Classical Mechanics: Lagrangian for pendulum with oscillating support

Click For Summary
SUMMARY

The discussion focuses on deriving the Lagrangian for a simple pendulum with a horizontally oscillating support, as presented in Landau's Mechanics. The correct Lagrangian is established as L = (1/2) ml² ̇φ² + m l a γ² cos(γt) sin(φ) + m g l cos(φ). The user’s initial derivation differs from the book's answer due to an oversight in handling the total derivative terms and the interpretation of the oscillating support's contribution. Clarification on notation and proper use of TeX for mathematical expressions is also emphasized.

PREREQUISITES
  • Understanding of Lagrangian mechanics and its formulation.
  • Familiarity with the concepts of kinetic and potential energy in physics.
  • Proficiency in using TeX for mathematical notation.
  • Knowledge of oscillatory motion and its mathematical representation.
NEXT STEPS
  • Study the derivation of Lagrangians for systems with non-inertial reference frames.
  • Explore the application of the Euler-Lagrange equation in complex mechanical systems.
  • Learn about the implications of total derivatives in Lagrangian formulations.
  • Review examples of oscillating systems in classical mechanics for deeper insights.
USEFUL FOR

Students of classical mechanics, physicists working on dynamic systems, and anyone interested in advanced applications of Lagrangian mechanics.

Opticmist
Messages
3
Reaction score
0

Homework Statement



Greetings! This is an example problem at the end of Chapter 1 in Mechanics (Landau):
A simple pendulum of mass m whose point of support oscillates horizontally in the plane of motion of the pendulum according to the law x=acos(\gamma t).
Find the Lagrangian.


Homework Equations





The Attempt at a Solution


x = a cos(\gamma t) + l sin(\phi) ; y = l cos(\phi)

Where, \phi is the angle the pendulum makes with the vertical. Then:


<br /> \dot{x}= -a\gamma sin(\gamma t)+l cos(\phi) \dot{\phi} ; \dot{y} = -l sin(\phi) \dot{\phi}

<br /> L = T - U = \frac{1}{2} m ( \dot{x}^2 + \dot{y}^2) - U<br />

Which gives (ignoring full time derivative terms):
<br /> L = \frac{1}{2} ml^2 \dot{\phi}^2 - m l a \gamma sin (\gamma t) cos(\phi)\dot{\phi} + m g l cos(\phi)<br />

Answer given in the book:
<br /> L = \frac{1}{2} ml^2 \dot{\phi}^2 + m l a \gamma^2 cos (\gamma t) sin(\phi) + m g l cos(\phi)<br />


My solution is different from the one given in the book. Could someone please tell me where I'm going wrong?
 
Physics news on Phys.org
for sure, there is a total derivated connecting both.
 
Funny, the internet pdf (2nd edition) has a problem 2 and a problem 3 that show similarities. If I look at problem 2 (and remove m1) that's just what you find:
upload_2015-5-12_23-26-18.png

in your case x(t) is given, and ##(-a\gamma \sin\gamma t)^2 ## isn't interesting (can you tell why not?), so that leaves your expression !

Note that ##a sin\gamma t## is not as clear as ##a\sin\gamma t## ( a \sin \gamma t ) because some folks might interpret the ##asin## as ##\arcsin## ( \arcsin ).
In addition, TeX recognizes the \sin as a function and takes care of proper spacing !

--
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
10K
  • · Replies 0 ·
Replies
0
Views
2K
Replies
0
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
3K
Replies
1
Views
2K