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Step-by-Step Solution
Step 1: Understand the Physical Setup
We have an equilateral triangle EFG of side a . Three particles of equal mass m are located at its vertices E, F, and G. We need to find the moment of inertia of this system about the line EX, where EX passes through vertex E and is perpendicular to EG, remaining in the plane of the triangle.
Step 2: Choose a Suitable Coordinate System
• Place vertex E at the origin: E = (0, 0) .
• Place vertex G on the x -axis: G = (a, 0) .
• Since the triangle is equilateral, vertex F will be at F = \left(\frac{a}{2}, \frac{\sqrt{3}\,a}{2}\right) .
Step 3: Identify the Axis of Rotation
• The axis EX is through point E and perpendicular to EG in the plane.
• Since EG lies along the x -axis, a line perpendicular to EG through E is along the y -axis.
Step 4: Calculate Each Mass’s Perpendicular Distance from the Axis
The moment of inertia of a point mass m about an axis is I = m\,r^2 , where r is the perpendicular distance from the axis.
Mass at E: Located at (0,0) . Distance from the y -axis (the axis EX) is 0 . Thus, its contribution to moment of inertia is 0 .
Mass at G: Located at (a,0) . Distance from the y -axis is a . Thus, its contribution to moment of inertia is m\,a^2 .
Mass at F: Located at \left(\frac{a}{2}, \frac{\sqrt{3}a}{2}\right) . Distance from the y -axis is \frac{a}{2} . Thus, its contribution to moment of inertia is m\,\left(\frac{a}{2}\right)^2 = \frac{m\,a^2}{4} .
Step 5: Sum the Contributions
The total moment of inertia I about EX is the sum of contributions from all three masses:
I = 0 + m\,a^2 + \frac{m\,a^2}{4} \;=\; m\,a^2 + \frac{m\,a^2}{4} \;=\; m\,a^2 \Bigl(1 + \frac{1}{4}\Bigr) \;=\; \frac{5}{4}\,m\,a^2.
Step 6: Relate to the Given Form
We are told that the moment of inertia has the form:
I = \frac{N}{20}\,m\,a^2.
Since we found I = \frac{5}{4}\,m\,a^2 = \frac{25}{20}\,m\,a^2 , we match coefficients:
\frac{N}{20} = \frac{25}{20} \quad \Longrightarrow \quad N = 25.
Conclusion
Therefore, the integer N is 25 .
