Moment of inertia of a rod at an angle to the axis

[mcheung4]

suppose a uniform rod of mass M and length L is at an angle B to the x axis, one end of the the rod touching the axis. wish to find moment of inertia about x axis.

let the rod touches the axis at x=0. let D=density=M/L, and I will integrate along x axis, that means that at a distance x from the the origin, the little mass dM = D*dx is at a distance x*tanB away from the axis. then the integral I get is I = D*(x^2)*[(tanB)^2] integrated from x=0 to x=L*cosB. the answer I got is (1/3)*m*(L^2)*cosB*(sinB)^2.

I know this is different from the usual solution where one should integrate along the rod itself,but I don't understand which part of this argument went wrong?

Thanks!

 

[Nugatory]

If you integrate dM across the same range that you're using to calculate the moment of inertia, what value do you expect?

 

[Pippip19]

I know this reply is quite late but I saw the problem and wanted to answer in case anyone else was confused about this.

Integrating along the x-axis instead of the rod is, of course, absolutely fine, however in this case your infinitesimal mass element dm should be dm = D*(dx/cosB), as each infinitesimal distance element dL along the length of the rod has a projection dx/cosB onto the x-axis.

With this in mind, the unwanted cosB factor of your result disappears, leaving the correct result of (1/3)*m*(L*sinB)^2