Save the rod

Initially the rod rotates about the edge. Using principle of conservation of energy , $0+ 0 = \dfrac{1}{2} I \omega^2 + \Bigg( - M g \dfrac{a}{3} \sin \theta \Bigg) .....(1)$ Where , $I = \dfrac{M}{12} (2a) ^{2} + M(^a/_{3})^2$ After putting the value of $\displaystyle{I}$ in $\displaystyle{(1)}$ ,we get $\omega^2 = \dfrac{3 g \sin \theta}{2 a}...................(2)$ Diff. wrt. $\displaystyle{\theta}$ ,we get, $2 \omega \dfrac{d \omega}{d \theta} = \dfrac{3g}{2a} \cos \theta$ or $\omega \dfrac{d \omega}{d \theta} = \dfrac{3g}{4a} \cos \theta .......(3)$ Since $\alpha = \dfrac{d \omega}{ dt } = \dfrac{d \omega}{d \theta} \times \dfrac{d\theta}{dt} = \omega \dfrac{d \omega}{d \theta}$ Equations for motion are $M g \cos \theta - N = M \alpha (^a/_3) = \dfrac{M a \omega }{3} \times \dfrac{d \omega}{d \theta}$ Using $\displaystyle{(3)}$ we get $N = \dfrac{3}{4} \times M g \cos \theta ......................(4)$ Also $\mu N - Mg \sin \theta = M(^a/_3) \omega^2$ $\Rightarrow \mu N - Mg \sin \theta = \dfrac{Ma}{3} \Bigg( \dfrac{3g \sin \theta}{2a}\Bigg) = \dfrac{Mg \sin \theta}{2}$ Hence using equation $\displaystyle{(4)}$ we get $\boxed{\mu = \dfrac{2}{1} \tan \theta }$