Three point sized bodies each of mass M are fixed at three corners of light triangular frame of length L about an axis perpendicular to the plane of frame passing through CORNER of frame the moment of inertia of three bodies is
Ans:2MLsquare
LOL - You just moved the axis over so you now have two masses at (L/2)sqrt 3 from the axis
To calculate the moment of inertia of three point-sized bodies around an axis perpendicular to the plane of the frame passing through one corner of the frame, we can use the parallel-axis theorem.
According to the parallel-axis theorem, the moment of inertia of a system of particles about an axis is equal to the sum of the moment of inertia of the particles about their center of mass plus the total mass of the system multiplied by the square of the distance between the center of mass and the axis of rotation.
In this case, since the three bodies are fixed at three corners of a light triangular frame, we can consider the center of mass of each body to be located at its respective corner. Therefore, the distance between the center of mass and the axis of rotation is the length of the frame, L.
The moment of inertia of a point mass M about an axis passing through its center of mass is given by MR^2, where R is the distance between the mass and the axis of rotation. In this case, since the masses are point-sized, R is equal to zero.
Therefore, the moment of inertia of each body is MR^2 = M(0)^2 = 0.
Using the parallel-axis theorem, the total moment of inertia of the three bodies about the axis passing through the corner of the frame is given by:
I_total = I_1 + I_2 + I_3 + M_total * L^2
Since the moment of inertia of each body is zero, we can rewrite the equation as:
I_total = 0 + 0 + 0 + 3M * L^2 = 3ML^2
Therefore, the moment of inertia of the three bodies is 3ML^2.
To calculate the moment of inertia of three point-sized bodies fixed at the corners of a light triangular frame, we can use the parallel axis theorem.
The moment of inertia of a single point-sized body of mass M about an axis passing through its corner can be calculated as MR^2, where R is the distance from the corner to the center of mass of the body. In this case, since the bodies are located at the corners of an equilateral triangle, each side of the triangle will be L.
Now, we need to find the moment of inertia about an axis perpendicular to the plane of the frame passing through a corner of the frame. We will consider the axis passing through one of the corners of the frame as the reference axis.
Let's consider one of the point-sized bodies located at a corner. The distance from the corner to the center of mass of this body is L/√3 (due to the equilateral triangle). So, the moment of inertia of this body about the reference axis is M(L/√3)^2 = ML^2/3.
Since all three bodies are identical in mass and shape, the moment of inertia for each body will be the same. Therefore, the total moment of inertia of the three bodies about the reference axis can be calculated as:
Total moment of inertia = 3 × (ML^2/3) = 2ML^2
Hence, the moment of inertia of the three bodies fixed at the corners of the light triangular frame about an axis perpendicular to the plane of the frame passing through a corner is 2ML^2.