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Lagrangian of a Rotating Mass, Spring and Hanging Weight

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The mass m1 moves on a smooth horizontal plane, m2 moves vertically under the force of gravity and the spring. Using polar coordinates r, theta for m1, l for m2 and taking b for the total length of the string plus the unstretched length of the spring, find: (diagram attached in file)
a. the Langrangian of the system
b. the equations of motion for mass m1 in terms of the radial coordinate r, and for m2 in terms of l
c. at any given angular velocity, theta, there will be 'equilibrium' values for the positions of m1 and m2. FInd these values r zero and l zero.

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Solution Summary

The solution shows in detail how to construct the system's lagrangian, obtains the equations of motion and calculates equilibrium values.

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In polar coordinates we have:
(1.1)
Therefore:
(1.2)
The square of the speed is:

(1.3)
The kinetic energy of is
(1.4)
The speed of is , thus the kinetic energy of is:
(1.5)
The total kinetic energy is:
(1.6)
If we set the zero potential energy plane as the table, then has no potential energy and has a potential energy of:
(1.7)
The length of the string and the stretched spring is , while the length of the ...

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