Question

A uniform square lamina of side $2 \,m$ 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

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 $COG$,
$I _{ g }= m \left[\frac{( a )^{2}+( a )^{2}}{12}\right]=\frac{1}{6} ma ^{2}$
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} $
$=1.41+0.47$
$=1.88 \,m$