Two identical spheres each of mass 1.20 kg and radius 10.0 cm are fixed at the ends of a light rod so that the separation between the centers is 50.0 cm. Find the moment of inertia of the system about an axis perpendicular to the rod passing through its middle point.

Explanation

Consider the diameter of one of the spheres paralel to the given axis. The moment of inertia of this sphere about the diameter is
`I=2/5mR^2=2/5(1.20kg)(0.1m)62`
`=4.8xx10^-3kg-m^2`
Its moment of inertia about the given axis is obtained by using the parallel axes theorem. Thus,
`I=I_(cm)+md^2`
`=4.8xx10^-3kg-m^2+(1.20kg)(0.25m)^2 ltbr. =4.8xx10^-3 kgm^2+0.075kg-m^2`
`=79.8xx10^-3kg-m^2`
The moment of inertia of the second sphere is also the same so that the moment of inertia of the system is
`2xx79.8xx0^-3kg-m^2~~10.160kg-m^2`