• # question_answer 22) A uniform rod of length 2L is placed with one end in contact with the horizontal and is then inclined at an angle a to the horizontal and allowed to fall without slipping at contact point When it becomes horizontal, its angular velocity will be A) $\omega =\sqrt{\frac{3g\sin \alpha }{2L}}$B) $\omega =\sqrt{\frac{2L}{3g\,\sin \alpha }}$C)      $\omega =\sqrt{\frac{6g\sin \alpha }{L}}$D)      $\omega =\sqrt{\frac{L}{g\sin \alpha }}$

By the law of conservation of energy, P.E. of rod = rotational K.E. $mg\frac{l}{2}\sin \propto \frac{1}{2}{{\omega }^{2}}=\frac{1}{2}.\frac{m{{l}^{2}}}{3}{{\omega }^{2}}$ or         $\omega =\sqrt{\frac{3g\sin \propto }{l}}$ Putting $1=2L,\,\omega =\sqrt{\frac{3g\sin \propto }{2L}}$ Hence, the correction option is (a).