Question

Three particles of each mass \( m \) are kept at the three vertices of an equilateral triangle of side \( 1 \). The moment of inertia of the system of the particles about any side of the triangle is:

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

Hint:

When calculating the moment of inertia for a system of particles, use the parallel axis theorem and ensure you account for the correct distance of each particle from the axis of rotation.

Answer 1

The Correct Option is C

We are given that three particles of mass \( m \) are placed at the vertices of an equilateral triangle with side length \( l \). We are asked to find the moment of inertia of the system of particles about any side of the triangle. 
Step 1: Moment of inertia formula. The moment of inertia of a system of point masses is given by: \[ I = \sum m_i r_i^2, \] where \( m_i \) is the mass of the \( i \)-th particle and \( r_i \) is the perpendicular distance from the axis of rotation. 
Step 2: Geometry of the problem. The particles are placed at the vertices of an equilateral triangle. The distance from each particle to the axis (side of the triangle) is \( \frac{\sqrt{3}}{2} l \). 
Step 3: Moment of inertia calculation. The moment of inertia of the two particles not on the axis is: \[ I = 2 \cdot m \cdot \left( \frac{\sqrt{3}}{2} l \right)^2 = 2 \cdot m \cdot \frac{3}{4} l^2 = \frac{3}{2} m l^2. \] 
Step 4: Final answer. Thus, the moment of inertia is \( \frac{3}{4} m l^2 \), corresponding to Option 3.

Answer 2

Given:
Three particles each of mass \( m \) are placed at the vertices of an equilateral triangle with side length \( l = 1 \).

We need to find the moment of inertia \( I \) of the system about any side of the triangle.

Step 1: Moment of inertia about an axis passing through one side of the triangle involves calculating the perpendicular distances of each mass from that axis.

Step 2: Consider the equilateral triangle \( ABC \) with side length \( l \).
Choose the side \( AB \) as the axis of rotation.

Step 3: The particles at vertices \( A \) and \( B \) lie on the axis, so their perpendicular distance is zero and they do not contribute to the moment of inertia.

Step 4: The particle at vertex \( C \) is at a perpendicular distance equal to the height \( h \) of the equilateral triangle from side \( AB \):
\[ h = \frac{\sqrt{3}}{2} l \]

Step 5: Moment of inertia about side \( AB \):
\[ I = m \times h^2 = m \times \left(\frac{\sqrt{3}}{2} l\right)^2 = m \times \frac{3}{4} l^2 = \frac{3}{4} m l^2 \]

Therefore, the moment of inertia about any side of the equilateral triangle is:
\[ \boxed{\frac{3}{4} m l^2} \]