Pendulums, total energy?, and Mathematica

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SUMMARY

The discussion focuses on deriving the total energy equation for a simple linear pendulum using Mathematica. The total energy (E) is expressed as E = (1/2) m l^2 (d(theta)/dt)^2 + (1/2) mgl (theta)^2, which simplifies to E = (1/2) mgl(theta0)^2 at the initial angle theta0. The user struggles with applying the correct equations for kinetic and potential energy and seeks guidance on using Mathematica for solving differential equations and graphing the period versus initial angle.

PREREQUISITES
  • Understanding of classical mechanics, specifically kinetic and potential energy equations.
  • Familiarity with the simple harmonic motion of pendulums.
  • Basic knowledge of Mathematica, including functions like dsolve and manipulate.
  • Proficiency in calculus, particularly in differentiation and solving differential equations.
NEXT STEPS
  • Learn how to use Mathematica's dsolve function for solving ordinary differential equations.
  • Research the derivation of energy equations for pendulums in classical mechanics.
  • Explore Mathematica's graphing capabilities to plot functions and analyze data visually.
  • Study the relationship between angular displacement and period in simple harmonic motion.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and pendulum motion, as well as individuals learning to use Mathematica for solving and visualizing mathematical problems.

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Homework Statement


A)Show that for a non-frictional, simple linear pendulum (Sin(theta) ~ theta) the total energy of the pendulum (K + U) or kinetic plus potential is given by

E = (1/2) m l^2 (d(theta)/dt)^2 + (1/2) mgl (theta)^2

and therefor E = (1/2) mgl(theta0)^2

theta0 = theta(t=0)


Homework Equations


F=ma , delta K = 1/2mv^2 , delta U = mgh


The Attempt at a Solution



Alright, so I'm essentially lost in this problem, and my last calculus class was approximately 2 years ago.
I understand that the total energy should be the sum of the potential and kinetic energies of the pendulum, so it seems that E = 1/2mv^2 +mgh. But it seems that i am stuck here. I observe that the change between the kinetic energy portion of the equation is different in that v^2 is now l^2 (d(theta)/dt)^2, and the potential mgh now appears as 1/2mgl (theta)^2, but I cannot think of how to determine how to get to that point. And therefor I am unable to get to the main portion of the problem.

{a} One more problem. use mathematica to solve (d(theta)/dt)^2 +g/lsin(theta)=0 . And, show a graph of period vs. (theta 0).

Before this course I have not used mathematica, and now I am facing difficulties. I tried using the dsolve and manipulate functions many times over the past week in an attempt to graph the problem but I have been unsuccessful.

Please note that I am NOT asking for any answers just for guidance.
 

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I managed to figure out, after drawing a simple diagram (of course!), that h in the potential equation has got to be equal to L-Lcos(theta). Also, the velocity is going to be equal to L(dtheta/dt), the length multiplied by the rate of change of the center angle. Knowing this,
k= 1/2 m (L(dtheta/dt))^2, and
U= mgh= mg (L-Lcos(theta))
so
E= 1/2 m L^2 (dtheta/dt)^2 + mg(L-Lcos(theta)), which appears to be very similar to the original equation that is given, but I am again stuck. I can't seem to find why the first given equation is equal to the second.
 
So... anybody understand how to graph in mathematica or have any words of wisdom on the above problems? Anything would be useful... :D
 

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