A) 2 \[\omega \alpha t\]
B) \[\omega \alpha t\]
C) \[\frac{3}{2}\,\omega \alpha t\]
D) \[\frac{\omega \alpha t}{2}\]
Correct Answer: A
Solution :
Since, the moment of inertia of thin rod about an axis passing through its centre of mass and perpendicular to its geometrical axis is \[{{I}_{CM}}=\frac{M{{L}^{2}}}{12}\] M.I. of rod about an axis perpendicular to its geometrical axis and passes through one of the ends \[{{I}_{1}}={{I}_{CM}}+M{{(L/2)}^{2}}\] Theorem of parallel axis \[{{I}_{1}}=\frac{M{{L}^{2}}}{12}+\frac{M{{L}^{2}}}{4}\] \[{{I}_{1}}=\frac{M{{L}^{2}}}{3}\]
If initial angular speed of rod \[=\omega \] When room temperature increases by t and coefficient of linear expansion of rod is \[\alpha \] then New M.I. of rod \[{{I}_{2}}=\frac{M}{3}{{(L+l)}^{2}}\]where \[l=L\times \alpha \times t\] (increment in length) No external torque is acting so angular speed will decrease. Let now angular speed \[=(\omega -\Delta \omega )={{\omega }_{2}}\] (let) According to conservation of angular momentum (Since, torque \[\tau =0\]) \[{{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}\] \[\frac{M{{L}^{2}}}{3}\omega =\frac{M}{3}{{(L+l)}^{2}}.(\omega -\Delta \omega )\] \[{{L}^{2}}\omega ={{(L+L.\alpha .t)}^{2}}(\omega .-\Delta \omega )\] \[{{L}^{2}}\omega ={{L}^{2}}{{(1+\alpha .t)}^{2}}(\omega -\Delta \omega )\] or \[\frac{\omega -\Delta \omega }{\omega }=\frac{1}{{{(1+\alpha t)}^{2}}}\] \[1-\frac{\Delta \omega }{\omega }={{(1+\alpha t)}^{-2}}\] Using binomial theorem \[{{(1+x)}^{-n}}\] \[=1-nx+....\] (leaving higher power terms) \[1-\frac{\Delta \omega }{\omega }=1-2\alpha t\] \[\frac{\Delta \omega }{\omega }=2\alpha t\] \[\Delta \omega =2\alpha t\omega \] \[\therefore \] Change in angular speed \[=2\alpha t\omega \]
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