Question

A uniform cylindrical rod of mass M and length L is rotating with an angular speed $ \omega $ . The axis of rotation is perpendicular to its axis of symmetry and passes through one of its end faces. If the room temperature increases by t and the coefficient of linear expansion of the rod is $ \alpha $ , the magnitude of the change in its angular speed is:

• A. 2 $ \omega \alpha t $
• B. $ \omega \alpha t $
• C. $ \frac{3}{2}\,\omega \alpha t $
• D. $ \frac{\omega \alpha t}{2} $

Answer 1

The Correct Option is A

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 $