What's wrong with this derivation of the Moment of Inertia of a disc?

Question

I'm currently in AP Physics, and we're doing our moment of inertia unit. I'm also currently in multivariable calculus, so I've been trying to apply some of those concepts to physics. I've run into a problem with calculating the moment of inertia of a solid disk. Here's how multivar does it:

$$ I=\int\int_R\text{(distance to axis)}^2\delta dA $$

$$ I=\int_0^{2\pi}\int_0^R r^2rdrd\theta = \int_0^{2\pi}\int_0^R r^3drd\theta $$

$$ I = \frac{\pi R^4}{2} $$

And here's the problem—everything in physics (including lists of moments of inertia) says that the moment of inertia of a disk is $$ I = \frac{1}{2}MR^2 $$

I take it that this is because I've missed a division by $\pi R^2$, but where?

Thanks!

Answer 1

The mass of the disk $D$ is obtained integrating $1$ on it, so it is $$ M = \iint_{D}dx\; dy=\pi R^2\ . $$

Answer 2

$I = \int\int_{Area}(\sigma).(r)^2.dA$

$\sigma$: density of the disk.(constant)

$I =\sigma\times (\int_0^{2.\pi}\int_0^R r^2.r.dr.d \theta)=\sigma\times(\int_0^{2.\pi}d\theta.\int_0^R r^3.dr)=\sigma\times2.\pi\times\frac{1}{4}.R^4=\frac{1}{2}.\pi.\frac{M}{\pi.R^2}.R^4$$=\frac{1}{2}.M.R^2$

Answer 3

$I=\int_M{r^2 dm}$. You’re missing the density and the thickness of the disc. I assume $\delta$ is the thickness which you have in the initial integral but you forget about, while density you never took into account.

$\frac {\pi R^4}{2} \cdot \rho \delta =\pi R^2 \cdot \delta \cdot \rho \cdot \frac{R^2}{2}=\frac{M\cdot R^2}{2}$