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I have just recently studied Euler's Equations for Rigid body dynamics. Following through the proofs and equations, there is one thing that I can't seem to make a sense of.

in

$$ \dot{\textbf{L}} + \boldsymbol{\omega} \times {\textbf{L}} = \mathbf{\Gamma} $$

Where angular momentum $\textbf{L}$ is calculated in a rotating reference frame with coordinate axes as principal axes. $$ \textbf{L} = (\lambda_1\omega_1 ,\lambda_2\omega_2, \lambda_3\omega_3) $$

My question is if we measure the angular velocity in the rotating frame or the 'body frame' (i.e., attached to the body). How can the angular velocity of the body in the 'body frame' be anything but zero? And if $\boldsymbol{\omega}$ is zero, so should be the Angular momentum in the rotating frame.

Qmechanic
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  • Please see my answer here. ( I couldn't make $\omega$ bold either ). – Kurt G. Jul 06 '21 at 17:28
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    \boldsymbol is the way to go – Quantum Mechanic Jul 06 '21 at 18:12
  • @KurtG. I might take some time to completely understand what you are saying there, but really saying that $\dot{\textbf{Q}}$ is not equal to $\textbf{V}$ in the body frame just doesn't sound very convincing. Isnt Velocity 'defined' this way? – Gaurang Agrawal Jul 07 '21 at 05:58
  • @KurtG. Given that this is my first course in Mechanics, I might get troubled understanding the likes of Arnold's. However, I found one more thread on the same question couple of hours ago and It does sound convincing. It may very well be the case that you and he are both saying the same thing, can you please confirm it. – Gaurang Agrawal Jul 07 '21 at 06:08
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    @gaurang: looking briefly at the other thread you found I am sure Arnold is saying exactly the same thing albeit his book combines minimalism with elegance. One sentence in the other thread I can confirm enthusiastically: "I remember being so confused by this when I first took analytic mechanics." Advice: try as hard as you can looking for the jewels in Arnold. When you are tired relax reading this. – Kurt G. Jul 07 '21 at 06:25

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