Originally Posted by **McKenziemc799** Yes. |

Unless I'm missing some other information, I disagree with both your calculation for period and the "actual answer" value.

Using the parallel axis theorem,

$I = I_{com} + md^2$

for a uniform ring of mass $m$, $I_{com} = mr^2$ and $d = r$

so, $I = mr^2 + mr^2 = 2mr^2$

$T = 2\pi \sqrt{\dfrac{I}{mgd}} = 2\pi \sqrt{\dfrac{2mr^2}{mgr}} = 2\pi \sqrt{\dfrac{2r}{g}}$

For $r= 1 \text{ m } \implies T = 2\pi \sqrt{\dfrac{2}{g}} \approx 2.84 \text{ sec}$