Question

A system is shown in the figure. Find the moment of inertia about an axis passing through its $COM$ and perpendicular to the plane. (Assume masses are connected with light inextensible strings)

• A. $\frac{M{L}^2}{18}$
• B. $\frac{17M{L}^2}{6}$
• C. $\frac{25M{L}^2}{18}$
• D. $\frac{7M{L}^2}{6}$

Answer 1

The correct option is B $\frac{17M{L}^2}{6}$

As we know,
$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
Taking origin at mass $M$,
$x_{com}=\frac{M\left(0\right)+3M\left(L\right)+2M\left(2L\right)}{M+3M+2M}$
$x_{com}=\frac{7L}{6}$ from mass $M$.

Now, moment of inertia about $\text{COM}:$
$I=\Sigma M{r}^2$


$I=M{\left(\frac{7L}{6}\right)}^2+3M{(\frac{7L}{6}-L)}^2+2M{(2L-\frac{7L}{6})}^2$
$I=\frac{49M{L}^2}{36}+\frac{3M{L}^2}{36}+\frac{50M{L}^2}{36}$
$I=\frac{17}{6}M{L}^2$