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?
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in an equilateral triangle the radius of circumcircle passing through the three vertices is given by :-.
so M.I of the three bodies is
so M.I of the three bodies is
rexus:
sry if you know this you can finish this
sry I can't understand this could you send me clearly
the formula i have given you will help you find the perpendicular distance of the three identical masses from the axis taken through the middle of the centre. now the M.I of the three masses will be (3* mr^2) as in this case three masses are identical.
kk thank u
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Answer:
ml^2
Explanation:
when we draw three medians of the triangle we get the center. the line is passing through the center therefore divides the equilateral triangle perpendicularly. Sum of angles in equilateral triangle is 180 therefore each angle is 60 degree. If you draw the diagram we get a right angle triangle at the base. One angle is 90 and the other is 30(60÷2). therefore cos 30 = l÷2/r(radius)
r = l÷2/root 3/2
2,2 gets canceled
r = l/ root3
substitute r value in formula
mr^2
= 3m×l/ root3 whole square
3,3 gets canceled
ml^2 is the answer
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