Question

Two identical solid spheres each of mass $1.20\text{kg}$ and radius $10.0\text{cm}$ are fixed at the ends of a light rod so that the separation between the centres is $50.0\text{cm}$. Find the moment of inertia $\left({\text{kgm}}^2\right)$ of the system about an axis perpendicular to the rod passing through its middle point.

Answer 1

From given, we can visualize as shown
for a single body,
Moment of inertia $\left(I_{cm}\right)$
$I_{y{y}^{\prime }}=I_{cm}+M{x}^2{I}_{y{y}^{\prime }}=\frac{2}{5}M{R}^2+M{x}^2$

putting all values,
$I_{y{y}^{\prime }}=\frac{2}{5}1.2\times {0.1}^2+1.2\times {0.25}^2{I}_{y{y}^{\prime }}=0.08kg-m^2$

Because here two identical sphere, so
$I_{y{y}^{\prime }}=0.08\times 2I_{y{y}^{\prime }}=0.16kg-m^2$