Considering a pulley or belt I understand the derivation of the difference in tension(capstan equation) due the resultant torque needed for the pulley to rotate.

My question is for the tiny section of rope, in the rotating frame it is indeed in static equilibrium and hence we can derive the capstan equation. But, in this rotating frame why are there no centrifugal forces?Is the normal reaction force $dN$ the fictitious force?

It seems to make sense that in the inertial frame, the vertical components of $T$ and $T+dT$ would not be cancelled out as we need some centripetal force for the tiny rope section move in a circle while in the rotating(non-inertial frame) the normal reaction force gives the pseudo-force to keep forces working in the rotating frame.

  • $\begingroup$ Why would we view it from the rotating frame? Can't the equation be derived from this scenario but kept in an inertial frame? Are you sure that is not the case? $\endgroup$ – Steeven Aug 27 '18 at 14:45
  • $\begingroup$ I realize that way is simpler but to employ the use of co-efficient of static friction and non-slippage I've seen derivations everywhere to (supposedly) use the rotating frame where everything is in equilibrium for the tiny section of the rope. So my question is merely, is the normal reaction force used there in a rotating frame, a real force? $\endgroup$ – Tausif Hossain Aug 27 '18 at 14:48
  • $\begingroup$ If it is indeed an inertial frame, how can the section of rope be in equilibrium if it is indeed moving in a circle along with the pulley surface. An example of a derivation(they are all done in a similar way) is this video: youtube.com/watch?v=H0T7m537YT0 $\endgroup$ – Tausif Hossain Aug 27 '18 at 14:49
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    $\begingroup$ Why is this equilibrium a requirement? As far as I can see, all you need is to know that there is static friction between circle and belt. If you do view it from the rotating frame, then I agree that a fictitious centrifugal force must be included - since it isn't, then the frame used is not rotating. $\endgroup$ – Steeven Aug 27 '18 at 15:56
  • $\begingroup$ I understand, but is it true that the normal reaction force is real (in an inertial frame) or is it fictitious(only present in non-inertial frame)? Also, what causes the angular acceleration of the small sections of the rope? $\endgroup$ – Tausif Hossain Aug 27 '18 at 17:15

The capstan equation is derived on the assumption that the system is in static equilibrium - ie it is not rotating. In this case there is no centrifugal force.

It is necessary for there to be a normal force in order for there to be a friction force which prevents the rope from slipping against the bollard or pulley which it is wrapped around. And if there is tension in the rope, and the angle $\theta$ is non-zero, then there must be a normal force in order to balance forces on the section of rope in contact with the bollard or pulley.

Contact forces are always real, never fictitious. They are exerted by real objects, and have a physical origin.

If the rope is actually rotating, as when it is wrapped around a pulley, then there is a fictitious centrifugal force acting on it. This reduces the normal reaction between the rope and the pulley and thereby also reduces the friction force between the rope and pulley. However, the rope is usually assumed to have negligible mass and the angular velocity of the pulley is relatively low, so the centrifugal force is insignificant.

A hoop which is rotating in space has a real tension in it. The resultant tension on each segment is radially inward. In an inertial frame, this tension is necessary to explain the centripetal acceleration of the hoop. In a rotating frame the resultant tension force on each segment is balanced by the fictitious centrifugal force.

In reply to your comment

If there is a constant torque on the pulley this causes angular acceleration of the pulley and rope. Each segment of the rope accelerates tangentially - ie its speed increases while the torque is maintained. While the rope is moving in a circle each segment is also accelerating centripetally.

Before the pulley has started to rotate each segment of rope is at rest so it has no centripetal acceleration. The resultant tension force on it has a tangential component causing tangential acceleration. It also has a radial component, which is balanced by the outward normal contact force from the pulley.

When the pulley is moving each segment of rope now has centripetal acceleration so there is a net force inwards. The radial component of tension has not changed because the torque acting on the pulley has not changed. What has changed is that the normal contact force has reduced. This also reduces the static friction force which is accelerating the pulley.

As the angular velocity of the pulley increases so will the centripetal acceleration of each segment. At some point the normal contact force will become zero : the rope is rotating with the pulley but it is no longer pressing against the pulley. The rope loses contact with the pulley and moves outwards. Friction is no longer able to act so the pulley stops accelerating but the rope does not.

So the tension force in the rope does not increase as the angular velocity of the pulley increases, unless the torque is increased. Centripetal acceleration increases as tangential velocity increases. When the tension force can no longer provide sufficient centripetal acceleration the rope lifts off the pulley.

In the rotating frame we see the centrifugal force gradually increasing as the angular velocity increases. At first the fixed tension force is sufficient to keep the rope pressed against the pulley, but at some point the centrifugal force has grown so big that the tension force cannot prevent the rope from lifting off the pulley and continuing to move outwards.

If we did wish to ensure that the rope kept in contact with the pulley as its angular velocity increased then Yes the radial component of the tension force would have to increase also to provide more centripetal acceleration. This means that the difference between $T_2$ and $T_1$ would have to increase. But this increase in torque would also increase the tangential speed which would increase the centripetal force which is required, which means the difference between $T_2$ and $T_1$ would have to increase even further, in a vicious cycle.

The Role of Friction

Friction is required to transmit some of the torque from the rope to the pulley in order to make it accelerate. As with friction on the ground, this requires a normal force between the rope and pulley, otherwise the friction force is zero. There also needs to be a tangential force ($T_2-T_1 \gt 0$).

Suppose we push a small box which is resting on top of a large box which is resting on a frictionless floor. (In our case the rope is the small box, the pulley is the large box and the axle of the pulley is the frictionless floor.) Friction between the two boxes transmits some of the force which we apply from the small box to the large box to accelerate it at the same rate, because friction opposes relative motion.

It is the same with the rope and pulley. Friction from the rope causes the pulley to rotate. Without it the rope would slide around the pulley.

The normal contact force is a reaction force. As the pulley rotates some of the radial component of tension is needed to provide centripetal force. What is left over is opposed by the normal contact force.

As angular velocity increases more centripetal force is needed so there is less left over and the normal force gets smaller. This means that the friction force also gets smaller. At some point the friction force is not big enough to keep the pulley rotating at the same rate as the rope.

Going back to the boxes, if the small box is heavy enough when we push on it with a fixed force $F$ then the large box moves also. But if we remove some weight from the small box this reduces the friction force. There comes a point at which the friction force is so weak that if we continue to push with the same force $F$ the small box slides over the large box. If we want to keep the large box moving we have to reduce $F$ or put weight back into the small box.

In the same way with the rope if we want to keep providing friction to accelerate the pulley then we have to either increase the sum of the 2 tension forces $T_1, T_2$ while keeping the difference the same (this increases the normal contact force - like adding weight to the small box) or we reduce the difference between the tension forces while keeping the sum the same (this reduces the torque - like reducing the force $F$ we apply to the small box).

  • $\begingroup$ This is a great answer, but there is just one thing I still don't understand. In the inertial frame, the pulley has a resultant torque on it so it has an angular acceleration so its angular speed increases so for there to be no slipping, each segment of the rope must also have an angular acceleration, is this thinking right? If so, does that mean the tension must also increase to provide more and more centripetal force, so how does this work physically, and what am I missing if it's something else? $\endgroup$ – Tausif Hossain Aug 28 '18 at 12:21
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    $\begingroup$ I have updated my answer to address your comment. $\endgroup$ – sammy gerbil Aug 28 '18 at 13:35
  • $\begingroup$ I think I understand much better now, thank you so much. But can you please include how friction is playing a role in the inertial frame. How will its magnitude be adjusted for this to occur. $\endgroup$ – Tausif Hossain Aug 28 '18 at 14:07
  • $\begingroup$ In particular I am asking about the friction at the start when the pulley and rope start from rest. $\endgroup$ – Tausif Hossain Aug 28 '18 at 14:20
  • $\begingroup$ Your analogy is brilliantly written. Thank you once again. $\endgroup$ – Tausif Hossain Aug 28 '18 at 16:58

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