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Homework Help: Find the speed of a knot on rope

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data
    image.png
    http://s12.postimage.org/q2c8bqw97/image.png

    http://i.imgur.com/6STfr.png

    The part a is easy and i get the ans, the required speed should be v/cos30
    but i really have no clue about how to do the part b
    i know that the knot will move down and left but i just can't list any equation for the motion

    2. Relevant equations
    the length of string is constant



    3. The attempt at a solution
    i use √X^2 + Y^2 = constant is part a
    and get the v/cos 30

    but i really have no idea about b part

    any clue please >??
     

    Attached Files:

    Last edited: Oct 28, 2012
  2. jcsd
  3. Oct 28, 2012 #2

    PeterO

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    Not sure if you have noticed, but the image has not come through? Not here anyway.
     
  4. Oct 28, 2012 #3
    Sorry for my bad english

    you can't see the image???
     
  5. Oct 28, 2012 #4

    PeterO

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    Until I hit quote, I didn't even realise you had tried to include one. The imbedded image code showed up among your text.

    EDIT: the image reference just appeared in the original post.

    EDIT 2: But the image is too small to read, and has insufficient resolution to be enlarged ??
     
  6. Oct 28, 2012 #5
    i have now giving the website of image

    can u see now?

    maybe i can upload again

    I am now updating the zip file

    is it ok???

    thx for your help
     
    Last edited: Oct 28, 2012
  7. Oct 28, 2012 #6
    i am now uploading another image
    hope it is all right
     
  8. Oct 28, 2012 #7

    PeterO

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    Image works, but it is 2:00 am here and I am about to go to bed.
     
  9. Oct 28, 2012 #8
    thank you and have a good dream =D
     
  10. Oct 28, 2012 #9

    PeterO

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    OK,

    I see that the pulley is supposed to be infinitely small, so is to be represented as a point on the bottom rail matching the other geometry restraints.

    Can you calculate how far K is from B [or from the pulley for that matter, since it is in the middle of the rope] at the "30o" situation.
    btw: you should be able to calculate that distance without reference to a pen, paper or calculator. [hint: 2,1,√3 ]

    You can get a guide to the answer by establishing where K was when the angle was 29o and 31o and calculating the time taken for that change.
     
  11. Oct 28, 2012 #10

    PeterO

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    Just to let you know how I am visualising the situation.

    Clearly as the string is drawn in, the knot, K, gets closer to the pulley, O, and at the same time the angle, θ, is going to get larger, until it reached 90o.

    Lets imagine those changes separately.
    Considering only the angle change - I see K tracing out a circular arc, going up and left. [I can draw that in my mind or sketch it on paper]
    I then imagine a series of radii drawn to that arc.
    Considering that K is actually getting closer to O, I then see that K would actually be spiraling down towards O.

    I am then faced with the calculus to describe that situation!!
     
  12. Oct 28, 2012 #11
    yes. i share the same idea with u. i think we can use polar coor. i am on my way back home and i will upload my calculation. thank you
     
  13. Oct 29, 2012 #12
    I have figure out the solution ( maybe)

    I use polar coor. to find the speed

    the only problem is that i have to find the angular velocity

    let the angle be θ and let the radius of the Sector be L

    i found that if i consider the arc length and use
    Ldθ = dx that would give a wrong ans

    if i consider the area
    so that (1/2) L^2 dθ = (1/2) dx H where H is the height

    i will get the right ans

    the attached file is my steps

    plz help me and thank you for your clue, It proved that my original idea is right , i just lack of confidence =D
     

    Attached Files:

  14. Oct 29, 2012 #13

    PeterO

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    have you noticed that the geometry of the 30o position gives a very special relationship between the distances OK or BK (they are equal of course since K is the midpoint) and H
     
  15. Oct 29, 2012 #14
    similar trian?

    i am not good at geometry =(

    do i get the right ans in my pdf? thank you

    also, the horizontal distance between k and b is √3 /2 h

    the vertial is h/2 ???

    the angle 30 gives sin 30 =1/2 cos 30 = √3 /2 <-- this special relation ?
     
    Last edited: Oct 29, 2012
  16. Oct 29, 2012 #15

    PeterO

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    I believe at this position, the distance you are calling L is equal to H, so there is one less variable than you have.
    I can't follow your calculation: looks to me to be more of an advanced maths problem than Introductory Physics one.
     
  17. Oct 29, 2012 #16
    so i need to post to the advanced mechanics ?
    i am moving now plz del this post

    and thank you for your help! =D
     
    Last edited: Oct 29, 2012
  18. Oct 29, 2012 #17
    1. The problem statement, all variables and given/known data[/b]
    image.png
    http://s12.postimage.org/q2c8bqw97/image.png

    http://i.imgur.com/6STfr.png

    this post is moved from the introductory physics as this actually involves advanced mechanics

    The part a is easy and i get the ans, the required speed should be v/cos30

    but the part b is the thing that i can't understand

    i actually use polor coor. to get b ans and it seems right as when i assume the speed v be a certain number, and the find the speed after 0.000001 , the ans is equal that ans calculated from from my equation


    2. Relevant equations
    the length of string is constant
    the area of the triangle


    3. The attempt at a solution
    i use √X^2 + Y^2 = constant is part a
    and get the v/cos 30

    in part b

    it's difficult to explain and i just upload my steps

    the problem is that if i use s=rθ the equation will be wrong

    but if i use the area of sector = 1/2 l^2 θ = 1/2 b x h where b is the base and h is the height and l is the radius

    i will get correct ans

    can anyone help ?

    the attached pdf is my steps
    http://pdfcast.org/pdf/dsada-dsad-as-das [Broken] <-- this is the steps that plugging the valve v and assure my equation is correct
     
    Last edited by a moderator: May 6, 2017
  19. Oct 29, 2012 #18
    the admin said that it doesnt involve advanced.math
    can anyone provide a ans to me? my answer is right or wrong?
     
  20. Oct 29, 2012 #19

    haruspex

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    You have the speed and direction of the string at B. Think about the speed and direction of a point on the string OB near to O. Point K is halfway between the two. What do you think the relation would be between the three velocities?
     
  21. Oct 29, 2012 #20
    the speed of a pt near O is pointing downward??? with speed v ???
     
  22. Oct 29, 2012 #21

    haruspex

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    Speed v, yes, but not downwards exactly. Think again.
     
  23. Oct 29, 2012 #22
    along the string ?
     
  24. Oct 30, 2012 #23

    haruspex

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    Yes. So you know the speed and direction of points on the string near B and near O. K is halfway between B and O, and OKB remains a straight line. Can you work out the speed and direction of movement at K?
     
  25. Oct 30, 2012 #24
    Does it involves vectors? or sin cos tan?
    or just using a^2 + b^2 = c^2 ???
    i am really bad at this
     
  26. Oct 31, 2012 #25

    haruspex

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    Yes, it certainly involves vectors. Are you uncomfortable with that?
    For ease of discussion, let me give a name to a point on the string which is almost at O. I'll call it S. As discussed, it's moving towards O.
    I felt it was obvious that the knot, which is fixed for short time at halfway between S and B, would be moving at their vectorial average. If that that's not clear to you, I'll see if I can explain it another way.
    Work out the x and y components of the speeds at B and S. (Please put this in your next post.)
    Now imagine you are on the string at S. You are moving with the string, so you are staying a constant distance from both K and B (yes?). This means that they must appear to be going in a circle around you. Since K is half as far away as B, it would look like it is travelling at half the speed.
    How fast, in the x and y directions, would B appear to move relative to where you are at S?
     
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