1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    ybzk0dhm.iyi.png

    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
    Yes

    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

    Matterwave

    User Avatar
    Science Advisor
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted