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I know that on one side of the pulley the tension must be larger to let the torque do its job and rotate the pulley. But how does the relation look like?

$$T_1 = k \cdot T_1'$$

What's $k$?

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  • $\begingroup$ Are you asking what the torque is if the pulley axle has friction? Static friction between the rope and the surface of the pulley makes the pulley turn. Only friction at the axle would oppose it. $\endgroup$
    – Bob D
    Mar 30, 2019 at 20:26
  • $\begingroup$ Are you imagining a frictionless pulley with rotational inertia? One with a stick axle? One with a wheel that isn’t round? How is it not ideal?? $\endgroup$ Mar 30, 2019 at 20:39
  • $\begingroup$ there's friction between the rope and the pulley. it is also fixed, it can only rotate. what I mean by not ideal is that it has mass and radius $\endgroup$
    – H4NS
    Mar 30, 2019 at 21:02
  • $\begingroup$ There is no sufficient information to find an answer to this. Please give additional information like if the rope has mass, or any weights or whether the pulley has any friction at the axle. $\endgroup$
    – Starboy
    Mar 31, 2019 at 4:40

1 Answer 1

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By an ideal pulley we mean that it is massless, and there is no friction between pulley and string. We also assume the string to be massless irrespective of pulley(considering mass of string would make the situation extremely complex as tension would vary with height). So for an ideal pulley, the pulley actually does not rotate. The tension throughout a string is same. The pulley does not experience a net force or torque(m=0) and the string slips over the pulley(the pulley does not rotate). In case of a non-ideal pulley, for calculations we neglect friction at the axle. We assume pulley to have a mass m. And consider STATIC friction between pulley and string. That means pulley would now rotate on its axis. So we need the moment of inertia of pulley(radius and mass). Since the pulley rotates, the tension varies in the string wrapped over the pulley. The pulley rotates in sense of net torque and the relation between the tensions on two ends of wrap is given by $T_1=T_2 e^{\mu \theta}$ where $\mu $ is coefficient of static friction between string and pulley and $\theta$ is the angle of wrap of string around the pulley. You can apply $\sum F=ma $for the pulley in this case in case it has an acceleration. Note that the tension at two ends is different but it is same throught the string when it is not in contact with the pulley. So in this case the pulley rotates and if there is enough friction, we assume there is no slipping between the string and pulley(static friction). Hope this clarified your doubts!

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  • $\begingroup$ thanks a lot for the answer, this is what I was looking for $\endgroup$
    – H4NS
    Mar 31, 2019 at 7:34
  • $\begingroup$ To make it explicitly clear, this relation $T_2=T_1e^{\mu\theta}$ springs out of the requirement that the string be massless and thus, the net force on any element of the string must vanish. If the string is massive, the relation between tensions at two different points in the string would depend on the dynamics of a particular system of which the string is a part. $\endgroup$
    – ACat
    Mar 31, 2019 at 18:08
  • $\begingroup$ In particular, a back-of-the-envelope calculation suggests that $T_2=T_1e^{\rho R^3 \alpha+\mu \theta}$ should hold for a string with a uniform linear mass density $\rho$ which is wrapped around a disk of radius $R$ rotating with an angular acceleration $\alpha$. This is deceptively simple because an explicit solution for $\alpha$ requires the knowledge of the value of tension at any one point of string around the pulley--this will depend on the rest of the system. $\endgroup$
    – ACat
    Mar 31, 2019 at 18:18

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