Moment of inertia of a door?

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1. Nov 28, 2014

timnswede

1. The problem statement, all variables and given/known data
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. Relevant equations

I=∫x^2dm

3. The attempt at a solution
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?

2. Nov 28, 2014

Orodruin

Staff Emeritus
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. Nov 28, 2014

timnswede

Oh, well that makes sense, thank you. Sometimes I feel like it's the simplest things that I get stuck on.