Question

A uniform, thin, solid door has height 2.30 m, width 0.810 m, and mass 24.5 kg.

(a) Find its moment of inertia for rotation on its hinges.
kg

Answer 1

This is an integral, from the definition;

I = INTEGRAL[x^2dm]

Where "x" is the distance perpendicular to the vertical axis thru the hinges to an element of mass "dm". To get dm in terms of metrical quantities define a density "D" as the mass element divided by the area "dA" it occupies;

D = dm/dA

Note that for a uniform door "D" is the same everywhere and is the same as total mass/total area; D = M/A

Replace dm by DdA and in rectangular coordinates , dA = dydx;

I = DOUBLEINTEGRAL[Dx^2dxdy]

Integrate over "y" to get the height "H" ;

I = DHINTEGRAL[x^2dx]

integrate over "x" out to the width "W";

= DH(W^3/3)

Use D = M/A = M/HW

I = (1/3)MW^2

This is the general formula. Now just plug in W & M.

=5.30 kgm^2

(b)The height of the door is unnecessary.

Note: As long as the door is thin the thickness will not be important. For a very thick door the perpendicular distance to the mass element is (x^2 + z^2 ) and you would have to define a volume density D = dm/dxdydz and do a triple integral.