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The temperature of uniformrodoflength Lhavinga coefficientof linearexpansionalpha_(L) ischangedby Delta T . Calculatethe newmomentof inertia of the uniformrodaboutaxispassingthroughitscenterandperpendicularto anaxisof the rod. |
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Answer» Solution :MOMENT of inertia of a UNIFORM rod of MASS and length l about its perpendicular bisector. Moment of inertia of the rod `I=(1)/(12)ML^(2)`  Increase in length of the rod when temperature is increased by `DeltaT`, is given by `L.=L(1+alpha_(L)DeltaT)` New moment of inertia of the rod `I.=(ML.^(2))/(12)=(M)/(12)L^(2)(1+alpha_(L)DeltaT)^(2)` `I.=I(1+alpha_(L)DeltaT)^(2)` |
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