Four spheres each of diameter 0.02m and 0.10 kg are placed with their centres on the vertices of a square of side 0.05m. Calculate the moment of inertia of the system about one side of the taken as the axis of rotation.

Explanation

Let the axis of rotation be along AD. Moment of inertia of B and C about the axis XY `=I_0+Md^2`
i.e M.I`=2/5MR^2+Md^2=0.1[0.4xx(0.01)^2+(0.05)^2]`
M.I`=2.54xx10^(4)kgm^2`
Moment of inertia of A & D about an axis XY is `=2/5MR^2`
`=0.4xx0.1xx(0.01)^2`
`=4xx10^(-6)=0.04xx10^(-4)kgm^2`
Hence total M.l of the system`=2xx2.54xx10^(-4)+2xx0.04xx10^(-4)`
`=5.16xx10^(-4)kgm^2`