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

The problem asks for the moment of inertia of a system of three masses placed at the midpoints of the sides of an equilateral triangle. The axis of rotation passes through the centroid of the triangle and is perpendicular to its plane.

Concept Used:

The solution uses the formula for the moment of inertia of a system of discrete point masses and the geometric properties of an equilateral triangle.

1. Moment of Inertia of Point Masses: The moment of inertia (\(I\)) of a system of particles about a given axis is the sum of the products of the mass of each particle (\(m_i\)) 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 point where the medians of the triangle intersect.
   • In an equilateral triangle, the medians are also the altitudes and angle bisectors.
   • The distance from a vertex to the midpoint of the opposite side (length of the median) is \(h = \frac{\sqrt{3}}{2}s\), where \(s\) is the side length.
   • The centroid divides each median in a 2:1 ratio. The distance from the centroid to the midpoint of a side is \(\frac{1}{3}\) of the median's length.

Step-by-Step Solution:

Step 1: List the given parameters.

• Masses: \(m_1 = 2 \, \text{kg}\), \(m_2 = 4 \, \text{kg}\), \(m_3 = 6 \, \text{kg}\).
• Side length of the equilateral triangle, \(s = 2 \, \text{m}\).

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 plane of the triangle. The masses are located at the midpoints of the sides. Therefore, the perpendicular distance of each mass from the axis is simply its distance from the centroid in the plane of the triangle.

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

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

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

Step 3: Apply the formula for the moment of inertia.

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

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

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

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

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

First, calculate the total mass of the system:

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

Next, calculate the square of the distance \(r\):

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

Now, calculate 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.

Answer 2

For an equilateral triangle, the distance \( r \) from the center of each edge to the centroid \( C \) is:
\[r = \frac{1}{\sqrt{3}}\]
The moment of inertia \( I \) about point \( C \) and perpendicular to the plane is given by:
\[I = r^2 \left[ 2 + 4 + 6 \right]\]
Substitute \( r = \frac{1}{\sqrt{3}} \):
\[I = \left( \frac{1}{\sqrt{3}} \right)^2 \times 12\]
\[I = \frac{1}{3} \times 12 = 4 \, \text{kg} \cdot \text{m}^2\]