Question

Three point masses \(m_1, m_2, m_3\) are located at the vertices of an equilateral triangle of side 'a'. The moment of inertia of the system about an axis along the altitude of the triangle passing through \(m_1\) is:

• A. \((m_2 + m_3)a^2\)
• B. \((m_1 + m_2 + m_3)a^2\)
• C. \((m_2 + m_3)\frac{a^2}{9}\)
• D. \((m_2 + m_3)\frac{a^2}{4}\)

Hint:

Remember, the moment of inertia depends not only on the mass but also on the square of the distance from the axis of rotation.

Answer 1

The Correct Option is D

1. Visualize the Setup
We have an equilateral triangle with side 'a'. Masses m1, m2, and m3 are placed at the vertices. The axis of rotation passes through m1 and is perpendicular to the plane of the triangle.
2. Determine the Distances
m1: Distance from the axis is 0.
m2 and m3: Distance from the axis is a/2 (half the side length).
3. Calculate the Moment of Inertia
The moment of inertia (I) is given by: $I = \Sigma (m \times r^2)$ where:
m is the mass
r is the perpendicular distance from the axis of rotation
I1 (due to m1): $I_1 = m_1 \times (0)^2 = 0$
I2 (due to m2): $I_2 = m_2 \times (a/2)^2 = m_2 \times (a^2/4)$
I3 (due to m3): $I_3 = m_3 \times (a/2)^2 = m_3 \times (a^2/4)$
4. Sum the Moments of Inertia
$I = I_1 + I_2 + I_3$
$I = 0 + m_2 \times (a^2/4) + m_3 \times (a^2/4)$
$I = (m_2 + m_3) \times (a^2/4)$
Therefore, the moment of inertia of the system about the given axis is $(m_2 + m_3)(a^2/4)$.
The correct answer is (4) $(m_2 + m_3)(a^2/4)$.