Three point-masses $m_1, m_2$ and $m_3$ are located at the vertices of an equilateral triangle, having each…

Three point-masses $m_1, m_2$ and $m_3$ are located at the vertices of an equilateral triangle, having each side of length $L$. The moment of inertia of the system about an axis along an altitude of the triangle passing through $m_1$ is given by
  1. $I=\left(m_1+m_2+m_3\right) L^2$
  2. $I=\left(m_1+m_2\right) \frac{L^2}{2}$
  3. $I=\left(m_2+m_3\right) L^2$
  4. $I=\left(m_2+m_3\right) \frac{L^2}{4}$

Solution

The given situation is shown below
We know that moment of inertia be $I=m d^2$ where, $d$ is the perpendicular distance between body and axis of rotation, $\begin{aligned} \therefore \quad I & =m_2\left(\frac{L}{2}\right)^2+m_3\left(\frac{L}{2}\right)^2 \\ & =\frac{m_2 L^2}{4}+\frac{m_3 L^2}{4}=\left(m_2+m_3\right) \frac{L^2}{4} \end{aligned}$

Asked in: MHT CET Full Test 7

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