Find the moment of inertia `A` of a spherical ball of mass `m` and radius `r` attached at the end of a straight rod of mass `M` and length `l`, if this system is free to rotate about an axis passing through the end of the rod (end of the rod opposite to sphere).

Explanation

In the system shown in the figure, the total moment of inertia can be given as the sum of moment of inertia of the rod and that of the spherical.
Moment of inertia of the rod about the axis passing through an end is `I_("rod")=Ml^(2)//3.`
Moment of inertia of the spherical ball about the axis passing through its geometrical centre is `I=2m^(2)//5`.
Using parallel axis theorem, we get the moment of inertia about the axis passing through the end of rod as
`I_("AA"')=2/5 mr^(2)+ml^(2)`
Thus total moment of inertia of the system can be given as
`I_("total")=I_(rod)+I_("AA"')`
or `=(Ml^(2))/3+(2/5mr^(2)+ml^(2))`