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Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass MMM and radius RRR about an axis perpendicular to the hoop's plane at an edge.

Sagot :

Answer:

I = sum m *r^2    where m represents the (small) individual masses and r is the distance of that mass from center of rotation

Note:  sum m = M

For the hoop given all masses are at a distance RRR from the center of rotation

I = MMM * RRR^2

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