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) $(m_2 + m_3)(a^2/4)$.

Answer 2

To solve the problem, we need to calculate the moment of inertia of a system of point masses located at the vertices of an equilateral triangle along an axis passing through one vertex.

1. Understanding the Problem:
The system consists of three point masses \(m_1\), \(m_2\), and \(m_3\), located at the vertices of an equilateral triangle with side length \(a\). The axis of rotation passes along the altitude of the triangle through vertex \(m_1\).

2. Conceptual Approach:
The moment of inertia of the system about the given axis is the sum of the moments of inertia of each mass about the axis. The mass \(m_1\) is located at the axis of rotation, so its moment of inertia is zero. For the other two masses, the moment of inertia is given by the formula \(I = mr^2\), where \(r\) is the perpendicular distance from the mass to the axis of rotation.

The altitude of the equilateral triangle divides it into two 30-60-90 right triangles, and the perpendicular distance of the masses \(m_2\) and \(m_3\) from the axis of rotation is \(\frac{a}{2}\), which is half the length of the side of the triangle.

3. Calculating the Moment of Inertia:
The total moment of inertia is the sum of the moments of inertia of \(m_2\) and \(m_3\), as \(m_1\) contributes nothing. The formula is:

\[ I = m_2 \left(\frac{a}{2}\right)^2 + m_3 \left(\frac{a}{2}\right)^2 = \left(m_2 + m_3\right) \frac{a^2}{4} \]

4. Conclusion:
The correct expression for the moment of inertia of the system is:

Final Answer:
The correct option is (D) \((m_2 + m_3) \frac{a^2}{4}\).