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From this answer by Floris, we can understand that for an accelerating rope there is some gradient of tension with respect to distance/ rope length. With that as a given, consider the analysis of Atwood machine done in Wikipedia with inertia and friction (last edited on 31 Jan 2021), reading between the lines for the section where they write about the magnitude of tension which torques the pulley on each side is equal to the tension acting on the mass.

For instance, on the side of rope which has $m_1$ the torque provided is $T_1r$ , now this is exactly where my confusion begins. If the rope is accelerating then how can it be that the tension which pulls up the mass is same as the one which torques the pulley? In my understanding , this contradicts the answer by floris, if it is an approximation then how do we justify it?

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Suppose a light rope goes around a massive pulley. The phrase 'light rope' means, as usual, that the mass of the rope is sufficiently small that the force required to accelerate the rope is negligible compared to all the other forces in the problem.

If a massive pulley is rotating at constant angular velocity then the torques on it are balanced, so the tension in the lengths of rope hanging on either side are equal. If, on the other hand, a massive pulley is accelerating, i.e. it is rotating with a changing angular velocity, then if the only torque on it is coming from the rope, then the two lengths of rope on either side of the pulley will have different tensions. This difference in tension, multiplied by the radius of the pulley wheel, gives the torque which is equal to $dL/dt$ where $L$ is the angular momentum. One has $$ (T_2 - T_1) r = \frac{dL}{dt} = I \dot{\omega} $$ where $\omega$ is the angular velocity of the pulley wheel and $I$ is its moment of inertia. The velocity of the rope, at any instant of time, is $$ v = r \omega $$ so the acceleration of the rope (and therefore also of any objects it is attached to, assuming it is inextensible), is $$ \frac{dv}{dt} = r \dot{\omega} = \frac{r^2}{I} (T_2 - T_1) . $$ By combining this with Newton's second law applied to masses suspended on the ends of the rope, one can solve the whole system. A typical case would be where $T_2$ and $T_1$ and the acceleration are initially unknown.

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  • $\begingroup$ Hmm I was not really looking for the analysis of it bcoz it is already contained in wiki, but rather an understanding how tension varies with the rope $\endgroup$ Apr 12, 2021 at 11:28
  • $\begingroup$ @Buraian the tension is constant along any section of rope that is itself light and is not interacting with anything else. To understand this, just consider a small section of rope and apply Newton's second law. If the section has negligible mass then the forces pulling it from the two ends must be equal. $\endgroup$ Apr 12, 2021 at 11:39
  • $\begingroup$ I have a heard time believing this, consider a section rope from first mass to second mass (which is near the mass but doesn't touch each), now it maybe noted that on end of the rope the tension is $T_2$ and the other one it is $T_1$ $\endgroup$ Apr 12, 2021 at 11:40

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