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Grade 11Mechanics

A rod of uniform cross-section of mass M and length L is hinged about an end to swing freely in a vertical plane.However,its density is non-uniform and varies linearly from hinged end to the free end doubling its value. The moment of inertia of the rod,about the rotation axis passing through the hinge point is

Profile image of samudhbhav prabhu
12 Years agoGrade 11
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1 Answer

Profile image of neeraj agarwal
12 Years ago
here, * As rod doesn`t have uniform density. * We have to integrate it. * I= r^2 dm dm= mass of a section at a distance x from the hinge = ( d+ x/L ) dx I= x^2 (d+ x/L) dx = (x^3/3 + x^4/4L) for 0 to L = L^3/3+ L^3/4 = 7 L^3/12