Two uniform solid of masses `m_(1)` and `m_(2)` and radii `r_(1)` and `r_(2)` respectively, are connected at the ends of a uniform rod of length `l` and mass `m`. Find the moment of inertia of the system about an axis perpendicular to the rod and passing through a point at a distance of a from the centre of mass of the rod as shown in figure.

Explanation

Draw the axis `"bb'", dd'` and `"cc"'` passing through the centres of the system and the centre of mass of the rod as shown in figure.

For `m_(1):I_("bb"')=2/5m_(1)r_(1)^(2)`
and` (I_("aa"'))_(I)=I_("bb"')+m_(1)(r_(1)+1/2-a)^(2)=2/5m_(1)r_(1)^(2)+m_(1)(r_(1)+1/2-a)^(2)`
[from paralel axis theorem]
For `m: I_("cc"')=(ml^(2))/12, (I_("aa"'))=I_("cc"')=ma^(2)=(ml^(2))/12+ma^(2)`
For `m_(2):I_("dd"')=2/5m_(2)r_(2)^(2)`
`implies(I_("aa"'))_(3)=I_("dd"')+m_(2)(r_(2)+1/2+a)^(2)=2/5m_(2)r_(2)^(2)+m_(2)(r_(2)+1/2a)^(2)`
The moment of inertia of the given system about `"aa"'`.
`I_("aa"')=(I_("aa"'))_(2)+(I_("aa"'))_(3)`
`={2/5m_(1)r_(1)^(2)+m_(1)(r_(1)+1/2-a)^(2)}`
`+{2/5m_(2)r_(2)^(2)+m_(2)(r_(2)+l/2-a)^(2)}`