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There is a thin rod of uniform cross-section of mass M and length L. If this rodis is bent at 90^(@) from the mid-point, then the moment of inertia about the axis passing through mid-point and perpendicular to the plane which includes both parts of rod is ......... |
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Answer» `(ML^(2))/(24)` After the ROD bent at right angle mass of its energy part = `(M)/(2)` Length of every part = `(L)/(2)` Now total moment of inertia, I = sum of moment of inertia of both the part `therefore I=(1)/(3)((M)/(2))((L)/(2))^(2)+(1)/(3)((M)/(2))((L)/(2))^(2)` `=2((1)/(3)((M)/(2))((L)/(2))^(2))` `therefore I=(ML^(2))/(12)` |
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