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Mechanics problem -- 2 masses and a string in motion

  1. Oct 5, 2016 #1
    [Mod Note: moved to homework forum, template missing but homework-type problem]

    There are two masses (m) tied at two ends of a string of length 'l'. Whole system is placed on a friction-less surface and force 'F' is applied at the centre of string in direction perpendicular to the string. Find the time taken by masses to come together and nature of path followed.
     

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  2. jcsd
  3. Oct 5, 2016 #2

    davenn

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    HI there
    welcome to PF :smile:


    is this homework ?
    what have you done so far yourself to work towards an answer ?

    also you shouldn't have used the advanced tag

    Dave
     
    Last edited by a moderator: Oct 5, 2016
  4. Oct 5, 2016 #3
    Basically I worked over it by taking some angle at some moment and resolving the force then writing down velocity and acceleration equation for x and y axis but I am getting entangled in calculus part.
     
  5. Oct 5, 2016 #4
    Ok.

    I am new here so bit unfamiliar with the prevailing norms.
     
  6. Oct 5, 2016 #5

    davenn

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    that's OK :smile:
    so is it homework ? if so, start a new thread in the homework section and use the template method supplied there to lay
    out your question and what work you have done so far :smile:

    as a note, you will notice at the top of most of the forum sections including the one you posted in ... a "Please don't post homework in here" thread


    Dave
     
  7. Oct 5, 2016 #6
    Certainly not homework. I am not somebody's teacher to give someone some homework. I thought this is a forum for Physics lovers to exchange the ideas and shared problem solving. :cry:
     
  8. Oct 5, 2016 #7

    davenn

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    That's OK ... it's just the way you worded you question it looked like the normal homework Q that the forum gets 1000's of
    whether you are a teacher or not isn't the issue :wink:

    it is, but a major part of PF is that we like people to think for themselves and rather than just putting out answers, we encourage the poster ... you ....
    to think about your problem, do some research, maybe come up with a formula or 2 that may be appropriate in helping to solve you query, then members here can guide you towards an answer, with good effort from you :smile:

    Dave
     
  9. Oct 5, 2016 #8
    Ok. Got it :smile:
     
  10. Oct 5, 2016 #9

    davenn

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    sooooooooooooooooo :wink:

    with all that behind us, what things do you think are relevant in working towards a solution ?
     
  11. Oct 5, 2016 #10
    Ok. This is what I did.

    I have assumed whole case at some angle θ with vertical and then resolved the force

    http://filesystem:https://web.telegram.org/temporary/851516233_42736_5351388760177376947.jpg [Broken]

    so,
    F = 2Tcosθ
    T = F/(2cosθ)
    Now, for an individual mass
    F(x) = T sinθ = (F tanθ)/2
    a(x) = (F tanθ)/2m ↔ ∫dv(x) = (F/2m) ∫tan ωt .dt [ I don't know what to do next at this point]
    F(y) = T cosθ
    a(y) = F/2m ↔ v(y) = Ft/2m

    Perhaps I can use one more equation

    x/y = tan θ
    x = y tanθ
    dx = y sec^2 θ. dθ
    dx/dt = d/dt ( y sec^2 θ. dθ) [Again I am stuck here..what next]

    Now I have equations for v(x) and v(y) but v(x) is function of θ/ω, and I don't know if angular velocity is really constant.

    again if i approach it from different method time for meeting can be calculating by assuming situation for motion along x axis

    l = (1/2)[{a(x) + a(y)} (t^2)]

    but here answer will come in terms of θ.
     
    Last edited by a moderator: May 8, 2017
  12. Oct 5, 2016 #11

    ehild

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    What do you call x and y? What is theta? Make a sketch and show, please. The picture in the post does not show.
     
    Last edited by a moderator: May 8, 2017
  13. Oct 5, 2016 #12
    Ok. This is the diagram I used to make equations..

    On the other hand I used some other method to solve it...

    Let the point of application moves distance d before masses meet, worked done on system would be F.d

    By work energy theorem
    F.d = 2 (1/2 m v^2)
    v^2 = Fd/m
    again for motion along y axis
    a(y) = F/2m, u = 0, s = d - l/2

    So using equation v^2 = u^2 + 2as

    Fd/m = 2 (F/2m) (d-l/2)

    well this equation is meaningless..reaches nowhere
     

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  14. Oct 6, 2016 #13

    ehild

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    v is the total speed, so v2=vx2+vy2 here.
    Along the y axis, you have to use the y component of velocity.
     
  15. Oct 6, 2016 #14
    Well...you see at the moment they meet, speed along x axis will be zero and as I am taking energy concept in use I have to care only about initial and final positions. initially speed along x as well as y axis was zero and finally speed is only along y axis.
     
  16. Oct 6, 2016 #15
    Finally whole speed is in y axis only, i hope.
     
  17. Oct 6, 2016 #16

    ehild

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    The horizontal components of velocity becomes zero when the balls collide. During their motion the x component of acceleration is towards the middle which makes the magnitude of the horizontal velocity increase, How can it become zero, if the acceleration never changes sign?
     
  18. Oct 6, 2016 #17
    Yes , you are right. But acceleration perpetually changes magnitude along x axis...and x component of acceleration / force will vanish ultimately. But what's the nature of acceleration and velocity along x axis..i really don't know. I feel calculus is involved here. Can you help?
     
  19. Oct 6, 2016 #18

    ehild

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    Sorry, I do not think this problem is properly set. You said, correctly, that F=2T cos(θ). The problem stated that initially a finite force F acted perpendicularly at the centre of the string. At the beginning, θ=pi/2, and cos (θ)=0. That would mean infinite tension in the string. I think, you should assume some initial angle between the strings and the vertical. Have you found this problem somewhere, or was it your idea?
     
  20. Oct 6, 2016 #19
    Sorry! question set up is right. A simpler version (only a(x)) has already been asked in IITJEE 2007. See here question number 3.

    http://www.kshitij-iitjee.com/IITJEE-Past-Year-Papers/IITJEE-2007-Paper-1

    This question has been taken from a test paper of a reputed coaching institute for IITJEE.
     
  21. Oct 6, 2016 #20

    ehild

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    This is quite different problem from that in the OP. No need to consider the motion of the balls. It simply asks ax, x component of the acceleration, at the moment when the separation between the balls is x, and the balls are connected to the ends of a string of length a and a force F acts vertically at the middle of the string. A fairly easy problem. The one in the OP needs calculus - a differential equation to solve. Use the equation you have already : ax = - (F tanθ)/2m, but write tan(θ) in terms of x and the length of the string.
     
    Last edited: Oct 7, 2016
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