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

Answer 1

The correct option is A $\frac{137}{15}M{R}^2$

Step 1:Draw diagram of the given problem
Step 2: Calculate moment of inertia of the balls and rod.
Given,
Mass of the identical spherical balls = $M$
Radius of the ball =$R$
Length of the rod =$2R$
Mass of the rod =$M$
Here, given that $L=2R$
For $I$ of the ball, using parallel axis
theorem.
$I_{ball}=\frac{2}{5}M{R}^2+M(2R{)}^2=\frac{22}{5}M{R}^2$

Considering both spheres at equal
distance from the axis, moment of inertia due to both spheres about this axis will be

$2I_{ball}=2\times \frac{22}{5}M{R}^2$
Moment of inertia of both the ball,
$I_{balls}=\frac{44}{5}M{R}^2$
Moment of inertia of the rod, $I_{rod}$
$\frac{\left(M\right(2R{)}^2)}{12}=\frac{\left(M{R}^2\right)}{3}$
Step 3:Calculate moment of inertia of the system.
$I_{system}=I_{balls}+I_{rod}$
$I=\frac{44}{5}M{R}^2+\frac{\left(M{R}^2\right)}{3}=\frac{137}{15}M{R}^2$
Final Answer: (c)