Determine the moment of inertia about the axis of the object shown in FIGURE P12.51.

The axis passes through the midpoint of the rod and is perpendicular to the rod.
Two balls of mass m₁and m₂ are connected as shown in the figure.
The mass of the rod is M and the length of the rod is L.

The moment of inertia of a rod of mass M and length L, where it is about an axis passing through the mid point of the rod and perpendicular to the rod is given as

$I=\frac{M{L}^2}{12}........................................\left(1\right)$

As radius of the mass is not given lets consider them as point mass.

The moment of inertia of point mass is given by I=mr²

Now find the moment of inertia of both point mass

Mass-1

$I_1=m_1{\left(\frac{L}{2}\right)}^2=\frac{m_1{L}^2}{4}..............................\left(2\right)$

Mass-2

$I_1=m_2{\left(\frac{L}{4}\right)}^2=\frac{m_2{L}^2}{16}..............................\left(3\right)$

Add (1) , (2) and (3) to get total inertia.

$I_{Total}=\frac{M{L}^2}{12}+\frac{m_1{L}^2}{4}+\frac{m_2{L}^2}{16}$