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I was reading my textbook and I encountered a section where it explains that Centripetal Acceleration is not constant, thus, I wonder if jerk exists in Uniform Circular Motion? The textbook states that it happens due to the continuous change on the vector's direction. So, pretty much. Does jerk manifest? or, if it does not, why?

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Jerk is the change in acceleration. If acceleration is changing, then there is jerk. Since there is a changing acceleration in circular motion, there is jerk in circular motion.

I'm not entirely sure what you mean by "manifest" though. It's simply something that can be measured.

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    $\begingroup$ I think he's using "manifest" to mean "to be present in this situation". I felt it was proper usage and didn't think twice about it. $\endgroup$ Oct 12, 2020 at 9:35
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    $\begingroup$ To clarify slightly: the magnitude of the acceleration is constant but its direction is constantly changing; therefore there is jerk. Likewise the direction of the jerk is constantly changing, the direction of snap (derivative of jerk) is constantly changing ... it's circular all the way down. $\endgroup$ Oct 12, 2020 at 14:15
  • $\begingroup$ The use of manifest is, IMO, slightly incorrect. When you are asking "is X there, or not" (perhaps gravity, electric charge, happiness) you don't, really, ask if it "manifests" which tends to mean "becomes apparent". So, you might say "the disease only manifests itself ..." (in winter, under dyes, whatever). If you're plain asking "is John Smith in the room or not" "does this cloud have electric charge or not" "is this ball experiencing jerk or not" you're not really asking if it "manifests", the answer is simply yes or no, it's there or not. $\endgroup$
    – Fattie
    Oct 12, 2020 at 18:54
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As other answers and comments said, yes there is jerk. But it doesn't mean what you think. In everyday use, a jerk is a sudden acceleration, like a sudden yank on a rope. That is not what centripetal acceleration is like.

If you tie a rope to a rock and swing it around your head, you must pull on the rope to keep it moving in a circle. That is centripetal force and centripetal acceleration. If the rock is circling at a constant radius and speed, the magnitude of the force does not change. The direction smoothly changes. You have to keep pulling toward yourself as the rock moves around you.

The other answers about jerk, snap, crackle, and pop are apparently technical terms for higher derivatives of velocity. Though I had never heard of them before this question. They are not in common use. And these technical definitions do not imply sudden accelerations either.

Jerk is the derivative of acceleration. It is a semi-good name. It makes more sense to apply it to the derivative of force than acceleration.

Suppose you are pulling with constant force on a rope, trying to move a stuck object. You give the rope a yank. The force momentarily rises and then returns to its former value. The derivative of force reasonably fits the idea of jerking the rope, even if the object is still stuck. If the object breaks free and there is a non-zero acceleration, then the derivative of acceleration also reasonably fits the idea.

Jerk is a non-zero vector in uniform circular motion. But the motion does not fit the everyday idea of jerking the rope.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Oct 12, 2020 at 20:20
  • $\begingroup$ G.Smith: yes it is entirely necessary to emphasize, when some. Many readers of Physics.SE will be HS grads or laymen, and as I illustrate when even 2nd-year Eng students find this issue counterintuitive, it's counterintuitive. @Fattie's comment "There is utterly no distinction between "technical" and "non-technical" uses, whatever that means" is factually incorrect, and a seriously damaging attitude to actually educating site users and correcting misunderstandings in learning. $\endgroup$
    – smci
    Oct 12, 2020 at 20:22
  • $\begingroup$ Other than the absolutely common, totally settled physics usage, the word "jerk" in daily English in non-physics means "tug" - it is utterly evident the OP is discussing the physics usage of "jerk" (note that the OP even gives a link for goodness sake). Note too that the sentence "Jerk is a non-zero vector in uniform circular motion." is completely, totally, absolutely wrong. The sentence "They are not in common use" is totally incorrect. The passages explaining what jerk "should" mean (??) and that the writer has never heard the term before are just bizarre. Delete. $\endgroup$
    – Fattie
    Oct 13, 2020 at 11:46
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Others have explained it well, but I wanted to add another point.

When I think of "jerk", I think of amusement park rides that "jerk" my head around. With constant acceleration (zero jerk), it's easy to keep your head steady with a constant application of your neck muscles. When acceleration changes quickly (non-zero jerk), it's hard to keep your head from moving relative to your body.

When I first read your question I thought "It's easy to keep your head steady on a carnival ride shown below, so there must be zero jerk". In this case, I think your body has zero jerk in the rotating reference frame centered at the ride's axis. In other words, since your body is rotating along with the reference frame, you don't feel the jerk.

enter image description here

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    $\begingroup$ you are correct. In the non-inertial rotating reference frame, the centrifugal force is constant in both magnitude and direction, and thus there is no jerk. Your answer most likely got downvoted because it is as much a question as an answer $\endgroup$ Oct 12, 2020 at 21:30
  • $\begingroup$ @thegreatemu: Thanks for the tip. I removed the question part of my answer. $\endgroup$
    – James
    Oct 12, 2020 at 21:33
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    $\begingroup$ Now that I think about it, this particular ride does have jerk when the ride is tilted, because the centrifugal force is aligned/anti-aligned with gravity as the ride spins. Flat spinning rides have no jerk $\endgroup$ Oct 13, 2020 at 2:36
  • $\begingroup$ @thegreatemu: +1 Good catch. $\endgroup$
    – James
    Oct 13, 2020 at 11:34
  • $\begingroup$ Literally every single use of "acceleration" and "jerk" in this answer is incorrect - surprising on a physics site $\endgroup$
    – Fattie
    Oct 13, 2020 at 11:47

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