Resulting motion of blocks attached to a pulley with mass 
two blocks of mass m and 3m are suspended by a massless-stretchless cord passing over a frictionless pulley of mass 2m in the shape of a uniform circular disk of radius R (as shown in the figure). The cord does not slip on the pulley.


The problem then asks to derive an expression for the speed $v_f$ of the $3m$ block after it has fallen from rest to a distance $h$ using only $m, g, R$ and $h$. 
I started by finding the sum of the forces acting on the pulley, $$F= 3mg - mg = 2mg$$ Then I found the moment of inertia of the pulley $$I=mr^2 = 2mR^2$$ Then I used this information to find the angular acceleration of the pulley $$\alpha = \frac{\tau}{I} = \frac{F\cdot R}{I} = \frac{2mgR}{2mR^2} = \frac{g}{R}$$
I converted the angular acceleration into linear acceleration
$$\alpha  = \frac{a}{r} \rightarrow a = \alpha r = \frac{g}{R}R = g$$
The linear acceleration of the pulley is equal to the acceleration of the block. So I finished with a kinematics equation to solve for $v_f$
$$v_f^2 = v_0^2 + 2ad \rightarrow v_f = \sqrt{2gh}$$
but my answer is incorrect and it does not tell me the correct answer. Can someone show me where I messed up?
 A: Using free-body diagrams and Newton's Laws one would proceed as follows:


*

*Do not assume that the forces acting on the pulley are the weights. The forces pulling on the pulley are the tensions in the ropes which must be different from the weights if the weights are accelerating.

*If the pulley rotates and has significant mass and size, the tensions on each side must be different.

*Choosing the positive direction of motion as the $3m$ mass falling we get
$$3mg-T_3 = 3 ma$$
$$T_1 - mg = ma$$
$$T_3R - T_1R = I\frac{a}{R}$$

*Solve this for $a$ and find the acceleration from the constant acceleration relation (be sure to pay attention to which direction is positive for the 3m mass)
$$V_f^2-V_i^2=2a(y_f-y_i).$$

A: This problem is best solved by Conservation of Energy because there's no friction (a non-conservative force) involved. As the pulley's centre of gravity doesn't change position and the cord is inelastic, the change in potential energy $U$ is:
$$\Delta U=3mgh-mgh=2mgh$$
This is converted to kinetic energy $K$, and that change is:
$$\Delta K=\frac12 (3m)v^2+\frac12 mv^2+\frac12 I\omega^2$$
where $I$ is the inertial moment of the pulley and $\omega$ its angular velocity.
Without slippage we have:
$$v=\omega R\implies \omega=\frac{v}{R}$$
Insert:
$$\Delta K=2mv^2+\frac12I\Big(\frac{v}{R}\Big)^2$$
Conservation of energy tells us:
$$\Delta K=\Delta U$$
So:
$$2mgh=2mv^2+\frac12I\Big(\frac{v}{R}\Big)^2$$
Calculate $v$ from there with the correct value of $I$.
