Calculating the Rotational Inertia — Question 1

Question 1

Moving the pivot along a rod produces a quadratic inertia

A uniform thin rod has mass $M$ and length $L$. A perpendicular axis crosses it at distance $a$ from its left end, with $0\le a\le L$.

1. Derive its inertia directly by integration.

2. Find the minimum inertia and compare an end pivot with a central pivot.

Question 1 - Solution

Moving the pivot along a rod produces a quadratic inertia

(a) Derivation and calculation

Choose $x$ from the left end, so $dm=(M/L)\,dx$ and the perpendicular distance is $|x-a|$: $$I(a)=\frac ML\int_0^L(x-a)^2\,dx =\frac{M}{3L}\left[(L-a)^3+a^3\right].$$ $$\boxed{I(a)=M\left(a^2-La+\frac{L^2}{3}\right) =\frac{ML^2}{12}+M\left(a-\frac L2\right)^2}.$$

(b) Result and interpretation

Minimize the nonnegative square, or differentiate: $$\frac{dI}{da}=2M(a-L/2)=0\Rightarrow a=L/2,\qquad \boxed{I_{\min}=\frac{ML^2}{12}}.$$ $$I(0)=I(L)=\frac{ML^2}{3},\qquad\boxed{\frac{I_{\mathrm{end}}}{I_{\mathrm{center}}}=4}.$$

Physical check

The expression is symmetric under $a\mapsto L-a$, as a uniform rod requires.