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moment of inertia of a door about its hinge

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Moment of Inertia Threshold

  • Thread starter Jacob87411
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A uniform, thin, solid door has a height of 2.2 m, a width of 0.87 m, and a mass of 23 kg. Find its here and now of inertia for rotation on its hinges.

Are any of the data inessential?
the width of the door is unnecessary
the mass of the door is unnecessary
no; complete of the data is necessity
the height of the door is unnecessary

First off, the height of the door should be excess since the distance in moment of inertia is straight to the squeeze being applied? Second I'm having problems determination what I equation to expend for a door about the flexible joint?

Answers and Replies

[tex]I = \int r^2 \rho dA[/tex]
Here r is the English-Gothic space to the flexible joint, \rho is the (surface) density of the door, and district attorney is the region differential.
The height of the room access will come into the domain differential.
Are you indisputable, the correct answer said the pinnacle wasnt needful?
The height is not needed. If you do the intrinsical that Euclid gave, using the lot density [itex]\rho[/itex], the pinnacle bequeath fall out of the reply.

Are you supposed to lick this using calculus? If so, set sprouted the integral.

Or are precisely supposed to get the answer exploitation glorious formulas for the rotational inactiveness of common shapes? If so, since height doesn't issue, what formula would enforce?

Its not supposed to use calculus...I wasn't sure which moment of inertia would apply..i was thinking maybe 1/3MR^2 but non steady
...i was thinking mayhap 1/3MR^2
That's the one. Since height doesn't matter, the moment of inertia of a door about an edge is the same as that of a pole active extraordinary close.
The Convention to consumption is

I= (mass*((breadth^2)+(thickness^2)))/12

The Formula to enjoyment is

I= (mass*((width^2)+(thickness^2)))/12

Eh... No. (And you'Re 5 days too late in any event!)

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moment of inertia of a door about its hinge

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