I Lagrangian mechanics - generalised coordinates question

[curiousPep]

I think I undeerstand Lagrangian mechanics but I have a question that will help to clarify some concepts.
Imagine I throw a pencil. For that I have 5 generalised coordinates (x,y,z and 2 rotational).
When I express Kinetic Energy (T) as:
$$T = 1/2m\dot{x^{2}}+1/2m\dot{y^{2}}+1/2m\dot{z^{2}} + I\dot{\theta^{2}} + I\dot{\phi^{2}}$$
and potential energy (V)
$$V=mgz$$
Then I use Lagrangian to find the EOM.
For x,y,z is fine but for $$\theta$$ and $$\phi$$ I have a question. I see how x,y,z can be a expressed as functions of $$\theta\;and\;\phi$$, but why should I do this. I mean in cases that something is less obvious, then I will get the wrong EOM.
Thank you, and I hope the latex code works.

 

[PeroK]

You need double dollars or double hashes to delimit Latex here.

Im not sure I understand your question. A point particle has linear KE. A extended rigid body has linear KE of the CoM plus rotational KE of the body about the CoM. You have to know to consider both.

 

[curiousPep]

You need double dollars or double hashes to delimit Latex here.

Im not sure I understand your question. A point particle has linear KE. A extended rigid body has linear KE of the CoM plus rotational KE of the body about the CoM. You have to know to consider both.

I will try to explain it better, cause I see it's a bit confusing.
When I have a rigid body like a pencil of length 2L, the generalized coordinates defined are x,y,z (COM relative to (0,0)) and
$$\theta, \phi$$ (Euler's angles).
However, x,y,z can be expressed as functions of $$\theta,\phi$$.
For example: $$x = X + Lsin\theta cos\phi$$. My question is that, why do I need to do this in oder to find the right EOM for $$\theta\;and\;\phi$$?
Or is this not needed?I mean in a more complex case the relationship mu not be that obvious, so I won't know if my EOM are right or not.

 

[PeroK]

If $(X, Y, Z)$ are the coordinates of the CoM, then that's what you need for $T$. The rotational motion involves the moment of inertia $I$, which encapsulates the position of every point mass in the body in terms of calculating rotational KE.

Your $x$ above seems to be just the position of one end of the pencil!

 

[vanhees71]

The right degrees of freedom are the center-of-mass position components $\vec{X}$ and the three Euler angles for the rotation of the rigid body around this center of mass. So what you look for is the Lagrangian for the (symmetric) spinning top in terms of Euler angles. See, e.g.,

https://hepweb.ucsd.edu/ph110b/110b_notes/node36.html

To simplify the equations somewhat, you may neglect the rotation around the axis of the pencil, because the corresponding moment of inertia is small compared to that around any axis perpendicular to it, i.e., you may set $I^{(3)}=0$.