Three point sized bodies each of mass M are fixed at three corners of light triangular frame . In the above problem about an axis perpendicular to the plane of frame and passing through a corner of frame the moment of inertia of three bodiea is ??
simple. EAch point will give 1/2 M distance^2. So one point will have zero distance(the axis is thru that point), so the other two will give 1/2 M (s1^2 + s2^2) where s1 and s2 are the distances form each M to the axis.
To find the moment of inertia of the three bodies about an axis perpendicular to the plane of the frame and passing through a corner of the frame, we can use the parallel axis theorem.
According to the parallel axis theorem, the moment of inertia about an axis parallel to an axis passing through the center of mass is given by the sum of the moment of inertia about the center of mass and the product of the mass and the square of the distance between the two axes.
In this case, since the bodies are point-sized and have the same mass, the moment of inertia of each body about its own center of mass is zero.
Therefore, we only need to consider the contribution from the distance between the center of mass and the corner of the frame.
Let's assume the side length of the triangular frame is 'a'. Then, the distance between the center of mass of each body and the corner of the frame is 'a/√3' (since it forms an equilateral triangle).
Now, the moment of inertia of each body about the given axis is given by:
I = m * (a/√3)^2
Since we have three bodies, the total moment of inertia is:
Total I = 3 * (m * (a/√3)^2)
Simplifying further:
Total I = m * a^2 / 3
Therefore, the moment of inertia of the three bodies about an axis perpendicular to the plane of the frame and passing through a corner of the frame is (m * a^2) / 3.