I was assigned a question asking: Find the moment of inertia of a circular cylinder with base $a$ and height $h$ about the diameter of the base. We are working through applications of multiple integrals currently so I am trying to solve this using cylindrical coordinates.
In my book it gives us the idea that Moment of Inertia $= \int \int \int D^2\cdot\text{density}\ dV$.
With $D$ being the perpendicular distance from the volume element $dV$ to the axis of rotation L.
I am confused at how to find what $D$ is in this question.
Can anyone help?
Best Answer
You are rotating the cylinder about an axis perpendicular to the axis of symmetry. In this case, polars may not be your best bet.
Align the $x$ axis with the axis of rotation. For each point on the axis of rotation, which extands between $x \in [-a,a]$, there is a corresponding rectangular cross-section of the cylinder. Each point of the cross section is at a distance $D$ from that point of the axis of rotation. That rectangle has dimensions $2 \sqrt{a^2-x^2} \times h$. A point within that rectangle has distance from that point on the axis $D$ equal to $\sqrt{y^2+z^2}$. The get the element of inertia from this rectangle, integrate $y^2+z^2$ over the rectangular area. Then integrate the collection of elements of inertia from rectangles like this over $x$.
Setting this up: the element of inertia $dI(x)$ is an integral over that rectangular area:
$$dI(x) = \rho dx \int_0^h dz \: \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} dy \: (y^2+z^2) = \rho dx \frac{2 h}{3} \sqrt{a^2-x^2} (a^2+h^2-x^2)$$
Once you determine this, integrate over $x$ to get the total moment of inertia:
$$I = \rho \frac{2 h}{3} \int_{-a}^a dx \: \sqrt{a^2-x^2} (a^2+h^2-x^2) = \frac{1}{12} M (3 a^2+4 h^2)$$
I leave the details of evaluating the integrals to the reader.