Moment of inertia of a door?

  • #1
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Homework Statement


A uniform, thin, solid door has height 2.2 m, width .87m and mass 23kg. Find the moment of inertia for rotation on its hinges.

2. Homework Equations

I=∫x^2dm

The Attempt at a Solution


[/B]The way I set it up was that I made into strips of length H (height) and width dr. dm=λHdr, where λ is mass over the length, H. But when I do the integral ∫r^2λHdr from 0 to r, I end up with ⅓mr^3, but it should be ⅓mr^2. Is lambda supposed to m/rH? If so, can someone explain why please?
 

Answers and Replies

  • #2
Orodruin
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You can check this, in order for dm to be a mass, ##\lambda## must have dimensions of mass/length^2. It is the mass density per unit area of door.
 
  • #3
101
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You can check this, in order for dm to be a mass, ##\lambda## must have dimensions of mass/length^2. It is the mass density per unit area of door.
Oh, well that makes sense, thank you. Sometimes I feel like it's the simplest things that I get stuck on.
 

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