91Ó°ÊÓ

(a) The moment of inertia of an object of mass $m$rotating about an axis at a distance $L$from the axis of rotation is given by

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

The rod is a low-mass rod, so we can neglect the contribution of the rod’s moment of inertia to the total moment of inertia of the object. The rod has two red balls each at its end at a distance $\frac{L}{2}$from the center of the rod, and two balls each at a distance$0.08m$from the center of the rod. So, the moment of inertia of the object is sum of the moment of inertias of each ball about the center of rod. Hence, the moment of inertia of the device is

$$\begin{gathered} I_{total}=2I_{red}+2I_{blue} \\ {I}_{total}=2m{L^2}_{red}+2m_{blue}{L}_{blue}^2......\left(2\right) \end{gathered}$$

On substituting the known values in the equation (2), the moment of inertia of the device can be calculated as

$I_{total}=2m{L}_{red}+2m_{blue}{L}_{blue}^2$

$=2\left(0.82kg\right){\left(0.16m\right)}^2+2\left(0.29kg\right){\left(0.08m\right)}^2$

$=0.0457kg·m^2$