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Conservation of energy

  1. Oct 28, 2013 #1
    1. The problem statement, all variables and given/known data

    A 2 kg ball is attached to the bottom end of a length of fishline with a breaking strength of 44.5 N. The top end of the fishline is held stationary. The ball is released from rest with the line taut and horizontal (theta = 90 degrees). At what angle theta (measured w/vertical) will the fishline break?


    2. Relevant equations

    none i can think of

    3. The attempt at a solution

    [tex] F=\dfrac{mv^2}{l} [/tex] where l is the length of the wire. then from conservation:

    [tex] mgh = \dfrac{1}{2}mv^2 [/tex]

    [tex] mg(l \sin \theta) = \dfrac{1}{2} ( F\cdot l) [/tex]

    [tex] F = 2mg\sin \theta = 44.5 [/tex]

    Which is incorrect as sin theta must be < 1.
     
  2. jcsd
  3. Oct 28, 2013 #2

    rock.freak667

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    Homework Helper

    At the angle θ, what would be the components of the forces radially and tangentially and in what direction would the tension T act?


    After that, how would you find the centripetal force in terms of these forces?
     
  4. Oct 29, 2013 #3
    Sorry I still can't get it with your information. Any other hints/solution?
     
  5. Oct 29, 2013 #4
    Account for the weight opposing tension force.

    (At the bottom, Fnet = 2mg = Tension - mg(weight), so max tension would have to = 60N for the line to never break.)
     
  6. Oct 29, 2013 #5

    Doc Al

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    Staff: Mentor

    Two comments:
    - You have sine and cosine reversed; note that θ is measured from the vertical.
    - What you call F is the radial component of the net force on the ball, not the tension in the string.

    Draw a free body diagram of the forces on the ball.
     
  7. Oct 29, 2013 #6
    Oops, is this correct now:

    [tex]F=T-mg\cos\theta \implies T-mg\cos\theta = \dfrac{mv^2}{l}[/tex]
    Where l is the length of the line.

    [tex] mgh = \dfrac{1}{2}mv^2\implies mg(l\cos\theta) = \dfrac{1}{2}(T-mg\cos\theta)\cdot l [/tex]

    [tex] 2\cdot 9.8\cos\theta = \dfrac{1}{2}(44.5-2\cdot 9.8\cdot\cos\theta) [/tex]

    [tex] \theta = 40.82^{\circ} [/tex]

    Please correct if it's wrong.
     
  8. Oct 29, 2013 #7
    It looks good to me
     
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