1. The problem statement, all variables and given/known data Calculate by direct integration the moment of inertia for a thin rod of mass M and length L about an axis located distance d from one end. 2. Relevant equations I = Ʃmir2i Which can be rewritten as int (r2)dm as Δm -> 0 Since we cannot integrate over a mass, we need to switch to dx, which is dm = M/L dx 3. The attempt at a solution The question is asking us to find the moment of inertia as it rotates on an axis d from one end. Normally, if we are trying to find the moment of inertia of an "infinitely thin" rod rotating on one end, the moment of inertia is (1/3)ML2 However, we have to manually integrate this problem because we have mass on both sides of the axis of rotation. I took two integrals- one on one side of d, and on the other. First side: (M/d) int(x2)dx (with bounds 0 to d) (M/d)(x3/3) (with bounds 0 to d) = Md2/3 The second side: (M/L-d) int(x2) dx (with bounds from d to L) (M/L-d) (x3/3) (with bounds from d to L) (M/L-d) ( (L3/3) - (d3/3)) Then I added the two moments of inertia together, to get: (Md2/3) + (M/L-d)((L3/3) - (d3/3)) This answer is not correct, and I was wondering where i went wrong conceptually/mathematically. Homework is due soon, so any immediate help is greatly appreciated.