Reason for body attached to a string being in free fall? This is a question I found in a book: 

A string is wrapped around a uniform cylinder as shown in diagram. When cylinder is released string unwraps without any slipping and the cylinder comes down. 

I assumed that an equation $T - mg = ma$ could be formulated and the tension does negative work. However the answer is that the tension does zero work. This I understand is because the cylinder is in free fall, and the equation will be $T - mg = mg$ and therefore $T = 0$. 
Is my assumption correct? If it is, why is this so? It doesn't make sense to me. Should the tension exert some upward force and the downward acceleration be at least a bit less than $g$?
 A: Your equation $T-mg=ma$ seems right to me. The reason the tension does not do any work is not because $T=0$, but rather because the point where $T$ acts does not move.
This is simply because of the "no slipping" condition : the point of contact of the cylinder has speed 0. Hence, the work produced by $T$ is simply $W =\vec{T}\cdot d\vec{x}$ where $d\vec{x}$ is the instantaneous displacement of the contact point. However, at any given instant, $\frac{d\vec{x}}{dt} = 0$ because of the no slip condition, hence $d\vec{x}=\vec{0}$, and so $W=0$.
This is a bit counterintuitive at first, because obviously the contact point does move along the string. However, at any given time the speed of the contact point is 0. In other words, the cylinder is not moving with respect to the string.
A: The cylinder is not in free fall. It rolls down the string without slipping. 
A cylinder which rolls without slipping down an inclined plane accelerates less than an identical cylinder which slides without rolling. In the same way the cylinder in your question accelerates less than an identical cylinder which slides down the string without rotating. 
See eg https://www.youtube.com/watch?v=tMdS8Ttx0XA.
