Calculating the Rotational Inertia — Question 10

Question 10

An in-plane axis samples perpendicular distances within a plate

A uniform rectangular plate of mass $M$ is centered at the origin, spans length $a$ along $x$ and $b$ along $y$, and lies in the $xy$ plane. An axis in that plane passes through the center at angle $\phi$ to $+x$.

1. Derive its inertia using perpendicular distance $d=-x\sin\phi+y\cos\phi$.

2. For $a=2b$, find the largest and smallest inertias among these axes.

Question 10 - Solution

An in-plane axis samples perpendicular distances within a plate

(a) Derivation and calculation

Square the given perpendicular distance before integrating: $$I=\int(-x\sin\phi+y\cos\phi)^2\,dm.$$ Symmetry gives $\int xy\,dm=0$. For uniform density, $$\int x^2\,dm=\frac{Ma^2}{12},\qquad\int y^2\,dm=\frac{Mb^2}{12}.$$ $$\boxed{I(\phi)=\frac M{12}(a^2\sin^2\phi+b^2\cos^2\phi)}.$$

(b) Result and interpretation

With $a=2b$, $$I(\phi)=\frac{Mb^2}{12}(1+3\sin^2\phi).$$ $$\boxed{I_{\min}=\frac{Mb^2}{12}\ \text{along }x},\qquad \boxed{I_{\max}=\frac{Mb^2}{3}\ \text{along }y}.$$ The longer dimension is perpendicular to the maximum-inertia axis.

Physical check

For a square, $a=b$ makes the expression independent of the in-plane axis angle.