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Tension in a string attached to a bob moving in horizontal circles

  1. Aug 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Calculate the tension in a 2m string attached to a 2kg bob that is moving in
    horizontal circles of 0.5m radius.

    2. Relevant equations

    I thought it was F=MA but the working out shows all sorts

    3. The attempt at a solution

    I know the answer is 78.4N but dont know how to get there.
  2. jcsd
  3. Aug 13, 2011 #2


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    Homework Helper

    have you worked out this is a conical pendulum?

    If your text book has nothing anout analysing them, you could check this wikipedia address if you want to check what it means.

  4. Aug 13, 2011 #3
    the book gives:

    θ = tan-1 0.5/2
    = 14º

    Tcosθ = mg

    T = mg / cosθ

    T = 2*9.80 / 0.25

    T = 78.4N.

    I dont know why its carrying out those particular calculations

    Surely this should not be using tan-1 either since we only have the hypotenuse and opposite values.
    shouldnt it be Sine?
    Last edited: Aug 13, 2011
  5. Aug 13, 2011 #4
    Also just noticed that cos 14 isnt 0.25, its 0.97
  6. Aug 13, 2011 #5


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    Homework Helper

    I agree, the angle should be worked out with a sine function, not tan.
  7. Aug 13, 2011 #6


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    Staff: Mentor

    It would seem then that your book has some errors in the solution provided. You should be working to a diagram similar to this:


    Attached Files:

  8. Aug 13, 2011 #7
    Would i be correct by saying the answer should be:

    20.2N and not the one stated?

    How come it changes to cosine once i have calculated the angle?

    The amount of errors i have found on this course is disgusting, every other page there is an error. I would never reccomend anyone to ICS after what i have seen on this course.

    Thanks a lot for the help so far:)
  9. Aug 13, 2011 #8
    How come it changes to cosine once i have calculated the angle?

    The reason it changes is because you're looking for a different side of the triangle.
  10. Aug 13, 2011 #9
    Im looking for another side of the triangle, am i? i didnt think i was
  11. Aug 13, 2011 #10
    Yes, because that's the only one equal to mg.
  12. Aug 13, 2011 #11


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    Staff: Mentor

    If you want to bypass the trig functions altogether you can use Pythagoras' theorem and similar triangles. Use Pythagoras to find the length of the vertical leg of the big triangle, then construct suitable ratios for the triangle sides.
  13. Aug 13, 2011 #12
    So is 20.2N correct just so i know i'm doing this right?

  14. Aug 13, 2011 #13


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    Staff: Mentor

    Yes, 20.2N is good.
  15. Aug 13, 2011 #14
    thanks a lot, i shall be having some nice words with ICS about this:)
  16. Aug 14, 2011 #15
    Next question i dont where to start.

    Calculate the angular speed for the bob in the previous question?

    the working out gives:

    cosθ = g/1w2

    ω2 = g/1cosθ

    ω2 = 9.8/2cos14

    ω2 = 9.8/2*0.97

    ω2 = 5.05
    ω = 2.25 rad s-1.

    Now i thought to get angular speed it is w=angular displacement/time

    But this doesnt appear to be doing any of that

    Thanks in advanced
  17. Aug 14, 2011 #16


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    Staff: Mentor

    They are equating the ratio of the gravitational acceleration of the mass to the centripetal acceleration due to its rotation, with the ratio of the length of the long leg of the large triangle to the radius leg:


    In the diagram, [itex]y = L cos(\theta)[/itex]. The centripetal acceleration of the mass as it rotates with radius r is given by [itex] a_c = \omega^2 r[/itex].

    Attached Files:

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