Battle between Rod and Cube

At the instant of separation, Let Cube has Velocity 'V' and rod has angular velocity $\omega$ about the point 'O' . Then Using Energy Conservation (System Rod + Cube+Earth)

$\displaystyle{mgL(1-\sin { \alpha } )=\cfrac { 1 }{ 2 } (m{ L }^{ 2 }){ \omega }^{ 2 }+\cfrac { 1 }{ 2 } m{ v }^{ 2 }\quad .\quad .\quad .(1)}$ .

using angular equation of motion of Rod ; $\displaystyle{N\cos { \alpha } +mg\sin { \alpha } =m{ \omega }^{ 2 }L\quad \quad .\quad .\quad .\quad (2)}$ .

Condition of Sepration (Normal reaction must be zero) $\displaystyle{N=0\quad \quad .\quad .\quad .\quad (3)}$ .

Using Constrained Equation : (Relative Velocities of Point of Contact b/w rigid Bodies along Common Normal Must be zero , Because they are Rigid So their Inter Particle Distance Must not be Changed )

$\displaystyle{v=\omega L\sin { \alpha } \quad .\quad .\quad .\quad (4)}$ .

You Have 4 equations and 4 variable , Eliminate Them . and we should get ;

$\boxed { \cfrac { M }{ m } =\cfrac { 2-3\sin { \alpha } }{ \sin ^{ 3 }{ \alpha } } }$ .

Let $V$ be the velocity of the Cube and $v$ be the velocity of the upper end of the rod at the moment they get separated.

So by conservation of energy

$mgl(1-sin\alpha )=\frac { 1 }{ 2 } \left( M{ V }^{ 2 }+m{ v }^{ 2 } \right)$

As they just loose contact so till then the velocities of points in contact , perpendicular to the contact surface must be equal

so $vsin\alpha=V$

Substituting this in our 1st equation we get

${ V }^{ 2 }=\frac { 2mgl{ sin }^{ 2 }\alpha (1-sin\alpha ) }{ (m+M{ sin }^{ 2 }\alpha ) }$

After the moment m loose contact with the M no force acts on it so

$\frac{dv}{dt}=0$ or $\frac{dv}{d\alpha}\frac{d\alpha}{dt}=0$

Now as $\frac { d{ V } }{ d\alpha }=0$

So $\frac { d{ V }^{ 2 } }{ d\alpha } =0$

So $\frac { d }{ d\alpha } \left\{ \frac { 2mgl{ sin }^{ 2 }\alpha (1-sin\alpha ) }{ (m+M{ sin }^{ 2 }\alpha ) } \right\} =0$

On solving this we get

$\frac { M }{ m } =\frac { 2-3sin(\alpha ) }{ { sin }^{ 3 }\left( \alpha \right) }$