Question

In an arrangement four particles, each of mass 2 gram are situated at the coordinate points (3, 2, 0), (1, -1, 0), (0, 0, 0) and (-1, -1, 0). The moment of inertia of this arrangement about the z-axis will be:

• A. $8\text{units}$
• B. $16\text{units}$
• C. $43\text{units}$
• D. $34\text{units}$

Answer 1

The correct option is D $34\text{units}$


MOI about
z-axis
$I_z=\sum m{r}^2=m(r_1^2+r_2^2+r_3^2+r_4^2)...\left(1\right)$
$r_1=\sqrt{(3{)}^2+(2{)}^2}=\sqrt{13}$
$r_2=\sqrt{1^2+1^2}=\sqrt{2}$
$r_3=0$
$r_4=\sqrt{(-1{)}^2+(-1{)}^2}=\sqrt{2}$
From equation (1),
$I_z=2[13+2+0+2]$
$=34\text{units}$