How to proof mass moment of inertia formula for a hoop with axis across the diameter?

Question

Moment of inertia of the hoop is given by:
$1/2mr^2$
How to proof that?

You may attach a hyperlink or write down the derivation from $I=mr^2$ .

Thanks

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Answer 1 · 2017-07-20T05:31:41

See the proof below

The volume is $=2piRt$

The thickness is $=t$

The radius of the hoop is $=R$

The density is $rho=M/(2piRt)$

The moment of inertia is

$I=intr^2dm$

As the axis is across the diameter

The distance from the differential mass $dm$ is $=Rsintheta$

$dm=rhoRt d theta$

$cos2theta=1-2sin^2theta$

$sin^2theta=1/2-1/2cos2theta$

Therefore, substituting in the integral, we integrate from $0$ to $pi$ and multiply by $2$

$I=2int_0^piR^2sin^2thetarhoRt d theta$

$=2R^3rho t int_0^pi sin^2thetad theta$

$=2R^3 rho t *[theta/2-1/4sin(2theta)]_0^pi$

$=2R^3 rho t(pi/2)$

$=M/(2piRt)*2R^3*t*pi/2$

$=1/2MR^2$