Which constraint would you use?

Let $V$ be the speed of the wedge and $v$ be the velocity of the CoM of the rod.

Clearly no external force act in the horizontal direction on the system

So $MV=mv_{x}$ ................(1)

By the conservation of the energy-

$\frac { mgl }{ 2 } (sin(60)-sin(30))=\frac { 1 }{ 2 } M{ V }^{ 2 }+\frac { 1 }{ 2 } m{ v }^{ 2 }+\frac { 1 }{ 2 } { I\omega }^{ 2 }$

Now here comes the main part of the solution-- img

This figure shows the velocity of point A, B ,C relative to earth and relative to point C. C is the CoM of the rod.

So ${ V }_{ A/C }=-\frac { l }{ 2 } \omega sin\theta \hat { i } -\frac { l }{ 2 } \omega cos\theta \hat { j }$

and ${ V }_{ B/C }=\frac { l }{ 2 } \omega cos\theta \hat { j } +\frac { l }{ 2 } \omega sin\theta \hat { i }$

As ${ V }_{ BC }={ V }_{ B }-{ V }_{ C }$

Using this in vertical direction we will get

$\frac { l }{ 2 } \omega cos\theta \hat { j }=0-{ V }_{ C }\hat { j }$

${ V }_{ C }\hat { j }=-\frac { l }{ 2 } \omega cos\theta \hat { j }$

So ${ V }_{ c }=\frac { MV }{ m } \hat { i } -\frac { l }{ 2 } \omega cos\theta \hat { j }$

(got this -- $\frac { MV }{ m } \hat { i }$ from conservation of momentum in X direction)

Also ${ V }_{ AC }={ V }_{ A }-{ V }_{ C }$

So ${ V }_{ A }=\left( \frac { MV }{ m } -\frac { l }{ 2 } \omega sin\theta \right) \hat { i } -l\omega cos\theta \hat { j }$

Also as the the point A of the rod remains in the contact with the wedge so the velocity of the point A and wedge perpendicular to the contact surface should be equal.

Now velocity of the wedge perpendicular to the contact surface is $Vsin(60)$

Now we have to find the velocity of the point A perpendicular to the contact surface. For this we will use dot product approach. For this we have to find the unit vector perpendicular to the contact surface.

fig

From the figure unit vector perpendicular to the contact surface is

$\vec { B } =-\frac { \sqrt { 3 } }{ 2 } \hat { i } -\frac { 1 }{ 2 } \hat { j }$

And $\vec { { V }_{ A } } =\left( \frac { MV }{ m } -\frac { l }{ 2 } \omega sin\theta \right) \hat { i } -l\omega cos\theta \hat { j }$

$\vec { { V }_{ A } } \cdot \vec { B } ={ V }_{ A }Bcos(\alpha )$ where $\alpha$ is the angle between $V_{A}$ and B)

or ${ V }_{ A }cos(\alpha )=\frac { \vec { { V }_{ A } } \cdot \vec { B } }{ B }$

Also ${ V }_{ A }cos(\alpha )$ is the component of the $V_{A}$ in direction perpendicular to the contact surface.

So $Vsin(60)=\left\{ \left( \frac { MV }{ m } -\frac { l }{ 2 } \omega sin\theta \right) \hat { i } -l\omega cos\theta \hat { j } \right\} \cdot \left\{ -\frac { \sqrt { 3 } }{ 2 } \hat { i } -\frac { 1 }{ 2 } \hat { j } \right\}$

$Vsin(60)$ in the LHS is the velocity of the wedge perpendicular to the contact surface.

On putting the values we will get $\frac { -5\sqrt { 3 } V }{ 2 } +\frac { 3\sqrt { 3 } \omega }{ 8 } =\frac { V\sqrt { 3 } }{ 2 }$

or $\omega =8V$

As ${ V }_{ c }=\frac { MV }{ m } \hat { i } -\frac { l }{ 2 } \omega cos\theta \hat { j }$

On putting $\omega =8V$ in this equation we will get

$V_{C}=\sqrt { 37 } V$

On putting this value in the equation that we got from the conservation of the energy we will get

$15\left( \sqrt { 3 } -1 \right) =142{ V }^{ 2 }$

or $V=\sqrt { \frac { 15\left( \sqrt { 3 } -1 \right) }{ 142 } }$

So $a+b=142+15=157$ .

The most difficult part of the question is to find relation between $V$ and $\omega$ .

I used 4 concepts to make 4 equations of 4 variables.

The variables are : vertical velocity of rod , horizontal velocity of rod , angular velocity of rod , horizontal velocity of wedge!

Concepts employed in my method are

1. Conservation of Momentum along horizontal direction .

2. The Right end of rod has no net vertical velocity .

3. The contact points have same velocity along common normal .

4. Energy Conservation! . :)

The problem becomes quite easy if we apply some simple geometry.Note that v=rate of change of displacement.so once we can get these displacements we can differentiate wrt time.these displacements can be easily found geometrically.but its better that we find the displacements in both x and y directions.note that if the wedge has a velocity v then the com of the rod has $v(x)=-5v$ .but then the technique discussed above(of differentiation)yields $(v+3wL/4)=v(x)$ .equating gives $w=-8v/L$ .Now we may use energy conservation KE(of system of wedge and rod)= $(1/2)Iw^2+(1/2)m(5v)^2+(1/2)m(8v)^2+(1/2)Mv^2$ and equating it to the loss of gravitational potential energy yields $v=√15/142(√3 -1)$ .I don't know how to post solutions as above but this way was quite simple.

To arrive at equation 2 I did the following.1)-Let's say that the wedge moves a distance $x$ .The leaning rod at that infant makes an angle $@$ with the horizontal.Let the bottommost end of the rod be at a distance $y$ from the wedge.2)Using geometry we can find the displacement of com in the $X$ direction.This is of the form $X=x-(y-L/2cos@+L/4)$ .where y can be easily computed by using sine law. $y/sin(60-@)=L/sin120$ .So $dX/dt=(dx/dt)-(dy/dt)-(L/2)sin@d@/dt$ .But $d@/dt=w$ .and we have $y$ so we can compute $(dy/dt)$ and $dx/dt$ is nothing but speed of wedge.And from momentum conservation $dX/dt=-5v$ .Now setting in the eq and $@=30$ yields $w=-8v/L$ .So the vertical velocity is $v(*)=8v$