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Circular motion questions

  1. Dec 27, 2011 #1
    a bead of mass m can slide without friction along a circular hoop of radius a which is fixed in a vertical plane. the bead is connected to the highest point of the hoop by a light elastic string of natural length a and modulus 3mg. initially the bead is moving with speed u through the lowest point point of the hoop. given that u^2=ag, show that the bead just reaches the highest point of the hoop. show that the speed was u at the instant when the string first went slack and find the reaction of the hoop on the bead at that instant.

    i think i have done the first part.

    i used energy to find the velocity at the top then resolved at the top to get

    T+mg=mg which gives T=0 so complete circle just,BUT i have no reaction. do i not have a reaction force here?if i put R in i could just have R=-T?

    for the second part i am lost. any help would be appreciated. what equations do i get to show when string goes slack speed is u? how do i put R into this?


    2)
    the ends of a light string are fixed to 2 points A B in the same vertical line with A above B,and the string passes through a small smooth ring of mass . the ring is fastened to the string at a point P and when the string is taut the angle APB is a right angle,the angle BAP is T and AB=5a AP=4a. the ring revolves in a horizontal circle with constant angular velocity w and with the string taut.

    if the ring is free to move on the string,instead of being fastened, show that it will remain in the same position on the string if the angular velocity,w,satisifies

    12aw^2=35g

    once again,have no idea how to set up the equations of motion. do i have only a reaction and no tension? how does the reaction work? what condition do i need for it not to move??
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 27, 2011 #2

    Redbelly98

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    Why are you doing this, you just need to find the speed at the top of the loop -- what did you get for the speed, by the way?
    The question does not ask for the reaction force at the top of the loop.
    You said you used energy conservation before to find the speed at the top, and you can use it to find the speed at the point where the string goes slack.
    To find the reaction force, you will need to first draw a free-body diagram.

    Please post separate questions by starting a separate discussion thread. If somebody comes along to help with your second question while you and I are discussing the first question, it would become very difficult to follow what is going on.
     
  4. Dec 27, 2011 #3
    If it reaches the highest point then why will it get slack? ... And what do you mean by modulus ?

    attachment.php?attachmentid=42243&stc=1&d=1325032240.png

    You meed to make sure that the ring at P does not move up of down ... What should be the condition for that?
    Hint: It has something to do with Tension in the string ...
     

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  5. Dec 28, 2011 #4
    ok,the way you say this makes me think im doing it all wrong.

    the reason i done this was before when i had to show complete circles for particles attached to a string, i found the speed at the top of the circle, then used f=ma which for this case was just T+mg=mv^2/r and complete circles was when T>0

    now as this particle is on the wire,im guessing this is no longer the approach. i found the speed to be u/2 so not 0. is that enough to say complete circle.








    ok,so i think i see how to do this now. i pick a point where string is slack,work out the energy equation to find the speed then use f=mv^2/r? but in the force diagram i have jsut R and not tension?

    it looks like im misunderstanding the set up of string and wire. am i correct to think the string need not have tension at the top of the circle for the particle to do complete circles but can/will go slack at other parts of the circle?



    re your last point: sorry i posted 2 but i have lots of these questions! would have filled the first page up :). I think you have given me enough to see what i was doing wrong on the first question so i will reply to the second post regarding the 2nd question.
     
    Last edited: Dec 28, 2011
  6. Dec 28, 2011 #5
    this is what i am unclear on. how does the tension work if the ring is no longer attached to the bead? suely all the forces acting on the bead now are reaction force and mg down?
     
  7. Dec 28, 2011 #6

    Redbelly98

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    If the speed is u/2 at the top, that means the bead began with more than enough speed to reach the highest point -- contradicting the instruction to "show that the bead just reaches the highest point of the hoop".

    If you show your calculation of the speed at the top, maybe then I can help you figure out where the problem is.

    (Note that whenever the distance from the bead to the top is less than the natural length a, then the string tension is zero.)

    .
    .
    .

    Slightly different: you pick a point where the string just becomes slack (i.e., it's length is the natural length, it is neither stretched out nor are the ends closer together than the natural length.)

    You're partially correct that f=mv2/r. Actually that refers only to the force component in the direction toward the center of the loop. And note they are using a, not r, for the radius.

    In fact the string does not have tension at the top, or anywhere that two ends of the string are closer together than the natural length of the string.
     
  8. Dec 28, 2011 #7
    i let PE=0 level be bottom of circle. then extension in string is a so

    PE+KE+EPE=0+u^2/2+3g(a^2)/2a=ag/2+3ag/2=2ag

    at top

    PE+KE+EPE=2ag+v^2/2+0

    so

    2ag=2ag+v^2/2+0

    so v^2=0

    that looks better :) i put u^2=2ag in doh!!

    yeah,got the point it goes slack, when the string forms equilateral triangle of side a so i can finish off this question. thanks.
     
  9. Dec 29, 2011 #8
    still unsure what your hint is getting at? can you explain what forces the ring are subjected to if it is not connected to the strings?

    how do i put in the strings into f=ma?
     
  10. Dec 29, 2011 #9

    Redbelly98

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    Looks good!
    You're welcome.
     
  11. Dec 29, 2011 #10
    really lose the thread with all the words in these questions, then get lost by what motion is occuring! still a bad slip writing the wrong value for u!! any thoughts on the other query? or my other threads?
     
  12. Dec 30, 2011 #11
    Hey ... sorry for late reply

    (I mistakenly deleted my registry :tongue: reformatted just now)

    The two tensions in string will act so that they can provide the necessary centripetal force ... It wont be wrong to assume the string to be ≈ 0 ...
    so you can say that the tension in upper and lower (of length 4a and 3a) to be ______.

    now all you need to do is that equate sum of their horizontal components to ______ and sum of vertical to ______.
     
  13. Dec 31, 2011 #12
    hi.
    still none the wiser on this im afraid.

    if i look at the ring,as its not attached to the strings surely i only have mg and reaction force acting on it?

    where does the reatcion force act? in your diagram you have it pointing in towards the centre,why is that? i would put it perp to the contact, but is it in contact with the top string or bottom string?

    my equation of motion would then be

    horizontal component of R=mv^2/ (radius of circle)

    where does the tension come into play?
     
  14. Jan 1, 2012 #13
    What, according to you, is the origin of this reaction force?
     
  15. Jan 1, 2012 #14
    now im worried. as you ask that question,i must be wrong.

    im assuming the ring touches the string so there is a reaction to the contact with the string?
     
  16. Jan 2, 2012 #15
    Well there will be a reaction from the string ... and that reaction will be in the form of tension
    Imagine it this way ...

    The ring when rotates about a center will e pushed out in tangential direction if there is no centripetal force ... Thus lets assume that initially ring does move out tangentially ...
    it will strike the string and will push it out ... not the string will get deformed as we saw in this case (making angle 90° with itself) ... now the different parts of string will stretch out and there will be a force (electrostatic) which will try to out string back in original shape ... And of course this force shall be tangential to that string's surface. We call this force TENSION in the string
    And because the string is light, this tension will be of same magnitude at all parts of the string ...
     
  17. Jan 2, 2012 #16
    thanks for your patience!
    are you saying,if i resolve vertically and horizontally i have
    i have
    x-component of T=v^2/r
    y-component of T=mg
     
  18. Jan 2, 2012 #17

    Redbelly98

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    I'm not sure you are quite understanding how the tension works here, since there are actually two forces, each of magnitude T, but acting in different directions.

    Here is how we can convince ourselves of what the forces are:

    Imagine a short section of the rope, the short curved or bent section that is in contact with the mass. Using cupid.callen's earlier figure, this would be a small section of rope located at point P:

    attachment.php?attachmentid=42243&stc=1&d=1325032240.png

    The forces acting on this small rope section are:

    1. Rope tension T from the upper straight section of rope.
    2. Rope tension T from the lower straight section of rope.
    3. The contact or reaction force of the mass acting on the short section of rope.

    Since this rope section is considered massless, the net force acting on it is zero (F=ma and m=0). Therefore, the reaction force due to the mass is equal in magnitude, but opposite in direction, to the vector sum of the two tension forces.

    Now consider the force exerted by the rope on the mass. By Newton's third law, it is equal in magnitude, but opposite in direction, to the force exerted by the mass on the rope. Therefore, it is equal to the vector sum of the two tension forces exerted by the two straight sections of rope -- i.e., same magnitude and direction as this vector sum.

    Hope that helps clarify things.
     
  19. Jan 3, 2012 #18
    thanks.

    no examples like this in my book,so i have not seen this idea. would never had got this!!
     
  20. Jan 3, 2012 #19
    well i thought i got it,but when i came to do the question still drew a blank.

    when you say vector sum of the 2 tension forces do you mean

    resolve the 2 tensions into x component i+y component j

    then add to get
    sum of x components of 2 tensions i+sum of y components of 2 tensions j

    and then when i resolve horizontally i just take

    sum of x components of 2 tensions=mv^2/r?

    sum of y components =mg?

    I assume T is not the same as T from the first part of the question?
     
  21. Jan 3, 2012 #20

    Redbelly98

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    Yes, that's right.
    As long as you account for the fact that one of those y-components is in the downward direction, therefore negative, then yes.
    I don't understand. Question 2 only has one part. If you mean that T is not the same as in Question 1, then yes -- they are separate independent questions and you should not think that something in one of the questions has any relevance to the other question.
     
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