Moment of inertia of a disc

Question

In my mechanics textbook there is a derivation of the moment of inertia of a disc of mass $m$ and radius $r$ about an axis through its centre and perpendicular to its plane surface, which goes something like this:

The mass per unit area is $\dfrac{m}{\pi r^2}$. Dividing the disc into concentric rings, the ring has inner radius $x$ and outer radius $x+\delta x$ and so its area is $\pi(x+\delta x)^2-\pi x^2\approx 2\pi x\delta x$. So the moment of inertia of this ring is $\dfrac{m}{\pi r^2}(2\pi x\delta x)(x^2)=\dfrac{2mx^3}{r^2}\delta x$ so the moment of inertia of the whole disc is $\displaystyle \lim_{\delta x\to0}\sum_{i=0}^n\dfrac{2mx_i^3}{r^2}\delta x=\int_0^r\dfrac{2mx^3}{r^2}\,\mathrm{d}x=\frac{1}{2}mr^2$.

But the step $\pi(x+\delta x)^2-\pi x^2\approx 2\pi x\delta x$ is hand-wavy. How do I know this is a valid approximation?

Answer 1

$\delta x$ is supposed to be small. Expanding, we get $\pi (x+\delta x)^2 - \pi x^2 = \pi(x^2+2x\delta x + (\delta x)^2) - \pi x^2 = 2x\delta x \pi + \pi (\delta x)^2$

For $\delta x$ small, the $(\delta x)^2$ term is negligible in comparison to the linear term.

Answer 2

If $0<\delta x<1$, then multiplying both sides of $\delta x<1$ by $\delta x$ yields $(\delta x)^2<\delta x$.

Answer 3

The real thing is the constant density $\rho = \frac{m}{R^2\pi}$. What you have to do is to compute the integral $$ \int_0^R r^2 dm = \int_0^R r^2 \rho dA = \int_0^R r^2 \rho 2\pi r dr = 2\pi \rho [\frac{1}{4}r^4]_0^R= 2\pi \rho \frac{1}{4}R^4 = \frac{1}{2} R^2 m $$ the hand-wavy part is just the standard way to avoid integration by handy-wavy-this-is-a-riemann-sum-approximation of the integral

Answer 4

A previous answer was close but had their integral wrong.

Remember density is just mass / volume. So for a disk of constant density $\rho=\frac{m}{\pi r^2}$.

Using this density, you can consider some unit of mass $dm$ will have the same mass/density ratio, so $\rho = \frac{dm}{\pi r dr}$ or $dm = \rho\pi rdr$

expand $\rho$, so $dm = \frac{m}{\pi r^2}\pi r dr$ reduces to $dm = \frac{m}{r}dr$

Plug that into the integral

$$\int_0^R r^2dm = \int_0^Rr^2\frac{m}{r}dr = \int_0^Rmrdr = \frac{1}{2}mr^2|_0^R = \frac{1}{2}mR^2$$