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A rod of mass M kg and length L m is bent in the form of an equilateral triangle as shown. The MI of traingle about a vertical axis perpendicular to the plane of triangle and passing through the centre is?
Let a be the side of the equilateral triangle.
Then, L = 3a or a = L/3
thus, M is the mass of the rod the then the three sides of the triangle will be of ''M/3 mass.
and also, by construction, O is the centre or Centroid of the triangle. and Centroid is the points of intersectin of all the medians of the triangle. therefore,
tan30'' = 1/root3 = 6p/l or p = l/6root3 where p is the perpendicular drawn on side AB from point O.
Moment of Inertia of traingle about a vertical axis perpendicular to the plane of triangle and passing through the centre is :
I = [M/3 x L/6root3]^2 x 3 (other three rods of same dimension.]
I = Ml^2/108
is the ansr.
Plz Approve!
Is it M.L^2 / 2???
ans is ML^2/54..
Sorry some calc mistake i am getting ML^2 / 18
I dont know then I checked twice but getting the same anr.
ml^2/108...
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