Rotational Motion At Its Best !

We will Do This Question By ICR Instantaneous Centre of Rotation

Since There is No Horizontal Force on Rod So Centre of mass of Rod will fall vertically Downward only (Does not have any Horizontal Velocity) So We Locate ICR of This Rod By Taking Intersection Point of perpendicular Lines drawn two two velocity as Shown in figure ! And In ICR frame Rod is in Pure Rotational Motion So we can use Energy Conservation in ICR frame!

$\cfrac { mgL }{ 2 } (1-\sin { \theta } )\quad =\quad \cfrac { 1 }{ 2 } { I }_{ icr }{ { \omega } }^{ 2 }\\ \cfrac { mgL }{ 2 } (1-\sin { \theta } )\quad =\quad \cfrac { 1 }{ 2 } m(\cfrac { { L }^{ 2 } }{ 12 } +\cfrac { { L }^{ 2 } }{ 4 } \cos ^{ 2 }{ \theta } ){ { \omega } }^{ 2 }\\ \\ \omega =\sqrt { \cfrac { 12g(1-\sin { \theta } ) }{ L(1+3\cos ^{ 2 }{ \theta } ) } } \\ { V }_{ cm }=\quad \omega \cfrac { L }{ 2 } \cos { \theta } \\ \boxed { { V }_{ cm }=\quad \sqrt { \cfrac { 3gL(1-\sin { \theta } )\cos ^{ 2 }{ \theta } }{ (1+3\cos ^{ 2 }{ \theta } ) } } }$ .

Q.E.D

We can get a equation from the conservation of energy.

$\frac{mgl(1-sin\theta)}{2}=\frac{mv^{2}}{2} + \frac{I \omega^{2}}{2}$

Here green arrow shows the velocity of the lower end of the rod due to the rotation.

Velocity of the lower end of the rod in vertical direction is zero .

So $V_{com}=\frac{l \omega cos\theta}{2}$

Using these two equations we get

$V=\sqrt { \frac { 3{ cos }^{ 2 }\theta (1-sin\theta )gl }{ 3{ cos }^{ 2 }\theta +1 } }$