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Sphere describing a circle, find tension

  1. Jun 29, 2009 #1
    A sphere of 3 kg describe a circle with a horizontal speed of 1,2 m/s. Knowing that L = 800 mm, determine (a) the angle theta (b) the tension in the wire

    Answer:
    (a) 24,2°
    (b) 32,3 N


    I find 2 equations

    ZFn = m . an
    T - Pcos(theta) = m . v^2/p

    ZFt = m . at
    Psin(theta) = m . at

    As calculated at? (tangential acceleration)
     

    Attached Files:

  2. jcsd
  3. Jun 29, 2009 #2

    tiny-tim

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    Hi Apprentice123! :smile:

    (I can't see the diagram yet, but I assume this is a string of length 800 mm with a sphere on the end rotating at a steady speed)

    i] there is no tangential acceleration … the acceleration is purely radial (centripetal)

    ii] you must balance the centripetal acceleration the weight and the tension, by taking components both horizontally and vertically …

    try again! :smile:
     
  4. Jul 1, 2009 #3
    Re: Sphere

    The two formulas that I found are:

    ZFn = m . an
    T - Pcos(theta) = m . v^2/p

    ZFt = m . at
    Psin(theta) = m . dv/dt
     
  5. Jul 2, 2009 #4

    tiny-tim

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    Apprentice123, there are no tangential forces, and no acceleration!!! :rolleyes:

    v is constant!

    You need vertical components, not tangential ones.

    What is the vertical acceleration, and what are the vertical forces? :smile:
    (btw, I've been wondering for ages … in what language does P stand for weight? :confused: :wink:)

    (oh, and have a theta for copying: θ :wink:)


    T needs a θ also, and you need to write p in terms of L.
     
  6. Jul 2, 2009 #5
    Re: Sphere


    Correct ?

    ZFn = m.an
    T - P = m . v^2/L
    T = 34,83 N

    ZFt = m.at
    at = 0
    Tcos(theta) - P = 0
    theta = 30,9°
     
  7. Jul 2, 2009 #6

    LowlyPion

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    Re: Sphere

    The centripetal acceleration is m*v2/r.

    But r in this case is not L. It is the radius about the point that it rotates, and that point is not where the string is tied.
     
  8. Jul 2, 2009 #7

    tiny-tim

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    Hi Apprentice123! :wink:

    Now I see where you got this from …
    Yes, that would be correct if the centripetal acceleration were along the string (the P has cosθ, and the T has no θ).

    But, as LowlyPion :smile: points out, it's horizontal.

    Centripetal acceleration has nothing to do with physics, it's just geometry

    never mind why it's going in a circle (in this case, because of the string) …

    all that matters is the shape of the circle, and the speed …

    it's a horizontal circle, so the centripetal acceleration is also horizontal. :smile:
     
  9. Jul 2, 2009 #8
    Re: Sphere

    Ok. Thanks
     
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