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#