Calculating the Rotational Inertia — Question 6

Question 6

A frame and a filled plate distribute equal mass differently

A uniform square wire frame has total mass $M$ and side length $L$. The wire thickness is negligible. Its axis passes through the square center perpendicular to its plane.

1. Find its inertia by treating the frame as four rods.

2. Compare with a uniform filled square plate of the same $M,L$, whose central inertia is $ML^2/6$.

Question 6 - Solution

A frame and a filled plate distribute equal mass differently

(a) Derivation and calculation

Each side is a rod of mass $m_s=M/4$ whose midpoint is $L/2$ from the axis: $$I_s=\frac{m_sL^2}{12}+m_s\left(\frac L2\right)^2 =\frac{ML^2}{48}+\frac{ML^2}{16}=\frac{ML^2}{12}.$$ $$\boxed{I_{\mathrm{frame}}=4I_s=\frac{ML^2}{3}}.$$

(b) Result and interpretation

The comparison and midpoint approximation are $$\boxed{\frac{I_{\mathrm{frame}}}{I_{\mathrm{plate}}}=\frac{ML^2/3}{ML^2/6}=2}.$$ $$I_{\mathrm{midpoint}}=4\frac M4\left(\frac L2\right)^2=\frac{ML^2}{4},\qquad I_{\mathrm{frame}}-I_{\mathrm{midpoint}}=\frac{ML^2}{12}.$$ The omitted term is the sum of the sides’ own central inertias.

Physical check

The frame places more mass near the perimeter than the filled plate.