Question

Three point masses m₁, m₂, m₃ 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 m₁ is
(a) (m₂ + m₃) a² (b) (m₁ + m₂ + m₃) a²
4
(c) (m₁ + m₂) a² (d) (m₂ + m₃) a²
4

Answer 1

Since m₁ is along the axis, its distance from the axis is 0.

The masses m₂ and m₃ each is at a perpendicular distance 'a/2' from the axis, since the axis is along the altitude of the triangle which is the perpendicular bisector of the side of the triangle joining m₂ and m₃.

Hence the moment of inertia of the system of particles is,

$\longrightarrow\sf{I=m_1(0)^2+m_2\left(\dfrac{a}{2}\right)^2+m_3\left(\dfrac{a}{2}\right)^2}$

$\longrightarrow\sf{\underline{\underline{I=\dfrac{(m_2+m_3)a^2}{4}}}}$