In a no-slip ideal Atwood's machine (string mass is negligible), does tension or $mg$ affect the torque?

Question

The conditions in this case are: string mass is negligible; string DOES NOT slip on pulley; the pulley's bearings are assumed to be frictionless. I understand that tension is the force that creates the torque when the string does not slip on the pulley. Although it is a fact that tension is not equal mg, my textbook claims that mg is the force that creates torque instead of the tension. From my understanding, the torques should be torque(g1)=Tension1*R and Torque(g2)=Tension2*R where Tension1 is not equal to Tension2. What is the correct interpretation in this case? Here is the whole context of the problem for those interested:

Answer 1

You you and the textbook are both correct.

The textbook is taking an alternative approach which considers only external forces and ignores the tensions in the string, which are internal forces. In this approach the linear momentum of the masses is counted as orbital angular momentum, so that the whole problem can be tackled using the rotational equation of motion $\tau=I\alpha$, instead of splitting the problem into separate but connected rotational and translational motions.

(I notice that the heading is "Atwood's Machine Revisited" - which suggests to me that (i) the solution using the tensions in the string was done previously , and (ii) this chapter deals with the equation torque = rate of change of angular momentum.)

The tensions in the strings depends not only on the masses but also their common acceleration $a$.

$T_1=m_1(g-a)$
$T_2=m_2(g+a)$
The torque on the disk is
$(T_1-T_2)R=[(m_1-m_2)g-(m_1+m_2)a]R$.

However, the torque on the whole system is $(m_1-m_2)gR$. This torque acts on the masses, increasing their orbital angular momentum, as well as increasing the spin angular momentum of the disk.