Question

Two identical spherical balls of mass $M$ and radius $R$ each are stuck on two ends of a rod of length $2R$ and mass $M$ (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is :

• A. $\frac{152}{15}M{R}^2$
• B. $\frac{17}{15}M{R}^2$
• C. $\frac{137}{15}M{R}^2$
• D. $\frac{209}{15}M{R}^2$

Answer 1

Answer: (C)

For Ball using parallel axis theorem.
$I_{\text{ball}}=\frac{2}{5}M{R}^2+M{\left(2R\right)}^2$
$=\frac{22}{5}M{R}^2$
$2$ Balls so $\frac{44}{5}M{R}^2$
Irod = for rod $\frac{M{\left(2R\right)}^2}{R}=\frac{M{R}^2}{3}$
$I_{\text{system}}=I_{\text{Ball}}+I_{\text{rod}}$
$=\frac{44}{5}M{R}^2+\frac{M{R}^2}{3}$
$=\frac{137}{15}M{R}^2$