Question

Two balls each of mass $1\text{kg}$ are connected to both ends of a uniform circular rod of mass $5\text{kg}$ and length $2\text{m}$. Find the moment of inertia of the system about an axis passing through the middle of the rod and perpendicular to the length of the rod.

• A. $\frac{5}{3}{\text{kg-m}}^2$
• B. $2{\text{kg-m}}^2$
• C. $\frac{13}{3}{\text{kg-m}}^2$
• D. $\frac{11}{3}{\text{kg-m}}^2$

Answer 1

The correct option is D $\frac{11}{3}{\text{kg-m}}^2$

Each ball of mass $m=1\text{kg}$ is at a distance of $1\text{m}$ from axis $\text{OO'}$
i.e $d=1\text{m}$.
So, moment of inertia of both balls is
$I_1=2\times m{d}^2=2\times 1\times (1{)}^2$
$I_1=2{\text{kg-m}}^2$...(I)

Moment of inertia of uniform circular rod of mass $M=5\text{kg}$ and length $l=2\text{m}$ about $\text{OO'}$ is
$I_2=\frac{M{l}^2}{12}=\frac{5\times (2{)}^2}{12}=\frac{5}{3}{\text{kg-m}}^2$ ...(II)
So, Moment of inertia of system about axis $\text{OO'}$ :
$I=I_1+I_2$
$=2+\frac{5}{3}=\frac{11}{3}{\text{kg-m}}^2$