# Classical Mechanics: Lagrangian for pendulum with oscillating support

1. Feb 14, 2012

### Opticmist

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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:

$$\dot{x}= -a\gamma sin(\gamma t)+l cos(\phi) \dot{\phi} ; \dot{y} = -l sin(\phi) \dot{\phi}$$

$$L = T - U = \frac{1}{2} m ( \dot{x}^2 + \dot{y}^2) - U$$

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

$$L = \frac{1}{2} ml^2 \dot{\phi}^2 + m l a \gamma^2 cos (\gamma t) sin(\phi) + m g l cos(\phi)$$

My solution is different from the one given in the book. Could someone please tell me where I'm going wrong?

2. May 12, 2015

### Juan Carlos

for sure, there is a total derivated connecting both.

3. May 12, 2015

### BvU

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:

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 !

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