Lagrangian of a Particle in Spherical Coordinates (Is this correct?)

[Xyius]

Homework Statement

a.) Set up the Lagrange Equations of motion in spherical coordinates, ρ,θ, $\phi$ for a particle of mass m subject to a force whose spherical components are $F_{\rho},F_{\theta},F_{\phi}$.

This is just the first part of the problem but the other parts do not seem so bad.

Homework Equations

Lagrangian equations of motion.
Since the problem doesn't state it is in any uniform conservative field from which it would have a potential function, I assume I use the more general form of the Lagrangian.
$$\frac{d}{dt} \frac{∂T}{∂\dot{q}}-\frac{∂T}{∂q}=Q_k$$

The Attempt at a Solution

So my three general forces would be..
$$Q_{\rho}=F_{\rho}$$
$$Q_{\theta}=\rho F_{\theta}$$
$$Q_{\phi}=F_{\phi}$$

The kinetic energy in Spherical coordinates is..
$$T=\frac{1}{2}m\left( \dot{\rho}^2 +\rho^2 \dot{\theta}^2+\rho^2 \dot{\phi}^2 sin^2\theta \right)$$
Thus the three equations of motion are..
$$\frac{d}{dt} \frac{∂T}{∂ \dot{\rho}}-\frac{∂T}{∂\rho}=Q_k=F_{\rho}$$
$$m \ddot{\rho}-m \rho \dot{\rho} \dot{\theta}^2-2 \rho \dot{\rho} \dot{\phi}^2sin^2 \theta=F{\rho}$$
and..
$$\frac{d}{dt} \frac{∂T}{∂ \dot{\theta}}-\frac{∂T}{∂\theta}=Q_k= \rho F_{\theta}$$
$$\frac{d}{dt}\left( m \rho^2 \dot{\theta}^2 \right)-m \dot{\theta} \rho^2 \dot{\phi}^2sin \theta cos \theta$$
$$2m \rho \dot{\theta} \dot{\rho}+m \rho^2 \ddot{\theta}-m \rho^2 \dot{\phi}^2sin\theta cos\theta$$
and finally..
$$\frac{d}{dt} \frac{∂T}{∂ \dot{\phi}}-\frac{∂T}{∂\phi}=Q_k=F_{\phi}$$
$$\frac{d}{dt}\left( m \rho^2 \dot{\phi}^2 sin^2 \theta \right)$$
$$2m \rho \dot{\rho} \dot{\phi}sin^2 \theta+m \rho^2 \ddot{\phi}sin^2 \theta +2m \rho^2 \dot{\phi}\dot{\theta}sin \theta cos \theta$$

Would these be correct? The rest of the problem says to substitute $\dot{\theta}$ for $\omega$ and then find the corresponding centrifugal and Coriolis forces. I do not recall how I would recognize these terms. I know these are both fictitious forces that are only a product of mass times acceleration. Their form is more obvious is polar form though.. If anyone could help me I would very much appreciate it. :]
Thanks!

 

[Liquidxlax]

use the spherical gradient when taking derivatives in spherical coordinates

http://en.wikipedia.org/wiki/Gradient

 

[Liquidxlax]

man i hate not being able to edit some posts...

i said that because it does not look like that you differentiated in spherical coordinates

 

[Xyius]

What do you mean? I started out in Spherical coordinates from the Kinetic energy didn't I? From there on out it was just the chain rule a bunch of times.