Why can't we use conservation of angular momentum in this question? A rod of negligible mass and length is pivoted at its centre. A particle of mass $m$ is fixed to its left end and another particle of mass $2 m$ is fixed to the right end. If the system is released from rest and after some time becomes vertical, the speed $v$ of the two masses and angular velocity at that instant.
My initial angular momentum is zero and when writing the final "$w$" is coming $0$ how is this possible?
$$I \left( 0 \right) + I \left( 0 \right) = I_{w} + I_{w}$$
And why we are unable to use equation of motion by finding $\alpha$ and then acceleration with displacement $x = \frac{l}{\sqrt{2}}$
$$ V^{2} = \frac{2 g}{3} \frac{l}{\sqrt{2}}$$
 A: As Marko points out, angular momentum is not conserved when there is an external torque (from gravity).  If there is no friction you can find the angular velocity (at any angle from the release point) using conservation of energy.  Since the angular acceleration is not constant, using that gets complicated.
A: 
Why can't we use conservation of angular momentum in this question?

Agular momentum is conserved only when net torque due to external forces is zero. In your example there are three external forces acting on the system:

*

*downward gravitational force on mass $m$ at distance $r$ from the pivot produces $\tau_1 = m g r$

*downward gravitational force on mass $2m$ at distance $r$ from the pivot produces $\tau_2 = 2mgr$

*upward normal force on the rod at distance $0$; does not produce torque about the pivot $\tau_0 = 0$
Taking positive rotation direction to be clockwise, force on mass $m$ rotates in negative direction and force on mass $2m$ rotates in positive direction. The net torque on the system is
$$\tau_\text{net} = -\tau_1 + \tau_2 = +mgr$$
This tells us that the system of particles will start rotating in the positive direction (clockwise). Angular momentum is not conserved because there is some net torque $\tau_\text{net}$ due to external forces acting on the system.


And why we are unable to use equation of motion by finding $\alpha$ and then acceleration with displacement $x = \frac{l}{\sqrt{2}}$

The moment the rod is no longer horizontal, the gravitational forces on two particles are no longer perpendicular to the distance from the pivot and the torque is not simply $mgr$ and $2mgr$ but
$$\tau_1 = +mgr \sin\theta, \qquad \tau_2 = -2mgr \sin\theta$$
where $\theta$ is angle from the equilibrium position (for $\theta = 0$ the particle $2m$ is at vertical down position), and the positive rotation direction is clockwise. The moment of inertia of the system about the pivot is $I = 3mr^2$ and the equation of motion is
$$I \alpha = \tau_\text{net}, \qquad \boxed{\ddot\theta = -\frac{g}{3r} \sin\theta}$$
The above equation of motion actually equals that of a simple pendulum. However, we cannot use the small-angle approximation $\sin\theta \approx \theta$ for your example because the system starts at $\theta = -\frac{\pi}{2}$. The (nonlinear) equation of motion of the form $\ddot\theta = -\kappa \theta$ does not have a closed-form solution and can be only solved numerically.
