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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

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

Grade:11

1 Answers

neeraj agarwal
34 Points
8 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

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