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 centre of frame the moment of inertia of three bodies is
Ans : MLsquare
Oh come on, I already did two of these. Now all three masses are L/sqrt 3 from the axis
3 M (L/sqrt3)^2 = 3 ML^2 /3 = M L^2
damon could u plss tell how all three masses are l /sqr 3
To calculate the moment of inertia of the three point-sized bodies fixed at the corners of a light triangular frame, we need to consider the moment of inertia of each point-sized body individually and then sum them up.
The moment of inertia of a point-sized body about an axis perpendicular to its plane and passing through its center is given by the formula:
I = MR^2
Where I is the moment of inertia, M is the mass of the body, and R is the distance of the body from the axis of rotation.
Since all three point-sized bodies have the same mass M, the moment of inertia of each body is MR^2.
Now, let's consider the given configuration. The three point-sized bodies are fixed at the corners of a light triangular frame. We can assume the length of the frame to be L, and the distance of each body from the axis of rotation (which passes through the center of the frame) to be R.
In a triangular frame (equilateral in this case), we can use the height of an equilateral triangle to calculate the distance R. The height of an equilateral triangle is given by the formula:
h = (sqrt(3) / 2) * L
Since the axis of rotation passes through the center of the frame, the distance R is equal to the height of the equilateral triangle divided by 2:
R = (sqrt(3) / 2) * L / 2
= (sqrt(3) / 4) * L
Now, we can calculate the moment of inertia of a single point-sized body:
I = MR^2
= M * ((sqrt(3) / 4) * L)^2
= M * (3 / 16) * L^2
= (3 / 16) * ML^2
Since there are three point-sized bodies, the moment of inertia of the system is:
Total Moment of Inertia = 3 * (3 / 16) * ML^2
= 9 / 16 * ML^2
Therefore, the moment of inertia of the three bodies fixed at the corners of the triangular frame is (9 / 16) ML^2.