Question

Two uniform solid spheres of masses $m_1$ and $m_2$ and radii $r_1$ and $r_2$, respectively, are connected at the ends of a uniform rod of length $l$ and mass $m$. Find the moment of inertia of the system about an axis perpendicular to the rod and passing through a point at a distance of $a$ from the centre of mass of the rod as shown in figure.

Answer 1

Draw the axes $b{b}^{\prime },d{d}^{\prime }$ and $c{c}^{\prime }$ passing through the centres of the spheres and the centre of mass of the rod as shown in figure

For $m_1:I_{b{b}^{\prime }}=\frac{2}{5}{m}_1{r}_1^2$

and ${\left(I_{a{a}^{\prime }}\right)}_1=I_{b{b}^{\prime }}+m_1{(r_1+\frac{1}{2}-a)}^2=\frac{2}{5}{m}_1{r}_1^2+m_1{(r_1+\frac{1}{2}-a)}^2$ (From parallel axis theorem)

For $m:I_{C{C}^{\prime }}=\frac{m{l}^2}{12},{\left(I_{a{a}^{\prime }.}\right)}_2=I_{c{c}^{\prime }}+m{a}^2=\frac{m{l}^2}{12}+m{a}^2$

For $m_2:I_{d{d}^{\prime }}=\frac{2}{5}{m}_2{r}_2^2\Rightarrow {\left(I_{a{a}^{\prime }}\right)}_3=I_{d{d}^{\prime }}+m_2{(r_1+\frac{1}{2}+a)}^2=\frac{2}{5}{m}_2{r}_2^2+m_2{(r_1+\frac{1}{2}+a)}^2$

The moment of inertia of the given system about $a{a}^{\prime }$

$I_{a{a}^{\prime }}={\left(I_{a{a}^{\prime }}\right)}_1+{\left(I_{a{a}^{\prime }}\right)}_2+{\left(I_{a{a}^{\prime }}\right)}_3=\left\{\frac{2}{5}{m}_1{r}_1^2{(r_1+\frac{1}{2}-a)}^2\right\}+\left\{\frac{2}{5}{m}_2{r}_2^2{(r_2+\frac{1}{2}-a)}^2\right\}$