Question

Two particles having masses $3\text{kg}$ and $6\text{kg}$ are situated in a plane perpendicular to line $AB$ at a perpendicular distance of $2\text{m}$ and $4\text{m}$ respectively as shown in figure. Find the moment of inertia about an axis $A^{\prime }{B}^{\prime }$ parallel to $AB$.

• A. $216{\text{kg-m}}^2$
• B. $108{\text{kg-m}}^2$
• C. $300{\text{kg-m}}^2$
• D. $200{\text{kg-m}}^2$

Answer 1

The correct option is A $216{\text{kg-m}}^2$

Given,
$m_1=3\text{kg},m_2=6\text{kg}$
$r_1=2\text{m};r_2=4\text{m}$ (perpendicular distances of $m_1$ and $m_2$ from axis $AB$)
Let us consider $3\text{kg}$ is at the origin along axis $A^{\prime }{B}^{\prime }$. Then $r_1^{\prime }=0,r_2^{\prime }=r_1+r_2=6\text{m}$ moment of inertia of two particle system about $A^{\prime }{B}^{\prime }$ is
$I_{A^{\prime }{B}^{\prime }}=m_1{r}_1^{\prime 2}+m_2{r}_2^{\prime 2}$
$=3\left(0\right)+6(6{)}^2=216{\text{kg-m}}^2$