Question

Three balls of masses $2 \, \text{kg}$, $4 \, \text{kg}$, and $6 \, \text{kg}$ respectively are arranged at the centre of the edges of an equilateral triangle of side $2 \, \text{m}$.The moment of inertia of the system about an axis through the centroid and perpendicular to the plane of the triangle, will be ____ $\text{kg} \, \text{m}^2$

Answer 1

Correct Answer: 4

This problem requires the calculation of the moment of inertia for a three-mass system positioned at the midpoints of an equilateral triangle's sides. The rotation axis is defined as passing through the triangle's centroid and being perpendicular to its plane.

Concept Used:

The methodology relies on the formula for the moment of inertia of discrete point masses and the geometric characteristics of an equilateral triangle.

1. Moment of Inertia of Point Masses: For a system of particles, the moment of inertia (\(I\)) relative to an axis is computed by summing the product of each particle's mass (\(m_i\)) and the square of its perpendicular distance (\(r_i\)) from the axis of rotation. \[ I = \sum_{i=1}^{n} m_i r_i^2 \]
2. Geometry of an Equilateral Triangle:
   • The centroid is the intersection point of the triangle's medians.
   • In an equilateral triangle, medians also serve as altitudes and angle bisectors.
   • The length of a median (distance from a vertex to the midpoint of the opposite side) is given by \(h = \frac{\sqrt{3}}{2}s\), where \(s\) is the side length.
   • The centroid divides each median in a 2:1 ratio. The distance from the centroid to a side's midpoint is one-third the median's length.

Step-by-Step Solution:

Step 1: Identify the given parameters.

• Masses: \(m_1 = 2 \, \text{kg}\), \(m_2 = 4 \, \text{kg}\), \(m_3 = 6 \, \text{kg}\).
• Equilateral triangle side length: \(s = 2 \, \text{m}\).

Step 2: Determine the distance of each mass from the axis of rotation.

The axis of rotation passes through the centroid and is perpendicular to the triangle's plane. The masses are located at the midpoints of the sides. Therefore, the perpendicular distance of each mass from the axis is its distance from the centroid within the triangle's plane.

First, calculate the median length (\(h\)) of the triangle:

\[h = \frac{\sqrt{3}}{2}s = \frac{\sqrt{3}}{2}(2 \, \text{m}) = \sqrt{3} \, \text{m}\]

The centroid of an equilateral triangle is equidistant from the midpoints of its sides. This distance (\(r\)) is one-third the median length:

\[r = \frac{1}{3}h = \frac{\sqrt{3}}{3} \, \text{m} = \frac{1}{\sqrt{3}} \, \text{m}\]

Consequently, all three masses are at an identical perpendicular distance \(r = \frac{1}{\sqrt{3}} \, \text{m}\) from the axis of rotation.

Step 3: Apply the moment of inertia formula.

The total moment of inertia (\(I\)) is the sum of the individual moments of inertia of the three masses:

\[I = m_1 r^2 + m_2 r^2 + m_3 r^2\]

As the distance \(r\) is constant for all masses, we can factor out \(r^2\):

\[I = (m_1 + m_2 + m_3) r^2\]

Step 4: Substitute the values and compute the final result.

Calculate the total mass of the system:

\[M = m_1 + m_2 + m_3 = 2 \, \text{kg} + 4 \, \text{kg} + 6 \, \text{kg} = 12 \, \text{kg}\]

Calculate the square of the distance \(r\):

\[r^2 = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \, \text{m}^2\]

Compute the total moment of inertia:

\[I = (12 \, \text{kg}) \times \left(\frac{1}{3} \, \text{m}^2\right)\]\[I = 4 \, \text{kg} \, \text{m}^2\]

The moment of inertia of the system about the specified axis is 4 kg m².