mathtrek Posted December 27, 2006 Posted December 27, 2006 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 tree Posted December 27, 2006 Posted December 27, 2006 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?
mathtrek Posted December 27, 2006 Author Posted December 27, 2006 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...
psi20 Posted December 29, 2006 Posted December 29, 2006 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.
EvoN1020v Posted December 29, 2006 Posted December 29, 2006 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.
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