Calculating the Rotational Inertia — Question 3

Question 3

An annulus approaches a hoop in the thin-ring limit

A uniform flat annulus has mass $M$, inner radius $a$ and outer radius $b$. Its axis passes through the center perpendicular to its plane.

1. Derive its inertia by summing circular rings.

2. Check the solid-disk and thin-hoop limits without confusing surface density with total mass.

Question 3 - Solution

An annulus approaches a hoop in the thin-ring limit

(a) Derivation and calculation

A ring of radius $r$ and width $dr$ has mass $dm=2\pi\sigma r\,dr$, where $$\sigma=\frac{M}{\pi(b^2-a^2)}.$$ $$I=\int_a^br^2\,dm=2\pi\sigma\int_a^br^3\,dr =\frac{\pi\sigma}{2}(b^4-a^4) =\boxed{\frac M2(a^2+b^2)}.$$

(b) Result and interpretation

The limiting cases at fixed total mass are $$a=0\Rightarrow\boxed{I=\frac12Mb^2},\qquad a,b\to R\Rightarrow\boxed{I\to MR^2}.$$ In the thin-annulus limit, $\sigma=M/[\pi(b^2-a^2)]$ increases as the area shrinks. Holding surface density fixed would instead change the mass.

Physical check

The mean squared radius lies between $a^2$ and $b^2$, so $Ma^2\le I\le Mb^2$.