Potential Energy of Modified Atwood Machine

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SUMMARY

The discussion focuses on calculating the total potential energy of a modified Atwood machine involving two equal masses connected by a spring and a heavier mass on the opposite side. The derived potential energy equation is confirmed as U = (1/2)kx², where k is the spring constant and x is the extension of the spring. The participants analyze the contributions of gravitational potential energy and spring potential energy, emphasizing the significance of ignoring constant terms in the context of Lagrangian mechanics. The conversation highlights the importance of understanding the center of mass motion in oscillatory systems.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with potential energy equations, specifically U_{grav} and U_{spr}
  • Knowledge of oscillatory motion and center of mass concepts
  • Basic proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study Lagrangian mechanics and its applications in oscillatory systems
  • Explore the derivation of potential energy in spring-mass systems
  • Learn about the center of mass and its implications in mechanical systems
  • Investigate the mathematical treatment of quadratic equations in physics
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Students and educators in physics, particularly those focusing on mechanics, as well as anyone interested in understanding the dynamics of spring-mass systems and potential energy calculations.

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Homework Statement


atwoodmachine.jpg


The two weights on the left have equal masses m and ar connected by a massless spring of force constant k. The weight on the right has mass 2m, and the pulley is massless and frictionless. The coordinate x is the extension of the spring from its equilibrium length; that is, the length of the spring is l+x where l is the equilibrium length with all the weights in position and with the 2m weight held stationary.

Show that the total potential energy is just U=\frac{1}{2}kx^{2}


Homework Equations


U_{grav}=mgh
U_{spr}=\frac{1}{2}kx^{2}


The Attempt at a Solution


Attempt 1 (ignoring all sorts of constants added onto potential energy, since they'll disappear in Lagrange's/Hamilton's equations):
U=2mgy-mgy-mg(y+l+x)+\frac{1}{2}kx^{2}
U=-mgx+\frac{1}{2}kx^{2}
(the l was dropped for the noted reason above)

Attempt 2 (preserving all constants incase I missed something):
let the length of the rope be L;
U=2mg(y-L)-mgy-mg(y+l+x)+\frac{1}{2}kx^{2}
U=\frac{1}{2}kx^{2}-mgx-mg(2L+l)

I turned up something on a Dutch forum about changing the quadratic in the bottom from the form Ax^2+Bx+C to something like Ax^2+D and then just letting D=0 (since constant adjustments of potential don't matter), but I have absolutely no idea how to do that and it seems kind of mathematically spurious that you can change a linear term into a constant.

Am I missing something here?
 
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Think center of mass. If you stretch the spring and let go, what kind of motion does the CM undergo?
 
Well, I worked through the whole problem working out the equations of motion, and for x it turns out that you get the oscillatory motion plus a constant (mg/k), and that constant is the displacement from equilibrium that occurs when you let the spring hang at rest, which is accounted for in the original problem statement by the definition of l, so it doesn't belong in the equation for x. I argued that you should ignore -mgx in the potential from this fact.

But from your question even, I still can't see how you can argue it from the beginning. The mass on the spring would oscillate about x=0. I'm not sure what CM you're talking about. The CM of the whole system? It would oscillate too.
 

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