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(B) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the

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$2ma^2$

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(B) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis of rotation.For the $Y$-axis,the perpendicular distance of a point $(x, y)$ is $|x|$.The coordinates of the four masses are $(a, 0)$,$(0, a)$,$(-a, 0)$,and $(0, 2a)$.The perpendicular distances from the $Y$-axis are $x_1 = a$,$x_2 = 0$,$x_3 = -a$,and $x_4 = 0$.Thus,$I_y = m(a)^2 + m(0)^2 + m(-a)^2 + m(0)^2$.$I_y = ma^2 + 0 + ma^2 + 0 = 2ma^2$.

Four point masses (each of mass $m$) are arranged in the $X-Y$ plane at coordinates $(a, 0)$,$(0, a)$,$(-a, 0)$,and $(0, 2a)$. What is the moment of inertia of this system about the $Y$-axis?

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