Answer:
[tex]I_1 = I_{left} + I_{right} = m(r/2)^2 + m(r/2)^2 = 2m(r^2/4) = \frac{1}{2}mr^2[/tex]
[tex]I_2 = mr^2[/tex]
The moment of inertia of the object with respect to an axis passing through one of the masses is greater than the moment of inertia with respect to an axis passing through the center of the rod.
Hence, [tex]I_2 > I_1[/tex]
Explanation:
Since the rod has negligible mass, it is straightforward to calculate the total moment of inertia of the object: the sum of moment of inertia of separate masses.
The moment of inertia of a point-like object is [tex]mr^2[/tex], and their distance to the axis is [tex]r/2[/tex].
In the second case, [tex]I_2[/tex], The axis passes through one of the masses, so that mass does not have a contribution to the total moment of inertia of the object. The only contribution comes from the other mass, whose distance to the axis is [tex]r[/tex].
