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Angular Momentum and Principal Axes of Inertia

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Mar29-10, 04:57 PM
P: 45

Iīm self-studying Alonso and Finnīs Mechanics and I have a question about this subject.

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

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

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 [tex] \vec L = I \vec \omega [/tex] for a body rotating about a principal axis of inertia but I donīt know how to derive this one.

Thank you
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Mar29-10, 06:00 PM
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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 [itex]\vec L = I \vec \omega [/itex] (for rotation about a principal axis).

The relation between [itex]\vec L[/itex] and [itex]\omega[/itex] 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:
\vec L = \vec u_{x} I_1 \omega_{x0} + \vec u_{y} I_2 \omega_{y0} + \vec u_{z} I_3 \omega_{z0}
It's really the only thing it can be if you know [itex]\vec L = I \vec \omega [/itex] holds for principal axes, there are three principal axes and the correspondence between [itex]\vec w[/itex] and [itex]\L[/itex] is linear.
Mar29-10, 06:51 PM
P: 45
Hello Galileo,

Thanks for the answer. Unfortunately, I couldnīt follow it because I donīt know what a tensor is. Iīm still a high school student. I guess Iīll just have to use it without knowing how to derive it. which is something I really hate.

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