$L=I\omega$ only holds when a rigid body is rotating about axis of symmetry. But I saw from multiple practice problems that calculates the angular momentum of a door rotating about the hinge using $L=Iw$. But the hinge is not the axis of symmetry of the door. How does $L=Iw$ hold in this case?
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$\begingroup$ You’re starting with an erroneous premise $\endgroup$– Bob DCommented Sep 21 at 8:05
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$\begingroup$ @BobD which premise is it? $\endgroup$– Ingun전인건Commented Sep 21 at 9:02
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$\begingroup$ Aha I see what you mean. I just found it doesn't have to be the axis of symmetry. THank you. $\endgroup$– Ingun전인건Commented Sep 21 at 9:13
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An object doesn't have just one moment of inertia. It has a moment of inertia for every possible axis of rotation, no matter if that's an axis of symmetry or not. It's just much easier to calculate if the axis of rotation and the axis of symmetry are equal.
If you have the moment of inertia $I_g$ for rotation along a given axis $g$ through the center of mass (for instance, an axis of symmetry), and if $h$ is an axis parallel to $g$ at distance $d$, then the moment of inertia for rotation along $h$ is $I_h=I_g+md^2$, where $m$ is the object's total mass. This is the parallel axis theorem.
In the door example, if we consider the moment of inertia along a vertical axis through the middle of the door, then we can shift the axis by half the door's width to obtain the moment of inertia for rotation along the hinges.
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$\begingroup$ I just found axis doesn't have to be of symmetry for $L=Iw$ to hold. $\endgroup$– Ingun전인건Commented Sep 21 at 9:14