Three point sized bodies each of mass M are fixed at three corners of light triangular frame of length L about an axist hrough any side of frame the moment of inertia of three bodies is
Ans:3MLsquare/4
Not nice it is not correct answer please check it again
thanx
Wrong answer plz put write ans
To find the moment of inertia of three point-sized bodies around an axis passing through any side of the frame, you can use the parallel axis theorem. The parallel axis theorem states that the moment of inertia of a system of point masses can be found by adding the sum of the individual moments of inertia of the bodies, along with a term that takes into account the distance of each body from the axis of rotation.
In this case, you have three point-sized bodies of mass M fixed at the three corners of a light triangular frame of length L. To calculate the moment of inertia, follow these steps:
1. Determine the moment of inertia of each individual body. Since each body is a point-sized mass, the moment of inertia can be calculated using the formula I = mr^2, where I is the moment of inertia, m is the mass, and r is the distance from the axis of rotation. In this case, the distance from each body to the axis of rotation is L/2 (half the length of the frame).
So, the moment of inertia for each individual body is I = M(L/2)^2 = ML^2/4.
2. Sum up the individual moments of inertia. Since there are three bodies, add up the moment of inertia of each body:
Total moment of inertia = 3 * (ML^2/4) = 3ML^2/4.
Therefore, the answer is 3ML^2/4 for the moment of inertia of the three bodies about an axis passing through any side of the frame.
Not nice
only one mass M is off the axis
How far?
half of triangle is L/2, (L/2)sqrt3 and L
so M is (L/2)sqrt 3 from axis
mass*distance squared
= M [(L/2)sqrt 3 ]^2