A uniform rod of length l is free to rotate in a vertical plane about a fixed horizontal axis through 0. The rod is allowed to rotate from its unstable vertical position. Then the angular velocity of the rod when it has turned through an angle $\theta$ is

Question

 

\[\begin {array} {1 1} (a)\;\sqrt {\frac{3g}{l}}. \sin \theta/2 & \quad (b)\;\sqrt {\frac{6g}{l}}. \sin \theta/2 \\ (c)\;\sqrt {\frac{3g}{l}}. \cos \theta/2 & \quad  (d)\;\sqrt {\frac{6g}{l}}. \cos \theta/2 \end {array}\]

Comments

I dint understand how did you write that height mgl/2(1-cos@) plzz explain by diagram .

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I also not understand it