Question

Three particles each of mass $2\text{kg}$ are placed at the corners of an equilateral triangle of side $6\text{m}$. What is the moment of inertia of the system about an axis passing through the height of the triangle on the same plane as the particles?

• A. $12{\text{kgm}}^2$
• B. $36{\text{kgm}}^2$
• C. $48{\text{kgm}}^2$
• D. $40{\text{kgm}}^2$

Answer 1

The correct option is B $36{\text{kgm}}^2$


Let the three particles be $A,B\&C$. $AD$ is the height of the triangle and the axis of rotation.
The height of an equilateral triangle passes through the mid-point of the base.
$\therefore BD=DC=3\text{m}$
Moment of inertia of the system
$=MI=m_A\times 0+m_B\times B{D}^2+m_C\times D{C}^2$
$\Rightarrow MI=0+2\times 3^2+2\times 3^2=36{\text{kg m}}^2$
$\therefore$ The moment of inertia of the system is $36{\text{kg m}}^2$.