Physics - Rotation Question with Solution | TestHub

From the given figure, the coordinates of the four point masses, each of mass $m$, are $(0,0)$, $(3a,0)$, $(a,a)$, and $(a,-\sqrt{2}a)$.

The moment of inertia of a system of particles about the X-axis is given by $I_x=\sum m_i{y}_i^2$, where $y_i$ is the perpendicular distance of the $i$-th mass from the X-axis.

Substituting the coordinates of the masses:

$I_x=m(0{)}^2+m(0{)}^2+m(a{)}^2+m(-\sqrt{2}a{)}^2$

$I_x=0+0+m{a}^2+2m{a}^2$

$I_x=3m{a}^2$

Answer: $3{\text{ ma}}^2$