Rotational Dynamics Problem-Rod slipping against block

Question

A uniform rod of mass m and length l is pivoted at point O. The rod is initially in vertical position and touching a block of mass M which is at rest on a horizontal surface. The rod is given a slight jerk and it starts rotating about point O. This causes the block to move forward. The rod loses contact with the block at θ = 30°. All surfaces are smooth.

1. The value of M/m?
2. The velocity of block when the rod loses contact with the block?
3. The acceleration of center of mass of rod, when it loses contact with the block?
4. The hinge reaction at O on the rod when it loses contact with the block?

Here is my attempt at the problem

The component of angular speed of the tip of the rod in horizontal direction is same as that of the block .

Let V be the speed of the block when rod loses contact with block. So,Horizontal component of angular speed of the tip of the rod = ωlcos60°=V

So,ωl/2=V Thus,ω = 2V/l

Applying Conservation of Energy ,

Loss in Potential Energy of Rod = Gain in Rotational Kinetic Energy of Rod + Gain in KE of Block

$\frac{Mgl}{2}(1-Sin60°) = \frac{1}{2}\frac{ml^2}{3}(ω ^2) +\frac{1}{2}MV^2 $

$\frac{Mgl}{2}(1-Sin60°) = \frac{1}{2}\frac{ml^2}{3}\frac{4V ^2}{l^2} +\frac{1}{2}MV^2 $

$\frac{Mgl}{4} = \frac{2mV^2}{3} +\frac{1}{2}MV^2 $

The forces on the rod are

Hinge force in x direction = $ H_{x} $ Hinge force in y direction = $ H_{y} $ Force due to gravity = mg Normal contact force from the block = N

The force on the block are

Normal contact force from the rod = N

Now, I am not able to proceed further .Kindly help me with the problem.