Calculating the Rotational Inertia — Question 4

Question 4

Radially increasing density weights the rim more strongly

A flat circular plate of radius $R$ has areal density $\sigma(r)=kr$ with $k>0$. Its mass is $M$.

1. Find the inertia about its central perpendicular axis.

2. Compare with a uniform disk of the same mass and radius.

Question 4 - Solution

Radially increasing density weights the rim more strongly

(a) Derivation and calculation

Normalize the areal density with the ring area element: $$M=\int_0^Rkr(2\pi r\,dr)=\frac{2\pi kR^3}{3} \Rightarrow k=\frac{3M}{2\pi R^3}.$$ $$I=\int_0^Rr^2(2\pi kr^2\,dr)=\frac{2\pi kR^5}{5} =\boxed{\frac35MR^2}.$$

(b) Result and interpretation

Compare at the same total mass and outer radius: $$\boxed{\frac{I}{I_{\mathrm{uniform}}}=\frac{3MR^2/5}{MR^2/2}=\frac65=1.20}.$$ The outward density gradient increases inertia by 20 percent because a larger share of the mass occupies large radii.

Physical check

The result lies between the uniform disk value $MR^2/2$ and the hoop bound $MR^2$.