Three point masses m_1, m_2,and m_3 are place | vedclass

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(A) 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

Vedclass · Accepted Answer

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

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(A) 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.The axis of rotation is the altitude passing through $m_1$. $1$. For mass $m_1$: The particle lies on the axis,so its perpendicular distance $r_1 = 0$.$2$. For mass $m_2$: The perpendicular distance from the axis is $r_2 = a \cos(60^\circ) = \frac{a}{2}$.$3$. For mass $m_3$: The perpendicular distance from the axis is $r_3 = a \cos(60^\circ) = \frac{a}{2}$.Now,calculating the total moment of inertia:$I = m_1(0)^2 + m_2 \left( \frac{a}{2} \right)^2 + m_3 \left( \frac{a}{2} \right)^2$$I = 0 + m_2 \frac{a^2}{4} + m_3 \frac{a^2}{4}$$I = (m_2 + m_3) \frac{a^2}{4}$

Three point masses $m_1, m_2$,and $m_3$ are placed at the vertices of an equilateral triangle of side $a$. Find the moment of inertia of this system about the altitude passing through $m_1$.

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