Three point sized bodies each of mass M fixed at three corners of light triangular frame of side length L about an axis perpendicular to the plane of frame and passing through centre of frame the moment of inertia of three bodies is
What is 3mL^2
To find the moment of inertia of the three bodies, we need to know the distribution of the masses within each body. Assuming each body is a point mass and considering the triangular frame as an equilateral triangle, we can calculate the total moment of inertia.
The moment of inertia of a point mass rotating about an axis passing through its center of mass is given by the formula:
I = m * r^2
Where:
I = moment of inertia
m = mass of the body
r = distance of the body from the axis of rotation
Since all three bodies are point-sized and fixed at the corners of the equilateral triangle, the distance of each body from the axis of rotation is L/√3 (dividing the triangle into two right-angled triangles).
The total moment of inertia of all three bodies can be calculated as the sum of the individual moment of inertia for each body:
I_total = I_1 + I_2 + I_3
I_total = (m * (L/√3)^2) + (m * (L/√3)^2) + (m * (L/√3)^2)
Simplifying the equation:
I_total = 3 * m * (L^2/3)
Therefore, the moment of inertia of three bodies fixed at the corners of a triangular frame of side length L is given by:
I_total = m * L^2/3