How to integrate partial differential eqn

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Homework Help Overview

The discussion revolves around integrating a partial differential equation related to the dynamics of a pendulum. The original poster expresses difficulty with the integration process due to the presence of a squared term in the equation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to integrate the equation by manipulating terms and considers various methods, including integrating a transformed version of the equation. Some participants question the origin of the equation and its assumptions regarding forces acting on the pendulum.

Discussion Status

Participants are exploring different interpretations of the problem, with some suggesting that the original equation may not fully account for all forces involved. There is a suggestion to integrate Newton's second law instead, indicating a potential shift in approach.

Contextual Notes

There is a mention of the need to incorporate the maximum amplitude of the pendulum and the relationship between angular displacement and tension, which may influence the integration process. The discussion reflects uncertainty about the completeness of the original equation.

agalea91
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Homework Statement



im not sure the elegant way to put it, i need to integrate this somehow (ie, throw over the dt and integrate each side) but that squared is really tripping me up. is there a trick i should be using?
-g(cos(x))=r(dx/dt)^2

Homework Equations





The Attempt at a Solution



my attempt was initially to integrate 1/(cos(x))^(1/2), that is not pretty

i heard classmates talking about multiplying though by (dx/dt), not sure if that applies

any help appreciated, thanks!
 
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Where did this equation come from?
 
Im solving for the tension in a pendulum, right now I am equating the radial accel with the force of gravity to try and integrate to something with a theta-not (representing the max amplitude of the pendulum) and theta (t). Hope that makes some sense
 
Your equation looks wrong since it suggests the only force in the radial direction is gravity, but the tension also acts radially.

You'd be better off integrating Newton's second law to find ##\theta(t)##, and then use F=ma to solve for the tension.
 
You're right, the full radial equation as I have it is this,

T=mgcos(theta)+mL(dtheta/dt)^2

My hope was that in integrating the right half independently of the tension then I could make a substitution (from the result of that integration, into the full equation) that would allow me to have an answer with theta-not component(s)
 

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