Question

A square hula-hoop of mass "m" and side "L" is placed on a hook. It is given a slight shove allowing it to oscillate as a physical pendulum about the hook. Given [m, L] Determine the period of oscillation.

Answer 1

The Time period for the physical pendulum is $T=\frac{2\pi }{\omega }= 2\pi \sqrt{\frac{I}{mgL}}$

where $\omega =\sqrt{\frac{mgL}{I}}$

Where I is moment of inertia, 'g' accelleration due to gravity.

The moment of inertia of square of length "L" is $I=\frac{ML^{2}}{6}$ about an axis passing through the center

but the in this example the axis is at the corner, So, according to the parallel axis theorem $I = \frac{MR^{2}}{6}+MR^{2} = \frac{7}{6}MR^{2}$

by substituting this value in the above formula for time period, we get

$T= 2\pi \sqrt{\frac{7L}{6g}}$ .