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Rotational Inertia

Three point m...

Question

Three point masses $m_{1}, m_{2},$ and $m_{3}$ are placed at three corners of a light equilateral triangle of side $a$ .The moment of inertia of the system about an axis coinciding with the altitude of the triangle passing through $m_{1}$ is:

m_{1} + m_{2} + m_{3}

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\frac{(m_{2} + m_{3}) a^{2}}{6}

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\frac{(m_{2} + m_{3})^{2}}{a^{2}}

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\frac{(m_{2} + m_{3}) a^{2}}{4}

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\frac{(m_{2} + m_{3}) a^{4}}{4}

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Solution

The correct option is D $\frac{(m_{2} + m_{3}) a^{2}}{4}$
The MI of $m_{2}$ is $m_{2} (a / 2)^{2} = m_{2} a^{2} / 4$
The MI of $m_{3}$ is $m_{3} (a / 2)^{2} = m_{3} a^{2} / 4$
The MI of $m_{1}$ is $m_{1} (0)^{2} = 0$
So total MI is $m_{2} a^{2} / 4 + m_{3} a^{2} / 4 + 0 = (m_{2} + m_{3}) a^{2} / 4$

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