Three point masses m_1, m_2, m_3 are placed a | vedclass

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

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$ (m_2 + m_3) \frac{a^2}{4} $

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(A) The moment of inertia $I$ of a system of point masses about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th mass from the axis.For the given axis passing through $m_1$ and bisecting the side $m_2m_3$:$1$. The distance of $m_1$ from the axis is $r_1 = 0$.$2$. The distance of $m_2$ from the axis is $r_2 = a/2$.$3$. The distance of $m_3$ from the axis is $r_3 = a/2$.Therefore,the total moment of inertia is:$I = m_1(0)^2 + m_2(a/2)^2 + m_3(a/2)^2$$I = 0 + m_2(a^2/4) + m_3(a^2/4)$$I = (m_2 + m_3) \frac{a^2}{4}$

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