2. The Lagrangian is defined as L = T - U, where T represents kinetic energy and U represents potential energy. The user intends to utilize the generalized coordinate phi to express the system's dynamics, employing the equations T = ½ I1w1^2 + ½ I2w2^2 and V = -m1h1 – m2h2 to calculate the energies involved. The challenge lies in correctly applying these principles to a system with two masses and a rigid connection.

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Homework Statement


two masses, m1 and m2, are hung from a point, P, by two string, L1 and L2, the masses are connected by a rod, length D, of negligible mass. The angle between the strings is theta, the angle between L2 and horizontal line through P is Phi. so essentially it looks like a hanger with a mass at both ends of it.

Homework Equations


The formula for the Lagrangian is simply L=T-U, where T and U are the Potential and Kinetic energy.
To find the energy I'm going to need velocity's which I think I can get by taking the Derivative of the position equations.

The Attempt at a Solution


I am going to use phi as my generalized coordinate, and if this were a simple pendulum I think this problem would be pretty easy, you take the derivative of your position equations to get velocity then use T=1/2MV^2 and U=mgh and then plug and chug. With this problem having two masses separated by a rigid rod I'm completely stumped on how to go about this. Any help at all is greatly appreciated.
 
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In the triangle, theta will remain constant. Draw the diagram. L1 and L2 are the lengths of the strings.

T = ½ I1w1^2 + ½ I2w2^2 = ½ m1*L1^2(dphi/dt) + ½ m2*L2^2(dphi/dt), since theta is a const.

V = -m1h1 – m2h2 = -mL1sin phi – mL2 sin (phi + theta).

Now you can set up L as phi as the only generalized co-ordinate.
 

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