question_answer 40)

                  Find out the increase 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 \[\Delta T\].                

Answer:

                  Moment of inertia of a uniform rod about its perpendicular bisector,                 \[l=\frac{M{{L}^{2}}}{12}.\]                 When rod is heated, its length becomes                 \[L+\Delta L,\] where \[\Delta L=L\alpha \Delta T\]                 \[\therefore \] New moment of inertia, \[I'=\,\frac{M}{12}\,{{(L+\Delta L)}^{2}}\]                 \[=\,\frac{M}{12}\,({{L}^{2}}+\Delta {{L}^{2}}+2L\,\Delta L)\]                 Since \[\Delta L\] is small, so neglect \[\Delta {{L}^{2}}\]                 \[\therefore \] \[I'=\,\frac{M{{L}^{2}}}{12}\,+\frac{M}{6}\,L\,\Delta L=\,I+\,\frac{ML}{6}\,\Delta L\]                 \[=\,I+\,\frac{ML}{6}\,\times \,L\,\alpha \,\Delta T=\,I+\,\frac{M{{L}^{2}}}{6}\,\alpha \,\Delta T\]                 \[\therefore \] Increase in M.I. \[=\,I'\,-\,I=\,\frac{M{{L}^{2}}}{6}\alpha \,\Delta T\]                 \[=\,2\,\left( \frac{M{{L}^{2}}}{12} \right)\,\alpha \,\Delta T=\,2I\,\alpha \,\Delta T\]