Lagrangian mechanics - rotating rod

[pj33]

Hello,

It might sound silly, but when I try to calculate the kinetic energy of a rotating rod to form the Langrangian (and in general), why it has both translational and rotational kinetic energy?

Is it because when I consider the moment of Inertia about the centre I need to include the translational since my "frame of reference" is the centre and it moves but when considering about end I only take into aacount the rotatioal since by "frame of reference" (end of rod) is stationary?

I am looking more for a physical interpretation/intuition.
I hope my explanation above is clear enough!

Thank you in advance!

 

[BvU]

You talk about a stationary end of rod, so you must have some specific configuration in mind. Can you divulge to us what it is ?

Note that $\int {1\over 2} {\dot r}^2 \; dm\$ is the kinetic energy in all cases. Often it is convenient/ useful/ practical to split it up in parts, e.g. as in your c.o.m plus rotation. A force like gravity works on all $dm$ but can comfortably be considered to work on the c.o.m. that way.

All in all nothing physical, just a mathematical approach

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[pj33]

You talk about a stationary end of rod, so you must have some specific configuration in mind. Can you divulge to us what it is ?

Note that $\int {1\over 2} {\dot r}^2 \; dm\$ is the kinetic energy in all cases. Often it is convenient/ useful/ practical to split it up in parts, e.g. as in your c.o.m plus rotation. A force like gravity works on all $dm$ but can comfortably be considered to work on the c.o.m. that way.

All in all nothing physical, just a mathematical approach

$\$

I was just thinking of a simple beam attached at one end at a stationary point.
If I understand this, it helps to tackle harder problems. Thank you!

 

[BvU]

I was just thinking of a simple beam attached at one end at a stationary point.

That is a system with a constraint, so already not all that trivial, but surely instructive.

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