Find out the increases in moment of inertia I of a uniform rod (coefficient of linear expansion alpha) about its perpendicular bisector when its temperature is slightly increased by DeltaT.

Find out the increases in moment of inertia I of a uniform rod (coeffi

1.	Find out the increases in moment of inertia I of a uniform rod (coefficient of linear expansion alpha) about its perpendicular bisector when its temperature is slightly increased by DeltaT.
Answer» SOLUTION :Suppose mass of rod is M and moment of INERTIA is I. Moment of inertia of rod about perpendicular bisector, `I=(Ml^(2))/(l^(2))` Increase in LENGTH due to increase in temperature `DeltaT`, `Deltal=l alphaDeltaT` . . . (1) `:.` Now, new moment of inertia of rod, `I.=(M)/(12)(l+Deltal)^(2)` `=(M)/(12)(l^(2)+2l+Deltal^(2))` But, `Deltal` is very SMALL, `Deltal^(2)` is neglected. `I.=(M)/(12)(l^(2)+2lDeltal)` `=(Ml^(2))/(12)+(MlDeltal)/(6)` `=I+(MlDeltal)/(6)` Increase in moment of inertia, `I.-I=(MlDeltal)/(6)` `:.DeltaI=2xx(Ml^(2))/(12)*(Deltal)/(l)""[because" MULTIPLYING and dividing by 2"]` `:.DeltaI=2Ialpha DeltaT""[because(Deltal)/(l)=alphaDeltaT]`