Question

Four particles each of mass $1\text{kg}$ are placed at the four corners of a square $ABCD$ with a side of length $2\text{m}$. What is the moment of inertia of the system about an axis passing through the mid-points of sides $AB$ and $CD$?

• A. $4{\text{kg m}}^2$
• B. $8{\text{kg m}}^2$
• C. $2{\text{kg m}}^2$
• D. $10{\text{kg m}}^2$

Answer 1

The correct option is A $4{\text{kg m}}^2$


Here, $PQ$ be the axis of rotation.
$P$ is the mid-point of $AB$ and $Q$ is the mid-point of $DC$.
$\therefore AP=PB=CQ=DQ=1\text{m}$
To calculate the moment of inertia, we consider the perpendicular distances of the particles from the axis of rotation.
$\therefore MI=m_A\times A{P}^2+m_B\times P{B}^2+m_D\times D{Q}^2+m_C\times C{Q}^2$
$=1\times 1^2+1\times 1^2+1\times 1^2+1\times 1^2=4{\text{kg m}}^2$
$\therefore$ The moment of inertia of the system is $4{\text{kg m}}^2$.