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)  \[\sqrt{\frac{6g}{l}}\cos \theta \]

Correct Answer: B

Solution :

                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 \[\frac{1}{2}I{{\omega }^{2}}=\frac{mgl}{2}(1-\cos \theta )\] \[\frac{1}{2}\frac{m{{l}^{2}}}{3}{{\omega }^{2}}=\frac{mgl}{2}(1-\cos \theta )\] \[{{\omega }^{2}}=\frac{6g}{2l}(1-\cos \theta )\] \[=\frac{6g}{2l}2{{\sin }^{2}}\frac{\theta }{2}\] \[\omega =\sqrt{\frac{6g}{l}{{\sin }^{2}}\frac{\theta }{2}}\] \[=\sqrt{\frac{6g}{l}}\sin \frac{\theta }{2}\]