Hi all,

Have been having some problems with the following math problem I was given last week:

 

A solid cube of mass M and side a has one corner at the origin and three sides along the positive x,y and z axes. The square of the distance of the mass element dM = pdV rom the z-axis is x^2 + y^2 so that the moment of inertia of the cube around the z-axis is given by

 

I = integral (x^2 + y^2) pdV

 

taken throughout the volume V of the cube.

 

If the cube has constant density, prove that I= (2/3) Ma^2

 

Any help/tips would be most appreciated!

The square of the distance of the mass element dM = pdV rom the z-axis is x^2 + y^2

the distance of the mass? that makes no sense. And what do you mean by pdV?

Badly worded question, isn't it?

 

pdV is p (ro - density) times the volume element dV.

 

The square of the distance of the mass element from the z-axis - i assume means the distance from the mass element to the z-axis, all squared...

Is x = y = a since this is a cube centered at the origin?

 

a^3 = V so 3a^2 da = dV. So the integral might be integral p(2a^2)(3a^2)da

Which becomes 6/5 pa^5 +C

M = pV = pa^3

Doing the calculations, I get 6/5 Ma^2 + C.

Integral of [math]x^2 + y^2[/math] is [math]\frac{1}{3}x^3 + \frac{1}{3}y^2 + C[/math]. Thought this might help.

Omit.