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

Answer: (A)

2 $\omega \alpha t$

Answer 2

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 $=2\alpha t\omega$