A thin rod mass M and length L is bent into a circular ring.the expression for moment of inertia of ring about an axis passing through its diameter is
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According to theorem of perp. axis, the m.i of ring about it,s diameter is equal to the sum of It's m.i about two perp. diameter suppose AB and CD
I.AB + I.CD= I.yy'
2I= MR2
I = MR2/2
I.AB + I.CD= I.yy'
2I= MR2
I = MR2/2
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