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So, this is how the problem looks:

Plus, the pulley is suspended on a cord at its center and hanging from the ceiling.

You're given masses of the objects, mass of the pulley and it's radius. And your assumptions are that the string is massless and inelastic and that there is no friction between the string and the pulley. You have to find the tension in the string suspending the pulley (it's not drawn in this particular figure).

My conclusion was that, since there is no friction between the string and the pulley, the tension would have to be equal all along the string's length and the pulley would not rotate, it would slip and there would be no way to transfer the linear motion of the masses to the rotational motion of the pulley. (The resulting torque would be zero.)

So, according to me, the tension in the string suspending the pulley would simply be the tension along the rope multiplied by two.

But the solution includes a rotating pulley and its rotational inertia, and it gives a different answer. Where did I go wrong? And why? If not, how can I prove I'm right?

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    $\begingroup$ your analysis is right, assuming no friction between the string and the pulley. It seems that the question is contradicting itself by saying that there's no friction, yet assumes that there is. why don't you try working out the answer in the case with friction so the pulley rotates w/o slipping? $\endgroup$
    – nervxxx
    Jan 28, 2013 at 14:10
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    $\begingroup$ Thank you, I just needed a sanity check from somebody! You're never too sure! Yeah, I've worked it out and it's the same answer... I don't even understand why would anyone give a problem of this kind, since the system is in such configuration that it could actually be answer painfully trivially by simply adding up the weights... $\endgroup$
    – user20250
    Jan 28, 2013 at 14:38
  • $\begingroup$ To re-iterate: Without friction the pulley cannot rotate. The rope will just slip towards the heavier mass. $\endgroup$ Jan 28, 2013 at 20:31
  • $\begingroup$ Raga Raga: yes the tension in the spring suspending the pulley in the frictionless case would be twice the tension in the rope, but that's not equal to the sum of the weights as your comment suggests... but since you got the correct expression for the case with friction I presume it's just a misunderstanding on my part $\endgroup$
    – nervxxx
    Jan 30, 2013 at 5:43
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    $\begingroup$ Perhaps your source meant to say 'no slip' instead of 'no friction'. $\endgroup$ Apr 9, 2020 at 8:08

2 Answers 2

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You're correct that if the pulley is frictionless the pulley wouldn't rotate. Most likely, the authors of the problem had in mind to say “this is an idealized physics problem” and didn't reason correctly about which parts of the system need this specified, or just wrote the wrong words or incorrectly simplified them during editing.

Here are some possible causes of energy dissipation that could be in the system that they arguably should have specified instead:

  • The pivot of the pulley needs to be frictionless.
  • A real cord winding around a real pulley will experience some friction as each portion of the cord pushes against the pulley and unwinds again later. This is the same sort of phenomenon as rolling resistance but with different geometry and materials.
  • To specify further (if you were trying to, say, create an real-world experiment with negligible error) you would want to specify that the objects are in a vacuum and there is no magnetic field or the materials are not electrically conductive. But it is generally reasonable to not discuss these things since they are "additional" physical phenomena that have not yet been discussed in a typical physics education.
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If your pulley has mass, you have to consider the rotational inertia of it. Inertia is the tendency to resist movement. Your rope does not have equal tensions. Consider $T_1$ and $T_2$ where $T_1$ is the tension on the side of $M_1$ and $T_2$ is the tension on the side of $M_2$. Assuming $M_1$ is greater than $M_2$, we have $M_1g-T_1 = M_1a$, $T_2-M_2g=M_2a$, and $(T_1-T_2)r=I\alpha$. Since it does not slip, $a=\alpha r$. We can now solve for the system's acceleration.

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