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Circular motion and tension

  1. Oct 23, 2006 #1
    [​IMG]
    A mass of 8.700 kg is suspended from a 1.490 m long string. It revolves in a horizontal circle.
    The tangential speed of the mass is 3.755 m/s. Calculate the angle between the string and the vertical (in degrees).


    There's the question I'm given. I cannot get terribly far with this one, despite being able to ace all the other questions on the assignment... Nonetheless, that is moot.

    There are 3 unknowns here for this problem... Fortunately, one of the unknowns can be written in terms of the other one. The radius of the lower circle can be written as (r)(sin(theta)) which is in turn (1.490)(sin(theta)). I believe tension can be written as T=mg/(cos(theta)), but I'm not entirely sure.

    I've attempted to do (mv^2)/(r(sin(theta)) = (mg)/(cos(theta)) and work it through to get ((v^2)/(gr)) = (tan(theta)), but it's not giving me the correct answer.

    I'm kind of stuck, not sure where I'm going wrong with this.
     
    Last edited: Oct 23, 2006
  2. jcsd
  3. Oct 23, 2006 #2
    Split the tension of the rope into horizontal and vertical components and then apply the second law in both directions.
     
  4. Oct 23, 2006 #3

    rsk

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    Did you get there yet? What is the answer? I tried it by eliminating all but theta and if I'm right, I'll give you some clues.
     
  5. Oct 23, 2006 #4
    Tx = (mv^2)/(sin(theta)r) and Ty=(mg)/(cos(theta))

    I can do that, sure, Neutrino. Just not sure what else to do with it from there. I have neither tension, theta, or the radius of the conical base. Only the length of the string, the mass, and the velocity.
     
  6. Oct 23, 2006 #5

    rsk

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    So eliminate all but one variable ....

    by resolving vertically you can get T in terms of theta and you already have R in terms of theta, so you now have just one variable.

    You say your answer isn't correct. What is the correct answer?
     
  7. Oct 23, 2006 #6
    I keep getting 44.0 degrees, but I think I know what I'm doing wrong, so I'll get back to you on that after playing around with the formula some more.
     
  8. Oct 23, 2006 #7

    rsk

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    If you know what the correct answer is I might be able to give you some clues.

    I have an answer, but would like to know it's correct before leading you up the garden path.
     
  9. Oct 23, 2006 #8

    radou

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    Me too, me too! :biggrin: :tongue2:
     
  10. Oct 23, 2006 #9

    rsk

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    GO on radou, what did you get? 51.1?
     
  11. Oct 23, 2006 #10

    radou

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    Umm...nope. :smile: I'd rather not get involved. :biggrin:
     
  12. Oct 23, 2006 #11

    rsk

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    oh well, bedtime.
     
  13. Oct 23, 2006 #12

    radou

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    Last off-topic post: well, it sure is long past bedtime where I'm from. :smile:
     
  14. Oct 23, 2006 #13
    According to the computer, that's the answer I should be getting... So, yes, 51.1 is the correct answer. I'm just not quite able to get that.

    I'm betting I'm just running into one little snag in the process that's messing it up.
     
  15. Oct 23, 2006 #14

    OlderDan

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    The problem is that one of these equations is not correct. Draw the FBD with the tension and gravity acting. Resolve the tension into components, and be careful about what you are equating. The mass is moving horizontally in a circular path, so the horizontal component of the net force is the centripetal force. The net vertical force is zero.
     
    Last edited: Oct 24, 2006
  16. Oct 24, 2006 #15

    rsk

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    OK venom. If it's not too late here's what I did. If is IS too alte - sorry - but I did offer this hours and hours ago!

    First I resolved vertically. That gives you an expression for T in terms of theta - in terms of cos(theta) to be precise - which I think you already wrote somewhere earlier.

    Then I resolved horizontally. You put R = Lsin theta (L = total length) and this gives you an expression for T in terms of theta - but it's got a sin^2(theta) in it.

    You now have two expressions for T - one in terms of cos(theta) and one in terms of sin^2(theta)

    Equate them, and remember that sin^2(theta) = 1 - cos^2(theta)

    Now you have a quadratic equation where cos(theta) is the variable

    Solve it and you can find theta.
     
  17. Oct 24, 2006 #16

    rsk

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    OK venom. If it's not too late here's what I did. If is IS too alte - sorry - but I did offer this hours and hours ago!

    First I resolved vertically. That gives you an expression for T in terms of theta - in terms of cos(theta) to be precise - which I think you already wrote somewhere earlier.

    Then I resolved horizontally. You put R = Lsin theta (L = total length) and this gives you an expression for T in terms of theta - but it's got a sin^2(theta) in it.

    You now have two expressions for T - one in terms of cos(theta) and one in terms of sin^2(theta)

    Equate them, and remember that sin^2(theta) = 1 - cos^2(theta)

    Now you have a quadratic equation where cos(theta) is the variable

    Solve it and you can find theta.
     
  18. Oct 24, 2006 #17
    Ah-ha! Perfect! Thanks guys. That really helped me out!
     
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