# Homework Help: Lagrangian Mechanics Problem

1. Oct 11, 2009

### pandagoat

1. The problem statement, all variables and given/known data
Consider a vertical plane in a constant gravitational feild. Let the origin of a coordinate system be located at some point in this plane. A particle of mass m moves in the vertical plane under the influence of gravity and under the influence of an aditional force f = -Ar^(a-1) directed toward the origin (r is the distance from the origin; A and a [does not = 0 or 1] are constants). Choose appropriate generalized coordinates, and find the Langrangian equations of motion. Is the angular momentum about thet origin conserved?

2. Relevant equations
$$L = T - U$$
$$\frac{\partial L}{\partial q_i} - \frac{d}{dt}\frac{\partial L}{\partial \dot q_i} = 0$$

3. The attempt at a solution
Choose the coordinate system (r', y') centered at the origin, so that the r' axis is in the direction of the origin to the particle of mass m and the y' axis is perpendicular to r'.

First, $$\arrow F = (-mg\cos{\theta}, -mg\sin{\theta}) + (-Ar^{a-1}, 0) = (-mg\cos{\theta} - Ar^{a-1}, -mg\sin{\theta})$$.

From here we can guess the potential:
$$U = mgr\cos{\theta} + \frac{A}{a}r^a + mgy\sin{\theta}$$.

(The preceding steps may be wrong, but the following is what I'm not sure about):
Notice that in our coordinate system y is always 0. So our potential is $$U = mgr\cos{\theta} + \frac{A}{a}r^a$$. For the kinetic energy, we get $$T = 1/2m(\dot r^2 + \dot \theta^2r^2) <ACCIDENT = 1/2m\dot \theta^2r^2>$$. So our Langrangian is $$L = 1/2m(\dot r^2 + \dot \theta^2r^2) - mgr\cos{\theta} - \frac{A}{a}r^a$$.

The rest is easy if I did the preceding correctly. You get that angular momentum is not conserved when you find $$\frac{\partial L}{\partial \theta} - \frac{d}{dt}\frac{\partial L}{\partial \dot \theta} = 0$$.

Last edited: Oct 11, 2009
2. Oct 11, 2009

### kuruman

Looks OK. You should get two equations of motion. What does this mean? Are you saying r-dot is zero?

$$T = 1/2m(\dot r^2 + \dot \theta^2r^2) = 1/2m\dot \theta^2r^2$$

Last edited: Oct 11, 2009
3. Oct 11, 2009

### pandagoat

That was an accident. In the Lagrangian, you can see I did not carry that part through. I will put the incorrect part in brackets. Thank you! My confidence in this material has been elevated.

Last edited: Oct 11, 2009
4. Oct 11, 2009

### kuruman

Just checking. Lagrangians are fun once you get used to them.