## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$I = \frac{1}{12}ML^2+\frac{1}{4}M_1~L^2+\frac{1}{16}M_2~L^2$
To find the moment of inertia of the object, we can add the moment of inertia of the rod, the moment of inertia of $M_1$, and the moment of inertia of $M_2$. Therefore: $I =\frac{1}{12}ML^2+M_1(\frac{L}{2})^2+M_2(\frac{L}{4})^2$ $I = \frac{1}{12}ML^2+\frac{1}{4}M_1~L^2+\frac{1}{16}M_2~L^2$