Moment of Inertia
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Consider a horizontal rod like element of thickness dx at a distance x from the base of the triangle Width of that element = 2(l-x) cot30 = 2 3 (l-x)Mass of that element = (4M/ 3 L2 ) 2 3 (l-x) dxMI of this elemental rod dI = (1/12)[(4M/ 3 L2 ) 2 3 (l-x) dx] [2 3 (l-x)]2 I = (8M/ 3 L2 )0 L (l-x)3 dx I = (2/ 3)ML2
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3 (l-x)
3 (l-x) dx
I = (8M/
L
I = (2/







