Answer to Question #154806 in Physics for Nashe

Question #154806

A slim rod 2.0 m long has two small 0.5 kg bodies attached at its two ends and a 0.25 kg body at its midpoint. Find the moment of inertia of the system of masses about an axis perpendicular to the rod and intersecting it at 0.50 m from one end.

Assume that the rod has negligible mass.


1
Expert's answer
2021-01-11T11:40:56-0500

I=0.50.52+0.5(20.5)2+I=0.5\cdot0.5^2+0.5\cdot(2-0.5)^2+


+0.25(2/20.5)2=1.313(kgm2)+0.25\cdot(2/2-0.5)^2=1.313(kg\cdot m^2) . Answer

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