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Rotational Inertia

Three point m...

Question

Three point masses $m_{1}, m_{2}, m_{3}$ are located at the vertices of an equilateral triangle of length 'a'. The moment of inertia of the system about an axis along the altitude of the triangle passing through $m_{1}$ , is:-

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

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(m_{1} + m_{2} + m_{3}) a^{2}

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(m_{1} + m_{2}) \frac{a^{2}}{4}

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

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Solution

The correct option is B $(m_{2} + m_{3}) \frac{a^{2}}{4}$
moment of inertia is the product of mass and square of separation between particle and axis of rotation .
e.g , $M . I = m r ²$
here, we see, separation of mass m1 and altitude $N N^{'} i s 0$ .
alteration between mass $m_{2}$ and $N N^{'}$ is $(\frac{a}{2})$ also for $m_{3}$ separation is $(\frac{a}{2})$
moment of inertia about altitude passing through $m_{1} = I_{1} + I_{2} + I_{3}$
where $I_{1}, I_{2}, a n d I_{3}$ are $M . I o f m_{1}, m_{2} a n d m_{3}$ respectively .
$M . I = m_{1} . (0) + m_{2} (\frac{a}{2})^{2} + m_{3} (\frac{a}{2})^{2}$
$= \frac{a^{2}}{4 \times (m_{2} + m_{3})}$

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