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One day I study the simple gravity pendulum, which an angle is less than $\frac {π}{2}$. It doesn't consider friction and air drag.

In the simple gravity pendulum, My textbook says ”assume the rope isn't slack“. And also wikipedia says “The rod or cord on which the bob swings is massless, inextensible and always remains taut.” https://en.wikipedia.org/wiki/Pendulum_(mechanics)

But when I tested pendulum, the rope didn't slacken. Why does this assumption exist?

I think we can't verify that the rope isn't slack.

Would you mind telling me?

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    $\begingroup$ "Why does this assumption exist ?" If it did not then we could start the pendulum bob from a position where the string is already slack, in which case the equation of motion derived for a simple pendulum would not work. $\endgroup$
    – gandalf61
    Jan 7 at 14:00
  • $\begingroup$ But when I tested pendulum, the rope didn't slacken.” Maybe not visibly, but a rope with mass held at its ends in a gravity field is never perfectly straight when nonhorizontal. See catenary. The problem is saying to ignore this curvature. $\endgroup$ Jan 7 at 15:04
  • $\begingroup$ Thanks both Mr.gandalf61 and Mr.Chemomechanics.My question is this assumption is necessary for a pendulum that starts with the thread taut at the beginning.I apologize I didn't write down the initial conditions. $\endgroup$
    – javy
    Jan 7 at 15:18
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    $\begingroup$ On a child's playground swing, it is possible to get the suspending rope or chain to go slack without starting from a slack position (been there, done that, a long time ago). It happens when the arc is nearly 180 degrees. $\endgroup$
    – John Doty
    Jan 7 at 15:31
  • $\begingroup$ Thanks Mr.John Doty. My test was 15,30,45,60 degree so I am surprised so that fact. $\endgroup$
    – javy
    Jan 7 at 15:47

2 Answers 2

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The simple pendulum problem is an idealization that like most textbook examples sacrifice real life details in order to study the dominant details in many familiar mechanical systems.

In the case of the pendulum, the problem is idealized with the assumption that there is no slack in the rope in order to restrain the pendulum to a semi-circular path, i.e it reduces the number of coordinates needed to describe the pendulum's motion to just one, viz. the angle of displacement. In the case where one did not assume zero slack, then the motion would be much more complicated. For example, zero slack cases might be represented by the following: picture a scenario where instead of releasing the pendulum cleanly from an angle $\theta$, one instead lifts the pendulum mass in such a way that the rope tension goes slack and drops the pendulum. In this case the motion is such that the pendulum will first free fall before the rope finally introduces tension as a constraint to alter the pendulum's free fall; only after the free fall is complete will the pendulum settle into the oscillatory motion typical of the simple pendulum problem. Essentially, without forces of constraint, like that provided by the no slack assumption, the motion of the pendulum could depend not just on the angle of displacement, but on additional coordinates as well, perhaps $(r,\theta )$, or $(r,\theta,\phi )$ if the motion is strictly confined to a plane.

Can we verify that the rope isn't slack? The condition for this should be the observation of tension on the rope. If gravity is able to keep the tension in the rope is equal to the weight of the pendulum, then the rope isn't slack. An assumption that is easily met under textbook ideals. In real life one might have pesky wind currents, earth quakes/tremors that transmit through the rope, or perhaps a cat that playfully paws the pendulum bob. In all of these cases the assumption could be violated.

Even under real life conditions, however, the simple pendulum ideal is rather accurate. This is the reason that the understanding of the dimple pendulum allowed for the construction of good clocks, which before the advent of quartz technologies, were based on period of a pendulum. 

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  • $\begingroup$ Thanks to answer my question. I hadn't thought about a pendulum with no initial thread tension :_ . And complex system that think friction and wind, etc is unmanage to me. My question is this assumption is necessary for a pendulum that starts with the thread taut at the beginning. If it was taut from the start, would this assumption be necessary?Or can we show that we don't need the assumption and there is always tension? $\endgroup$
    – javy
    Jan 7 at 15:13
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    $\begingroup$ Physics texts should start with "EVERY MODEL PRESENTED IN THIS TEXT IS WRONG IN REALITY. THESE MODELS MAY BE USEFUL, THOUGH, WHEN APPLIED WISELY." $\endgroup$
    – John Doty
    Jan 7 at 15:34
  • $\begingroup$ @ykk Technically the assumption must apply at all points of the motion, not only the initial conditions. For example, what happens if the rope breaks? The tension goes to zero and the pendulum's mass begins a parabolic journey towards the earth, a kind of motion no longer consistent with pendulum equation. $\endgroup$ Jan 7 at 15:46
  • $\begingroup$ Excellent point John Doty! When I first began in physics I expected that there would always be exact closed form solutions, in fact, I was a little appalled that a problem like the pendulum would have to be answered in terms of elliptic functions. It is amazing that simple physical principles form the support of immensely complex structures. $\endgroup$ Jan 7 at 15:50
  • $\begingroup$ If I can't think of the complex system, can't probe that the tension is always exist? $\endgroup$
    – javy
    Jan 8 at 1:21
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Another way the problem might have been states is "Assume the rope doesn't stretch." or "Assume the distance between the bob and the center of rotation is constant."

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