Angular Momentum and Principal Axes of Inertia

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The discussion focuses on deriving the formula for angular momentum \(\vec L\) in relation to principal axes of inertia for a body rotating about an arbitrary axis. The formula presented is \(\vec L = \vec u_{x} I_1 \omega_{x0} + \vec u_{y} I_2 \omega_{y0} + \vec u_{z} I_3 \omega_{z0}\), which is derived from the linear relationship between angular momentum and angular velocity, with the inertia tensor being diagonal along the principal axes. It is emphasized that while the relationship holds for principal axes, the angular momentum does not necessarily align with the rotation axis when the body rotates about an arbitrary axis. A participant expresses frustration about not understanding tensors, indicating a gap in knowledge that hinders their ability to derive the formula. The conversation highlights the complexity of the topic for those new to mechanics.
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Hi

I´m self-studying Alonso and Finn´s Mechanics and I have a question about this subject.

Let a body rotate about an arbitrary axis P having angular momentum \vec L.
Consider a referential with three perpendicular axes, X_{0} , Y_{0} , Z_{0} , which are also principal axes of inertia.
The book says we can write \vec L as

\vec L = \vec u_{x} I_1 \omega_{x0} + \vec u_{y} I_2 \omega_{y0} + \vec u_{z} I_3 \omega_{z0}

Does anybody how to derive this formula? The book usually explains things, but perhaps this is supost to be obvious.

By the way, I already know how to derive \vec L = I \vec \omega for a body rotating about a principal axis of inertia but I don´t know how to derive this one.

Thank you​
 
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Generally, if a rigid body is rotating about an arbitrary axis, the angular momentum need not point in the same direction as the rotation axis, as it does when \vec L = I \vec \omega (for rotation about a principal axis).

The relation between \vec L and \omega is still linear, and I is generally a tensor quantity (the inertia tensor).
An object always has three principal axes and in that coordinate system the inertia tensor is diagonal. This leads directly to:
<br /> \vec L = \vec u_{x} I_1 \omega_{x0} + \vec u_{y} I_2 \omega_{y0} + \vec u_{z} I_3 \omega_{z0} <br />
It's really the only thing it can be if you know \vec L = I \vec \omega holds for principal axes, there are three principal axes and the correspondence between \vec w and \L is linear.
 
Hello Galileo,

Thanks for the answer. Unfortunately, I couldn´t follow it because I don´t know what a tensor is. I´m still a high school student. I guess I´ll just have to use it without knowing how to derive it. which is something I really hate.
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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