Rope and rotational mechanics: why does my solution method work? 
i recently come across a physics problem involving rotation and rope with mass attached to it the (1) first way to solve this kind of problem that i come aross is to find the rope tension by using $$mg -T = ma$$  then torque would be equal to moment of inertia x angular acceleration however the second solution (2) of this kind of question i found very interesting is to think of the system as a mass that attached to the rotaional stuff and then think of torque as (moment of inertia of the rotational stuff +moment of inertia of the object using $mr^2$ when $r$ is radius ,not the length to bob) x angular acceleration . 
now my question is why does the second solution work?
 A: Actually 1st solution works.I think so if there is a condition given that there is no slipping then the acceleration of block is equal to r×(alpha) because that is the acceleration of the point at contact of string with that pulley.so now use a=r(alpha) you will get the answer hope it helped you.
A: In method 1 block A moves in a straight line. In method 2 block A is fixed to and rotates with the disk. In both cases you have applied $\tau=I\alpha$. 
Method 2 does not generally give you the correct answer. Your diagram suggests that block A is an extended object which is rigidly attached to the disk at one corner. However, as you have applied method 2, block A is a point particle concentrated at the point of attachment. Your method works only because you have treated block A as a point particle attached at the rim of the disk where the string makes contact. 
If the disk and block A have moments of inertia $I_D$ and $I_A$ about their own centres of mass, and block A is attached rigidly to the disk with its centre of mass at a distance $b$ from the axis of rotation, then using the Parallel Axes Theorem the total moment of inertia of the composite object is $I=I_D+I_A+mb^2$
where $m$ is the mass of block A. Your method assumes that $b=r$, the radius of the disk, and $I_A=0.$ That is, you are assuming that block A is a point particle.   
The method works for a point particle because, when such a particle is rotating in a circle it has the same kinetic energy as when translating in a straight line with the same speed. Your method does not work for an extended object which is rigidly attached to the disk because as it rotates around the centre of the disk an extended object also rotates about its own centre of mass. Unlike the block in method 1, it changes its orientation in space. In addition to its orbital KE is also has spin KE, so it has more KE than when translating at the same speed. Point particles have no spin KE even if they are spinning, because they have zero moment of inertia.
Method 2 would also give the same answer as method 1 if block A is attached to the rim of the disk at its centre of mass in such a way that block A is able to keep the same orientation in space as the disk turns. Then, like a point particle, it would have no spin KE, only orbital KE.
