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

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:

• Is the pulley massless? If not, it will have a moment of inertia. The solution method will differ a bit, depending on whether the pulley has mass or not. Commented Feb 19, 2017 at 23:58
• Please give details of the textbook. Commented Feb 19, 2017 at 23:58
• sorry for not clarrifying, but the pulley has mass Commented Feb 19, 2017 at 23:58
• the textbook is called "Physics for Scientists and Engineers (6th Edition)" Commented Feb 19, 2017 at 23:59
• Who are the Authors? Commented Feb 20, 2017 at 0:00

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.

• so the textbook is low-key wrong? Commented Feb 19, 2017 at 23:58
• "low-key" is just a little colloquial slang popular to the younger generations. Commented Feb 20, 2017 at 0:03
• I have added the context of the problem as requested. Hopefully this sheds more light on the textbook's error? Commented Feb 20, 2017 at 0:18
• Oh so in essence, my interpretation is only looking at the torque of the disk. Commented Feb 20, 2017 at 1:15
• Yes, you are only looking at torque on the disk. Tipler is looking at torque on the whole system, including the hanging masses. Commented Feb 20, 2017 at 1:49