Question

A uniform square lamina of side 2a is hung up by one corner and oscillates in its own plane which is vertical. Find the length of the equivalent simple pendulum.

Answer 1

When the lamina ABCD is at rest, its centre of gravity G must lie vertically below the corner A by which it is hung.

By geometry, the distance between centre of gravity and centre of suspension i.e.,

AG = √2 a = L

M.I. of lamina about horizontal axis passing through C.G. and perpendicular to its plane is given by

Ig = m (2a)2 + (2a)2/12 = 2/3 ma2

Radius of gyration about this axis is given by

K2 = 2/3 a2

Then length of the equivalent simple pendulum