Question

A uniform square lamina of side $2m$ is hung up by one corner and oscillates in its own plane which is vertical. What is the length (in meters) of the equivalent simple-pendulum? (Take $\sqrt{2}=1.41$ )

Answer 1

Answer: 1.88

When the lamina $ABCD$ is at rest, its centre of gravity $G$ must lie vertically below $A$ by which it is hung.
$AG=\frac{a}{\sqrt{2}}=L$
$MOI$ along axis passing through $COG$,
$I_g=m\left[\frac{(a{)}^2+(a{)}^2}{12}\right]=\frac{1}{6}m{a}^2$
For Radius of gyration,
$K^2=\frac{1}{6}{a}^2$
Length of the equivalent simple pendulum,
$=L+\frac{K^2}{L}$
$=\frac{a}{\sqrt{2}}+\frac{1}{6}\frac{a^2}{a}\sqrt{2}$
$=a\left[\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{6}\right]=2\left[\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{6}\right]$
$=\sqrt{2}+\frac{\sqrt{2}}{3}$
$=1.41+0.47$
$=1.88m$