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Vertical and horizontal circular motion: How to identify the centripetal force?

  1. Sep 28, 2014 #1
    I made notes on this in class, but I did not understand everything about it. So, what I know is that vertical motion is when the object goes around in a circle like this:
    cmrkq2.gif

    And horizontal I guess it is like this:
    ducm3a.gif

    But then I am very confused regarding what my teacher told me for horizontal motion. She told me there is the top and the bottom for horizontal motion which I completely do not get because it is always at the same distance to the ground, and so the tension is perpendicular to mg but the force diagram she given the class is just not like that. The 'experiment' was called ''rotating rubber on a string (horizontal) - assuming the string is horizontal''.

    For the 'top' (I still don't know how top and bottom exists here) it is mg + T = Fc and for bottom T - mg = Fc.

    Could someone please explain if I am misunderstanding what she means by horizontal motion or explain how top and bottom exist?
     
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  3. Sep 28, 2014 #2

    A.T.

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  4. Sep 28, 2014 #3

    CWatters

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    The "horizontal" case diagram you posted looks like pendulum swinging in a circle subject to gravity.

    However you said..

    The only way the string can be perfectly horizontal is if the effect of gravity is negligible compared to centripetal force - so perhaps that explains why the force diagrams are different? Gravity has been neglected?

    Perhaps post the diagram your teacher produced so we can see it?
     
  5. Sep 28, 2014 #4

    CWatters

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    Is it possible she got her vertical and horizontal mixed up when she was speaking? or was it in the written material? The reference to the tension being different top and bottom only makes any sense for the vertical case.
     
  6. Sep 28, 2014 #5
    Hello the reason I said I guess is because the teacher didn't actually draw diagrams for any of the two... She simply gave us the equations that show what creates the centripetal force in the two cases of ''rotating rubber on a string (horizontal) - assuming the string is horizontal'' which I she given as top being mg + T = Fc and bottom T - mg = Fc. And the vertical was ''rotating rubber bung on a string (vertical circle) - at the top and the bottom of the circle'', which she noted as top T + mg = Fc and bottom T - mg = Fc.

    I've placed the diagrams so you can see how I am considering the vertical circle rubber and the horizontal circle rubber... What I want to know if I am interpreting it correctly or not, whether the diagrams fit the cases, and what thereof makes the equations for top and bottom of the horizontal correct, because the vertical I can deal with, I do understand if the diagram I've drown for it is right.

    The core of my problem is that when I think of a vertical string is like when I spin a mass on the string in circles like this:
    28bv5eo.png
    (Sorry for the bad drawing, the bit that is joined with the body is the hand holding the string)

    But when I think of a rubber rotating on a HORIZONTAL string I thought that what is is the diagram given above, but it doesn't make sense to say there is top and bottom to that so I don't know what it means anymore for a circular motion to be horizontal...
     
    Last edited: Sep 28, 2014
  7. Sep 29, 2014 #6

    CWatters

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    I think you will have to ask her about the horizontal case. Your/her explanation about top and bottom makes no sense to me.

    Horizontal case
    Normally, in the string horizontal case (no gravity) the tension T is constant and T = mv2/R.

    It makes little sense to talk about a horizontal case with gravity because the mass would have to rotate infinitely fast to get the string horizontal (see conical pendulum below).

    Vertical case
    In the string vertical case (with gravity) the tension varies.. At the top the magnitude of the tension |T| = mv2/R - m*g. You can see from this that If the object rotates too slowly mv2/R will be smaller than mg, the tension will go negative and the string will go slack. At the bottom the magnitude of the tension |T| = mv2/R + m*g

    Conical Pendulum case
    The second diagram in your original post shows a third option with the string neither horizontal or vertical. This is called a conical pendulum. The tension is constant and there is no "top" or "bottom" to the problem. You can write two equations..

    In the vertical plane. The mass isn't accelerating so m*g = Tcos(θ) .

    In the horizontal plane.. mv2/R = TSin(θ).

    These two can be solved to give an equation for the angle θ for any particular velocity v. That's done here if you want to see the maths..
    http://www.a-levelphysicstutor.com/m-kinetics-circmotion.php
     
  8. Sep 29, 2014 #7

    TumblingDice

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    Is it possible that there's just a small terminology or language hurdle? Assuming the horizontal case is indeed what we expect, and looking at the horizontal diagram in the OP, it's clear there's a downward component of gravity indicated.
    Well, it makes sense if horizontal/vertical is only meant to differentiate between the planes of the centripetal force. When the teacher (or Tangeton) said top and bottom, could this be a reference to the relative altitude of the center/anchor location to that of the object in motion? There's a horizontal vector/component from circular motion and a vertical for gravity. Circular motion and gravity kind of allow someone to use top and bottom since the system is operating in three dimensions. The vertical case is two dimenional as all forces are operating in the same plane. :confused: (?)
     
  9. Sep 29, 2014 #8

    A.T.

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    How is "horizontal" defined without gravity?

    But it does make sense to talk about a horizontal case without gravity ?
     
  10. Sep 29, 2014 #9

    sophiecentaur

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    I would think the best way to resolve this confusion would be to look at a text book or Hyperphysics (there are many other sources) and see what they say. Whatever your teacher said - or you thought she said - is not worth worrying about. We can assume she knows what she is talking about (99% of the time, at least) so it must have been a misunderstanding.
    Also, when describing rotational motion, it is possible to refer to the plane of rotation or the axis of rotation. They are not the same; rotation on a horizontal axis will involve vertical motion at times etc. etc.
     
  11. Sep 29, 2014 #10

    CWatters

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    TumblingDice & AT..

    I explained why it makes little sense to talk about a horizontal case with gravity. The tension in the string would have to be infinite to get the string horizontal. If there is gravity then it's a conical pendulum problem.

    The OP only posted a drawing of a conical pendulum in the hope that's what the teacher was talking about. I doubt it as the equations she gave don't match that.
     
  12. Sep 29, 2014 #11

    A.T.

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    But not how you define "horizontal" without gravity.

    Why does the string have to be horizontal? The plane of the circular motion is horizontal. So it's called "horizontal case", to distinguish it from a circle in a vertical plane.

    Obviously, the horizontal vs. vertical distinction only makes sense with gravity, and doesn't make much sense without gravity. So I'm still puzzled why you think it's exactly the other way around.
     
  13. Sep 29, 2014 #12
    Okay seems fair enough I guess I will do more of my own research on this matter. Thanks all.
     
  14. Sep 29, 2014 #13

    TumblingDice

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    As I read through all of your posts in this thread, I read much of the confusion originates from your insistence that the OP wasn't using the right words. Starting from your first post:
    Yes, the force diagrams are different. The horizontal one includes gravity and the vertical one doesn't. What part of this didn't you understand?

    The teacher called the experiment, "rotating rubber on a string". That doesn't sound like advanced physics terminology. You wrote:
    I'm not disagreeing that more information from the OP and teacher would have been helpful. But being a snob about terminology and scoffing because things made "little sense" to you just seems a poor way to respond to someone seeking help, and a new member at that.

    I checked out Conical Pendulum on wiki. They have a diagram (I'm linking this direct from wiki):
    404px-Conical_pendulum.svg.png
    This looks like the horizontal diagram in the OP, no?

    I did a search on "horizontal circular motion rubber string" and found that whirling a rubber lab stopper on a string horizontally over your head is a common beginning physics experiment. In some places it's called a rubber bung. None of the experiments are called Conical Pendulum. They have names like, "Whirling a rubber bung on a string." That sounds an awful lot like what the teacher said. There's even a thread right here at PF with a variation.

    Often times, the string is threaded through a tube that's held vertically over one's head. Weights are attached to the bottom of the string, and the centripetal force is calculated by measuring the distance from the top of the tube to the rubber stopper. It think it would have been nice if the OP could have looked at some of this before leaving the thread. Even if it were not the exact experiment, I think it could have helped.
     
  15. Sep 30, 2014 #14

    CWatters

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    Probably best I just agree the OP needs to go back to the teacher for clarification. Little point in us trying to double guess what she was talking about.
     
  16. Sep 30, 2014 #15

    CWatters

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    I don't think it's the other way around. Clearly in the vertical case gravity is what makes the tension vary.

    If I was being picky.. Horizontal is usually defined as parallel to a horizon rather than orthogonal to gravity but time to let this go I think.
     
  17. Sep 30, 2014 #16

    A.T.

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    Try that in the mountains.
     
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