Question

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.

Answer 1

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

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