Moment of Inertia - Parallel Axis Theorem - Thin Rod

Physics Ninja looks at how to calculate the moment of inertia of a thin rod of mass M and length L about an axis through the center of mass and also an axis through the end of the bar. The parallel axis theorem is also review and applied to this problem.

[@kawanganyirenda452]

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[@longgnguyen9645]

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[@gad3386]

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[@guardbrawl9777]

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[@bariscelikW]

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[@mileslegend]

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[@gabrielarodezno8221]

I have never understood it and this short video just made it so clear, thank you

[@hadisardari1564]

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[@mileslegend]

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[@georgibaykalov8373]

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[@Papatya_2006]

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[@Martin07031]

What if I have a door on hinges with axis of rotation just like in your case number 2, and this door doesn’t have center of gravity in the exact middle? Should I use parallel axis theorem? So equation will be 1/3.m.L^2+m.d^2 ? Thank you.

[@PhysicsNinja]

If the center of gravity is not in the middle it means the density is not uniform. you should integrate assuming you know the density vs position.

[@Martin07031]

@@PhysicsNinja I mean, the door has the same density. It just doesnt have perfect rectangular shape. There are other plates and construction things on it. Do I have to calculate each construction separately and then sum all the moments of inertia of door itself and constructions?