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.
"I=0.5\\cdot0.5^2+0.5\\cdot(2-0.5)^2+"
"+0.25\\cdot(2\/2-0.5)^2=1.313(kg\\cdot m^2)" . Answer
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