Lagrangian mechanics: Bar connected to a spring

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SUMMARY

The discussion focuses on solving a Lagrangian mechanics problem involving two masses connected by a spring and a rod. The Lagrangian is defined as L = T - V, where T represents the total kinetic energy and V the potential energy. The equations of motion derived using the Euler-Lagrange equations are (m1 + m2) \ddot z = -kz + (m1 + m2) g for vertical motion and \frac{d}{dt}(m2 l^2 \dot \varphi) = 0 for rotational motion. The analysis confirms that the system behaves like a mass on a spring without damping, and the two degrees of freedom are independent.

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Homework Statement



Mass 1 can slide on a vertical rod under the influence of a constant gravitational force and and is connected to the rod via a spring with the spring konstant k and rest length 0. A mass 2 is connected to mass 1 via a rod of length L (forms a 90 degree angel with the first rod, and is otherwise rotatable).[/B]
1. Find the lagrangian
2. Find the equations of motion with the euler-lagrange equations
3. Find the solutions to the equations of motions

Homework Equations


L = T-V
\frac{d}{dt} \frac{\partial L}{\partial \dot z} - \frac{\partial L}{\partial z}
\frac{d}{dt} \frac{\partial L}{\partial \dot \varphi} - \frac{\partial L}{\partial \varphi}

The Attempt at a Solution


This is the first time I have to deal with lagrangian mechanics.
I thought we have 2 free coordinates. We have z for the up and down motion and \varphi for the rotation on the x-y Plane of mass 2.
The potential energy is: V = V_{Spring} + V_{grav1} + V_{grav2} = \frac{1}{2} k z^2 + m_1 g z + m_2 g z
The kinetic energy: T = T_1 + T_2 = \frac{1}{2} m_1 \dot z^2 + \frac{1}{2} m_2 (\dot L^2 + L^2 \dot \varphi ^2 + \dot z^2)
\dot L^2 must be 0 cause the length of the 2nd rod doesn't change over time
Also I thought, since mass 2 is moving on the edge of a cylinder, I would need to use the kinetic energy in cylinder coordinates for mass 2

Questions:
I plugged my lagrangian into my 2 euler-lagrange equations and got this:
for z : (m_1 + m_2) \ddot z = -kz + (m_1+m_2) g
for \varphi: \frac{d}{dt}(m_2 l^2 \dot \varphi) = 0

I am not sure about r. This looks like the equation of motion of a damped harmonic oscillator. It is not clear from the exercise if the spring is damped or not, so I am not sure if that is correct.
The 2nd equation just means, that angular momentum doesn't change with respect to time? Is there anything I could use from the conversation of angular momentum to solve the first equation or am I done here?[/B]

Edit:
Thanks
 
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Hi,
Paumi said:
I am not sure about z. This looks like the equation of motion of a damped harmonic oscillator.
No damping (usually that's a term ##\beta \dot z##). Jus a mass on a spring.
The 2nd equation just means, that angular momentum doesn't change with respect to time?
Correct !
Is there anything I could use from the conservation of angular momentum to solve the first equation or am I done here?
There is nothing in common, so: no. The two degrees of freedom are totally independent.
 
BvU said:
Hi,
No damping (usually that's a term ##\beta \dot z##). Jus a mass on a spring.
Correct ! There is nothing in common, so: no. The two degrees of freedom are totally independent.
Thank you very much! I'll try to solve it now
 

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