Moment of inertia about the vertex of a triangle of mass m and height h
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Let us assume the mass per unit area of the triangle be M/12ah.where a is the length of the base of the triangle.So moment of inertia I about the base of the triangle
I =1/2. M/1/2ah12ax . x dx,where x is the small strip along the the direction of h from the base..Therefore Moment of inertia about the base of the given triangle is,
I = 1/2.M/h.∫h0x2dx=M/2h.[x³/3]°h=M/2h×h²/3=mh²/6
The answer is 1/6mh².
I =1/2. M/1/2ah12ax . x dx,where x is the small strip along the the direction of h from the base..Therefore Moment of inertia about the base of the given triangle is,
I = 1/2.M/h.∫h0x2dx=M/2h.[x³/3]°h=M/2h×h²/3=mh²/6
The answer is 1/6mh².
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