Find the moment of inertia of a hoop (a thin-walled, hollow ring)...

Question

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

Accepted Answer

Hello everyone. So this problem a thin light cord is wound around a pulley of diameter centimeters and mass one kg. The pulley is considered to be a thin determine the moment of inertia of the pulley around an axis perpendicular to the police plane and passing through the court. So we have some polling it was considered a hoop and we have a cord wrapping around it And its diameter is 20 cm. So its radius We are is equal to 10 cm. It has a mass of one kg. Now the axis of rotation will be through this cord. You only recall that the moment of inertia for a hoop around its center of mass through the center is equal to M. R squared. But if you recall the parallel axis theorem, we can calculate this new moment of inertia as the moment of the show the through the center of that mass loss M times the distance from the center of mass to this new parallel axis which we want to find. So M. R. Squared. And now we can substitute this equation and get that the new moment of inertia is simply M. R squared plus M R squared. As the distance traveled from the center of mass to the court is simply the radius of the pulley. So we get that I prime is equal to two M squared. And we know are we know em so we can make the substitution one kg times two Times the radius which is 10 cm, which is simply just .1 m. We get that I prime is equal to 0. kg times meters square hands of choice. See Hope. This helps you have a great day.

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