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Qmechanic ♦
Qmechanic ♦
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How L=Iw$L=I\omega$ holds for an angular momentum of a rotating door when it's not about axis of symmetry?

$L=Iw$$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?

How L=Iw holds for an angular momentum of a rotating door when it's not about axis of symmetry?

$L=Iw$ 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?

How $L=I\omega$ holds for an angular momentum of a rotating door when it's not about axis of symmetry?

$L=I\omega$ 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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Ingun전인건
Ingun전인건
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How L=Iw holds for an angular momentum of a rotating door when it's not about axis of symmetry?

$L=Iw$ 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?