Three point masses m_1, m_2, m_3 are located | 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 t

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 of rotation.$1$. The axis passes through $m_1$ and the altitude of the equilateral triangle. The perpendicular distance of $m_1$ from this axis is $r_1 = 0$.$2$. In an equilateral triangle of side $a$,the altitude bisects the opposite side. Thus,the perpendicular distance of both $m_2$ and $m_3$ from the altitude passing through $m_1$ is $r_2 = r_3 = \frac{a}{2}$.$3$. The total moment of inertia $I$ is:$I = m_1(0)^2 + m_2(\frac{a}{2})^2 + m_3(\frac{a}{2})^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 $m_1, m_2, m_3$ are located at the vertices of an equilateral triangle of side length $a$. The moment of inertia of the system about an axis along the altitude of the triangle passing through $m_1$ is:

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