• # question_answer     A uniform rod of length$l$is free to rotate in a vertical plane about a fixed horizontal axis through point B. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle$\theta$its angular velocity co is given as A)  $\sqrt{\frac{6g}{l}}\sin \theta$                              B)  $\sqrt{\frac{6g}{l}}\sin \frac{\theta }{2}$ C)  $\sqrt{\frac{6g}{l}}\cos \frac{\theta }{2}$                          D)  $\sqrt{\frac{6g}{l}}\cos \theta$

The fall of centre of mass $h=\frac{l}{2}(1-\cos \theta )$ $\therefore$Decrease in potential energy$=mgh$ $=mg\frac{l}{2}(1-cos\theta )$ From law of conservation of energy, KE$o$ rotation = decrease in PE $\frac{1}{2}I{{\omega }^{2}}=\frac{mgl}{2}(1-\cos \theta )$ $\frac{1}{2}\frac{m{{l}^{2}}}{3}{{\omega }^{2}}=\frac{mgl}{2}(1-\cos \theta )$ ${{\omega }^{2}}=\frac{6g}{2l}(1-\cos \theta )$ $=\frac{6g}{2l}2{{\sin }^{2}}\frac{\theta }{2}$ $\omega =\sqrt{\frac{6g}{l}{{\sin }^{2}}\frac{\theta }{2}}$ $=\sqrt{\frac{6g}{l}}\sin \frac{\theta }{2}$