Is this a true statement regarding labeling tension & pulley?

[bobaustin]

If the pulley is massless then the tension T on one side of the pulley is the same T as on the other side, therefore it's the exact same T.
But if the pulley is not massless then you have to label one T1, say, and the other T2, even it's the same rope?
Same situation if the pulley is frictionless?

Thanks -

 

[rl.bhat]

Tension on both segments are the same. But when you want to find the acceleration of the system, you have to consider the torque on the pulley.

 

[bobaustin]

But if you can't ignore the pulley then the tension on one side is different than the other side, right? For example if you have a ("low friction") Atwood machine, the T is the same on both sides which is why you get a = (m2-m1)g/(m1+m2) etc, but in a problem where they talk about the pulley being a disk with I and r, and or if they specify that there is a frictional torque, then you have to label the tensions differently?

 

[Redbelly98]

Yes.

If the pulley has nonzero mass, I is not zero and there can be a net torque on it due to different tensions.

If the pulley is massless, I=0, and therefore the torque,

T = Iα​

is zero as well. In order to have zero torque, the two tensions must be equal.

 

[bobaustin]

Redbelly98: Thanks for a great explanation that makes perfect sense!