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Finding the tension in a circular pendulum without radius or angle.

  1. Apr 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Some kid is playing with a yoyo of mass m. The yoyo string is let out to length L, and is spun in a horizontal circle at a constant rate of ω. The yoyo string makes an angle of θ with the horizontal

    m = 39 grams = 0.039 kilgrams
    L = 46cm = 0.46m
    ω = 3 rads/sec

    Calculate the tension in the string in Newtons.

    2. Relevant equations

    Tx = Tcosθ = mω2r = mv2/r

    Ty = Tsinθ = mg = 0.3822N

    r = L cosθ

    h = L sinθ

    3. The attempt at a solution

    The vertical component of the tension was easy, the only force acting in this direction is gravity with a force of m*g newtons.

    The horizontal component is more confusing... Since the height, radius and length of string (hypotenuse) form a right triangle, the lengths of sides should correspond to the ratios on them.

    But I don't seem to know the vertical length, just the force's magnitude.
    I'm trying to solve for the hypotenuse's force's magnitude, but only know the length.
    And I don't seem to know anything at all about the horizontal length/radius of the circle has been formed or its force.

    So using Fx^2 + Fy^2 = T^2 or x^2 + y^2 = L^2 is out. The angular velocity seems hard to use without knowing the radius.

    Whats a step that takes me to finding the tension, radius, or θ?
  2. jcsd
  3. Apr 21, 2014 #2
    Welcome to PF!

    This should be sufficient to get tension in the string in terms of given parameters.
  4. Apr 21, 2014 #3
    Do you mean

    2Lcosθ ?

    That still leaves me with an unknown of T and an unknown of θ
  5. Apr 21, 2014 #4
    You do not require θ .

    Tcosθ = mω2Lcosθ .
  6. Apr 21, 2014 #5

    So T = mω2L...

    It seems to work, tough I'm still not sure how ω corresponds to 1/s^2. I'm so used to translating it linear units via the radius.

    And that the force along the hypotenuse is less than its y component... Blowing my mind.
    Last edited: Apr 21, 2014
  7. Apr 21, 2014 #6

    Sorry...I do not understand your question.The net force in the radial direction is Tcosθ and this force is equal to the centripetal force given by mω2r .Here the radius is Lcosθ .

    Could you reexpress your doubt ?

    No..The force along the hypotenuse is the tension T .The component of which in the vertical direction i.e the y-component is Tsinθ which is less than T .
  8. Apr 21, 2014 #7
    But the mass * gravity comes to weight of 0.3822 newtons, and the tension is only 0.1615 newtons.
  9. Apr 21, 2014 #8
    You haven't taken the normal force from the ground into account. The force equation in the y-direction will be Tsinθ + N = mg .
  10. Apr 21, 2014 #9
    It never even occurred to me that the yoyo was in contact with the grounds. Thanks!
  11. Apr 21, 2014 #10


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    Gold Member

    This is a horrible problem in my opinion. The problem never specified that there was contact with a ground, but the numbers they give will not work unless there is some additional force holding the yoyo up. There is, in general, a minimum angular velocity necessary for the yoyo to be suspended by the tension force of the string alone.
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