Question

At the corners of an equilateral triangle of side L, point masses, each of mass \(2\) kg are placed. The moment of inertia of the system about an axis along any one side is (in kg\(-\)m\({}^{2}\))

• A. \(\frac{3L^{2}}{4}\)
• B. \(\frac{L^{2}}{3}\)
• C. \(\frac{3L^{2}}{2}\)
• D. \(\frac{4L^{2}}{3}\)

Hint:

Masses on the axis contribute nothing; only the third mass matters.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Moment of inertia of point masses is \(I=\sum m r^{2}\), where r is the perpendicular distance from the axis.

Step 2: Distances
Two masses lie on the axis, so their distance is zero. The third mass is at the height of the triangle, \(r=\dfrac{\sqrt{3}}{2}L\).

Step 3: Calculation
\(I=2\times\left(\dfrac{\sqrt{3}L}{2}\right)^{2}=2\times\dfrac{3L^{2}}{4}=\dfrac{3L^{2}}{2}\)

Final Answer:
The moment of inertia about a side is 3L squared over 2. \[ \boxed{\text{(C) }\dfrac{3L^{2}}{2}} \]