Moment of Inertia of a rod at an angle

[kudoushinichi88]

A rod has mass $M$ and length $L$. Calculate the moment of inertia of the rod about an axis which is passing through its center of mass and forming an angle $\theta$ to the rod.

I drew a diagram on an xy-plane where the rod is on the x-axis and the center of the rod is at the origin. Chopping the rod up into small portions of $dm$, they have a distance of $x \sin \theta$ from the axis of rotation. Therefore,

$$I=\int r^2 dm$$

Assuming the rod is uniform, $\frac{dm}{M}=\frac{dx}{L}\Rightarrow dm=\frac{M}{L}dx$

Therefore,

$$I=\int x^2 \sin^2\theta dm$$
$$=\frac{M}{L}\sin^2\theta\int_{\frac{-L}{2}}^{\frac{L}{2}}x^2 dx$$
$$=\frac{1}{12}ML^2\sin^2\theta$$

Is this correct? I just need someone to check my work because I have no solutions to refer to for this question...

 

[Doc Al]

Looks good to me.

 

[kudoushinichi88]

Oh great, thanks!