Thin Rod

To derive Jalees Mughal's formula, we begin with the parallel axis theorem $I = {I}_{cm} + \frac{1}{2}m{r}^{2}$ where ${I}_{cm} = \frac{1}{3}m{r}^{2}$ . We let $r = \frac{L}{2}$ because the rod is uniform (thank goodness), where the rod experiences the most gravitational force. So we end up with $\frac{m{L}^{2}}{3}$ . Because the amplitude angle is only 10 degs, we may use the first order period formula $T=2\pi \sqrt { \frac { I }{ mgh } } .$ From here you can solve for L, and arrive at Mughal's formula.

very simple just use that formula, $L\quad =\quad \frac { 3g{ T }^{ 2 } }{ 8{ \pi }^{ 2 } } \quad =\quad 0.84$

Overrated!!!!!!!!!!

3gt^2/(8*pi^2)