Equation of motion for a pendulum using Lagrange method

Click For Summary
SUMMARY

The discussion focuses on deriving the equation of motion for a pendulum using the Lagrange method, specifically addressing the calculations of kinetic and potential energy for a system of spheres. Participants emphasize the importance of considering the vertical distance from the center of mass when calculating potential energy and the necessity of using the Pythagorean theorem for kinetic energy calculations. The potential energy is defined through the integral of gravitational potential and density, leading to the conclusion that the potential energy can be simplified by focusing on the center of mass. The correct formulation of these energies is crucial for accurate modeling of the pendulum's motion.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with kinetic and potential energy calculations
  • Knowledge of the Pythagorean theorem
  • Concept of center of mass in physics
NEXT STEPS
  • Study the derivation of Lagrange's equations for multi-body systems
  • Learn about the calculation of potential energy in gravitational fields
  • Explore the application of the Pythagorean theorem in physics problems
  • Investigate the role of center of mass in dynamics and energy calculations
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in advanced dynamics and the mathematical modeling of physical systems.

morpheus343
Messages
29
Reaction score
4
Homework Statement
Find the equation of motion of the pendulum
Relevant Equations
Lagrangian L=T-V
Kinetic-Potential energy
1706764977595.png


I move the system by a small angle θ . I am not sure if my calculations of kinetic and potential energy are correct.
1706765036261.png
 
  • Like
Likes   Reactions: DeBangis21
Physics news on Phys.org
What calculations? All you have given is the final result, which does not look correct to me.
 
  • Love
Likes   Reactions: Delta2
1706865003406.png

I do not understand if this just a problem treated like a pendulum with a simple sphere but just adding the other 2 sphere's masses. Is the potential energy supposed to have a different y for each sphere?
 
You are asserting that the speed of the outer spheres are the same as that of B? Is this reasonable?

morpheus343 said:
Is the potential energy supposed to have a different y for each sphere?
What would you say based on your second diagram in the OP? Are all spheres at the same height? Does the answer matter?
 
1706869811340.png

Then it would be like this. This was my first assumption but i got lost in the algebra and thought i was maybe wrong and it was supposed to be simpler
 
Well, for the potential energy, perhaps there is an argument to make it simpler? What can you say about the center of mass?

For the kinetic energy, I did not check the math, but yes, you indeed need to take the distance to C and D using the Pythagorean theorem. Note that, since the distance appears squared in T, you should probably keep the exact rational 1+4/9 = 13/9 in the expression.
 
Orodruin said:
Well, for the potential energy, perhaps there is an argument to make it simpler? What can you say about the center of mass?

For the kinetic energy, I did not check the math, but yes, you indeed need to take the distance to C and D using the Pythagorean theorem. Note that, since the distance appears squared in T, you should probably keep the exact rational 1+4/9 = 13/9 in the expression.
It's at sphere B which is the midpoint of CD. So i only need the vertical distance from the centre of mass when calculating the potential energy?
 
morpheus343 said:
It's at sphere B which is the midpoint of CD. So i only need the vertical distance from the centre of mass when calculating the potential energy?
Yes. By definition, the potential energy for any mass distribution is given by
$$
V = \int U(z) \rho(\vec x) dV
$$
where ##U(z)## is the gravitational potential and ##\rho## the density. Assuming ##U(z) = gz## (homogeneous field), you’ll find
$$
V = g \int z \rho(\vec x) dV \equiv m g \bar z
$$
where ##\bar z## is the z-coordinate of the center of mass by definition:
$$
\bar z = \frac 1m \int z \rho(\vec x) dV
$$
 
  • Informative
  • Like
Likes   Reactions: PhDeezNutz and Delta2

Similar threads

  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K