Question

In the diagram shown below, all three rods are of equal length $L=1\text{m}$ and equal mass $M=3\text{kg}$. The system is rotated such that rod $B$ is the axis. What is the moment of inertia of the system?

• A. $0.5{\text{kg-m}}^2$
• B. $0.9{\text{kg-m}}^2$
• C. $1.32{\text{kg-m}}^2$
• D. $2{\text{kg-m}}^2$

Answer 1

The correct option is A $0.5{\text{kg-m}}^2$

Moment of inertia of the system = MOI of $\text{A}$ + MOI of $\text{B}$ + MOI of $\text{C}$.
Moment of inertia of a rod of mass $M$ and length $L$ about an axis passing through its centre and perpendicular to its length $=\frac{M{L}^2}{12}$

So, MOI of rod $\text{A}=$ MOI of rod $\text{C}$ $=\frac{M{L}^2}{12}$
MOI of rod $\text{B}=0$
( $\because$ Moment of mass about an axis passing through its own position is zero)

Total MOI $I=\frac{1}{12}M{L}^2+0+\frac{1}{12}M{L}^2$
$=\frac{1}{6}M{L}^2$
$=\frac{1}{6}\times 3\times (1{)}^2=0.5{\text{kg-m}}^2$
$\therefore I=0.5{\text{kg-m}}^2$