How Do You Solve a Rotating Mass on a Spring Using Lagrangian Mechanics?

  • Thread starter Levi Tate
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This means that ε and ε'' can be eliminated, leaving a simple equation involving only x and x'', which has the form of a simple harmonic oscillator. Therefore, the frequency of small oscillations about the equilibrium point can be found using the same formula as for a simple harmonic oscillator, ω^2 = (k-mΩ^2)/m. This shows that the system has a natural frequency of oscillation, even though it is not a simple harmonic oscillator.
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Levi Tate
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Homework Statement



A Particle of mass m is threaded on a frictionless rod that rotates at a fixed angular frequency Ω about a vertical axis. A spring with rest length Xo and spring constant k has one of it's ends attached to the mass and the other to the axis of rotation. Let x be the length of the spring in it's stretched condition and thus the distance of the mass from the axis of rotation. Assume k is greater than mΩ^2

a. Construct a Lagrangian for this system

b. Find the equation of motion of the mass

c. Find Xeq, the equilibrium position of the mass

d. Find the frequency of small oscillations about this equilibrium point


Homework Equations



L= T - U

∂L/∂x = d/dt (∂L/∂x')



The Attempt at a Solution



I have my solution attached as a copy. I am not trying to find a solution, but to understand what the solution means, and how to make sense of it.

It is part c and part d that I am having trouble understanding..

A. L = 1/2m(x'^2 + x^2Ω^2) - 1/2k(x - Xo)

B. mΩ^2x - k(x-Xo) = mx''

(x' = dx/dt, ect..)

C. Xeq = (-kXo)/(mΩ^2-k) (because x' = x'' = 0)

D. Letting x = Xeq + ε → x'' = ε''

Then he writes here, mε'' +(k-mΩ^2)ε = 0

Which has the form of the simple harmonic oscillator, mx'' +kx = 0

So the frequency is just like in the simple harmonic oscillator, ω^2 = (k-mΩ^2)/m


My only questions concern the last two steps.

What is going on for part C I do understand that if the mass is in equilibrium than it does not have an acceleration or a velocity, so that follows naturally.

I do not understand part D, because in the equation of motion, kXo has vanished, which I do not understand.

x = Xeq + ε → x'' = ε''

I don't understand how this here implies that I can just ignore that Xo term. I'm just not really sure what's going on with this last part here. If anybody could explain this to me it would be of great use to me. Thank you.
 
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For part D, after substituting x = xeq + ε into the equation of motion, collect together all terms that do not involve ε or ε'' and show that these terms together equal zero.
 

Related to How Do You Solve a Rotating Mass on a Spring Using Lagrangian Mechanics?

What is a Typical Lagrangian Problem?

A Typical Lagrangian Problem is a mathematical problem that involves finding the minimum or maximum value of a function, subject to a set of constraints. It is named after Joseph-Louis Lagrange, who developed the method of Lagrangian optimization.

How is a Typical Lagrangian Problem solved?

A Typical Lagrangian Problem is typically solved using the method of Lagrange multipliers. This method involves setting up a system of equations known as the Lagrange equations, and then solving them to find the optimal solution.

What are the applications of Typical Lagrangian Problems?

Typical Lagrangian Problems have a wide range of applications in physics, economics, engineering, and other fields. They can be used to optimize the design of structures, maximize profits in business, and model physical systems.

What is the difference between a Lagrangian and an Euler-Lagrange equation?

A Lagrangian equation is a mathematical equation that describes a system in terms of its energy. An Euler-Lagrange equation is a type of differential equation that is used to find the extrema (maximum or minimum) of a function. The Euler-Lagrange equation is derived from the Lagrangian equation.

Can Lagrangian optimization be used for non-differentiable functions?

No, Lagrangian optimization is typically used for differentiable functions. If a function is not differentiable, it may not have a well-defined minimum or maximum value, making it difficult to apply the method of Lagrange multipliers.

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