Force/Torque done on Pulley Clarification

If we have an Atwood machine that has two masses, $M_1$ and $M_2$, where $M_2$ is laying flat on a table and $M_1$ is suspended by a rope connected through a pully to $M1$. The pulley can rotate, and for the sake of my question, there is no internal friction in the pulley. The table is frictionless as well.

Here is a quick diagram:

I tried to attempt to find all the forces, but what stumped me is what would be the effect of $M_1$ to the pulley. If we release the system, then $M_2$ will provide a force of gravity and thus create a torque on the pully. Because surface of the table is frictionless, there doesn't seem to be any force done by $M_1$ because of Newton's 3rd law. If the pulley was a point particle that had no effect on the system (and so no torque), I can easily see that $M_1$ moves because of $M_2$. When we go back to our original question at hand, even if $M_2$ moves $M_1$, how would $M_1$ end up affecting the pulley. I'm quite lost and I'm assuming this goes against my intuition of real life- that is with friction.

• Is the pulley massless? – garyp Aug 20 '16 at 2:32
• I suppose it has a mass of M. Are you trying to see if it has a moment of inertia? – Ian Limarta Aug 20 '16 at 2:33
• I'm trying to understand what torque concerns you. What do you mean when you say that there does not seem to be any force due to $M_1$ because of the third law? I'll say this: $M_1$ does not affect the pulley; it is not in contact with the pulley. The rope is in contact with the pulley. I have to say that I don't understand what's troubling you. – garyp Aug 20 '16 at 2:50

If the pulley did have a significant mass then the PE provided by the falling mass $M2$ would become translational KE of 2 blocks and rotational KE of the pulley.
However, unless the pulley is stated to have a mass $M$ and to be in a particular shape (eg disk with uniform mass distribution, or spoked wheel with mass concentrated at the rim), then probably the pulley is intended to be massless as well as frictionless. Its only purpose in the problem is to redirect the tension in the string from vertical just above $M2$ to horizontal just to the left of $M1$. The tension is the same in both places.