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Homework Help: Caluclating tension in string that is moving in horizontal circles.

  1. Dec 5, 2011 #1
    Hi all,

    First of all I'd like to say thank you to you all for providing a great forum which has been a great help to me working through my ics home learning course.

    As i have learnt on here and other places on the internet the information I have in my course materials is not always correct and can be very confusing causing hours of unnecessary extra work.

    The question I am very stuck on is: Calculate the tension in a 3m string attached to a 3kg bob that is moving in horizontal circles of 0.6m radius.

    All the information I have on working this out shows that tan should be used to work out the angle, I have questioned this with my tutor and he doesn’t seem to have an answer for me. (What chance have I got?)

    I have always been very interested in physics and have enjoyed learning throughout the course. I just can’t seem to find a simple clear formula for working this question out?

    Any help greatly appreciated
    Many thanks,Dave.
  2. jcsd
  3. Dec 5, 2011 #2
    Which forces do you think are acting on the bob?
  4. Dec 5, 2011 #3
    That would just be gravity?
  5. Dec 5, 2011 #4
    The pull of gravity is not the only force.

    Imagine you are the bob. Which pulls or pushes would you feel?
  6. Dec 5, 2011 #5
    I guess a force would be created with the momentum?
  7. Dec 5, 2011 #6
    There are only two forces:

    the pull of gravity mg and the tension in the string.
  8. Dec 5, 2011 #7
    Ok so I think I have got it:

    3x9.81=29.43N (force acting down on the bob)

    So now I need to work out the angle and get the force acting on the string?
  9. Dec 5, 2011 #8
    What kind of force does a particle need so that it can move in a circle?
  10. Dec 5, 2011 #9
    I think I finally have the answer.

    angle = tan-1 0.6/3
    angle = 11.30993247


    So the tension on the string will be 30N?

    The main thing that has confused me is that i read on here that tan shouldent be used also there is a mistake in my learning material.
  11. Dec 5, 2011 #10
  12. Dec 5, 2011 #11
    '...angle = tan-1 0.6/3...'

    note that tan(angle) is NOT 0.6/3

    0.6/3 = sin(angle)
  13. Dec 5, 2011 #12
    Ok i'm very confused now, thats the way a simular problem just different mass and string lengh is worked out in my learning material - is that wrong then?
  14. Dec 6, 2011 #13
    It all depends on which angle one takes.
    A diagram of the forces is very important in a problem like this.
  15. Dec 6, 2011 #14
    I have drawn a diagram that definitely helps.

    I have attached an example of the same question with different figures (notice the mistake cos14 is not 0.25?) I have also used this example to work out my problem and have written it to the side.

    Would you mind taking a look at this?

    Many Thanks

    Attached Files:

  16. Dec 6, 2011 #15
    I always start with a diagram of the forces acting so that I try to understand how the equations are obtained.

    Attached Files:

  17. Dec 6, 2011 #16
    Thats great.

    So I now have

    [itex]\theta[/itex] = sin-1 06/3

    [itex]\theta[/itex] = 11.53695903

    t = mg/cos[itex]\theta[/itex]

    t = 3x9.81/0.9797958971

    t = 30.03686797N

    That definitely clears up everything, Thankyou very much for your help.
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