Calculating the Rotational Inertia — Question 2

Question 2

A density gradient changes both the center and the inertia

A thin rod spans $0\le x\le L$ with density $\lambda(x)=cx$, where $c>0$. Its total mass is $M$.

1. Find $c$, the center of mass, and the inertia about a perpendicular axis through $x=0$.

2. Use the parallel-axis relation to find the inertia through its center of mass.

Question 2 - Solution

A density gradient changes both the center and the inertia

(a) Derivation and calculation

Normalize density and evaluate the first moment: $$M=\int_0^Lcx\,dx=\frac{cL^2}{2}\Rightarrow\boxed{c=\frac{2M}{L^2}}.$$ $$x_c=\frac1M\int_0^Lx(cx)\,dx=\frac{cL^3}{3M}=\boxed{\frac{2L}{3}}.$$ $$I_0=\int_0^Lx^2(cx)\,dx=\frac{cL^4}{4}=\boxed{\frac{ML^2}{2}}.$$

(b) Result and interpretation

Translate between parallel axes using the actual mass center: $$I_0=I_c+Mx_c^2\Rightarrow \boxed{I_c=\frac{ML^2}{2}-M\left(\frac{2L}{3}\right)^2=\frac{ML^2}{18}}.$$ The geometric midpoint is not the mass center and cannot replace $x_c$ in this subtraction.

Physical check

The mass center lies toward the denser end, beyond $L/2$, and the central inertia is positive.