Setting up a Double Integral for Moment of Inertia

I came across the following problem while browsing the Internet.

Problem:
Find the Moment of Inertia of a circular disk of uniform density about an axis which passes through the center and makes an angle of $\dfrac{\pi}{6}$ with the plane of the disc.

I wouldn't have posted this here since this is actually a Physics question. However, the method involved is purely a usage of Maths (seemingly Double-Integrals). If there are any issues, kindly tell me. I shall do whatever is needed.

Moment of Inertia ($I$) is $\int r^2dm$ where $r$ is the perpendicular distance from the chosen axis (and can vary) and $dm$ is an elemental mass.  I interpreted the disk as a 2 Dimensional structure, hence it has mass per unit area ($\sigma$)

$$dm=\sigma dA$$

Thus, $$I=\int r^2dm=\int \int_A \sigma dA$$  Using a change of variables to Cylindrical Co-ordinates,

$$dA=rdrd\theta$$  $$\Longrightarrow I=\int\int \sigma r^3 dr d\theta$$  However, this is the Integral for finding the Moment of Inertia for an axis perpendicular to the plane of the body. I cannot understand how to modify it for the given question.  I would be very grateful if somebody could please solve this question without resorting to differential equations. Many thanks in anticipation!

PS. The answer is $\dfrac{5}{16}MR^2$.

• Do you happen to know what the correct answer is? I get the value $\frac{5}{16} mr^2$ which "sounds" right...:) – David Quinn Sep 11 '15 at 19:11

I am doing this in two stages. The disc needs to be divided into elements in the form of infinitesimally thin and concentric hoops, which we then integrate. But first we have to find the MI of a hoop about an axis at angle $\frac {\pi}{6}$ to the plane of the hoop, through the centre of the hoop.

Consider a circle of radius $r$ in the $xy$ plane, centre $O$, and a point on the circumference with coordinates $P(r\cos\theta,r\sin\theta, 0)$.

We will write the axis, as specified in the question, as the line $$\underline{r}=t\left(\begin{matrix}\frac{\sqrt{3}}{2}\\0\\\frac 12\end{matrix}\right)$$

The distance $d$ from $P$ to the line is given by the magnitude of the cross product $$\overrightarrow{OP}\times\underline{\hat{r}}$$

A quick calculation gives $$d^2=\frac 14r^2(1+3\sin^2\theta)$$

Now let the circle be a hoop of mass $m$ and density per unit length $\rho$

The MI of an element of mass $\delta m$ at $P$ is $\delta I=\delta m d^2=\rho r \delta\theta d^2$

Therefore for the whole hoop, $$I=\frac 14\rho r^3\int_0^{2\pi}(1+3\sin^2\theta)d\theta=\frac 58mr^2$$

Now finally we can consider the disc of radius$R$, mass $m$ and density per unit are $\rho$ divided into concentric elements of thickness $\delta x$ and radius $x$

Then $$\delta I=\frac 58\delta mx^2=\frac 54\pi\rho x^3\delta x$$ $$\Rightarrow I=\frac 54\pi\rho\int_0^Rx^3dx$$ $$\Rightarrow I=\frac{5}{16}mR^2$$

• This is the equation of the axis in three dimensions – David Quinn Sep 11 '15 at 20:34
• Referring to your diagram, I am taking the x axis as the horizontal line and the z axis as perpendicular to the plane of the disc – David Quinn Sep 11 '15 at 20:40
• It's a vector giving the direction of the axis – David Quinn Sep 11 '15 at 21:03