Moment of inertia of the ring through the diameter

$\begingroup$

I know that the moment of inertia of the ring through the Diameter is $I_{x}=I_{y} = mr^2/2$.

But I cannot get this formula using the integral. Give me some hints how to do this. Thanks a lot!

$\endgroup$ 2

1 Answer

$\begingroup$

$I=\int r'^2 dm$

$dm={{M}\over{2\pi}}d\theta$

$r'=r\cos\theta$

$I=\int_0^{2\pi} r^2\cos^2\theta {{M}\over{2\pi}}d\theta$

$I={{Mr^2}\over{2\pi}}\int_0^{2\pi} \cos^2\theta d\theta$

$I={{Mr^2}\over{2\pi}}[{{\theta}\over{2}}+{{\sin 2\theta}\over{4}}]\biggl|_{\theta=0}^{2\pi}$

$I={{Mr^2}\over{2\pi}}[(\pi+0)-(0+0)]$

$I={{Mr^2}\over{2}}$

$\endgroup$