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Conservation of Energy - Swing

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data

    A child of mass M holds onto a rope and steps off a platform. Assume that the initial speed of the child is zero. The rope has length R and negligible mass. The initial angle of the rope with the vertical is [tex]\theta[/tex]0.

    [​IMG]

    (a)

    Using the principle of conservation of energy, develop an expression for the speed of the child at the lowest point in the swing in terms of g, R, and cos [tex]\theta[/tex]0.

    (b)

    The tension in the rope at the lowest point is 1.5 times the weight of the child. Determine the value of cos [tex]\theta[/tex]0

    (Known data)

    • Initial Speed = 0 (v0 = 0)
    • Rope length is R
    • Rope mass is negligible (0)
    • Initial angle from vertical is [tex]\theta[/tex]0.



    2. Relevant equations

    • [tex]\Delta[/tex]V + [tex]\Delta[/tex]K = (Vf - Vi) + (Kf - Ki)
    • Vf + Vk = Vi + Ki
    • PE = mgh
    • KE = (1/2)mv2 = mgh
    • v2 = 2gh
    • v2c = 2gha



    3. The attempt at a solution

    At point C, PE = 0
    At point C, child has lost mgha in PE, where ha = R (length of rope)
    In PEs place, child has gained KE = (1/2)mv2

    Results in V2 = 2gha = 2 (g)(R) = V2
    Above does not acount for different angle from 90 degrees.

    Therefore, we use h = R - Rcos[tex]\theta[/tex]0.

    a)

    PE[tex]\theta[/tex] = (M * g) * (R - Rcos[tex]\theta[/tex] = PE at [tex]\theta[/tex]).

    (1/2)(1/2)(v0max) = PE[tex]\theta[/tex].

    (1/2)(1/2)(v0max) = (M * g) * (R - Rcos[tex]\theta[/tex]).

    v2 = (M * g) * (R - Rcos [tex]\theta[/tex] ) / (1/2)(1/2)

    Solve for v (cant write out correctly with LaTex)

    So, does this look correct?

    b)

    I dont seem to understand how to solve this.
     
    Last edited: Oct 18, 2007
  2. jcsd
  3. Oct 19, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    Hmmm... a little confused by the 1/2 's

    this is right:

    PE at theta = (M * g) * (R - Rcostheta)

    so total energy at theta = (M * g) * (R - Rcostheta)

    at the bottom potential energy = 0.

    kinetic energy = (1/2)Mv^2

    set: (1/2)Mv^2 = (M * g) * (R - Rcostheta)

    solve for v...

    for part b)... try to get v at the bottom using centripetal motion ideas... then you can solve for theta using the formula from part a).
     
  4. Nov 24, 2007 #3
    at theta do you mean the lowest point? because

    "Using the principle of conservation of energy, develop an expression for the speed of the child at the lowest point in the swing in terms of g, R, and cos LaTeX graphic is being generated. Reload this page in a moment.0."
     
  5. Nov 24, 2007 #4
    as for part B it is easy after you get part a...

    Fnet = ma

    T - mg = ma

    1.5mg - mg = ma
    1.5mg - mg = (mv^2)/R

    plug the answer you got in part a for v (it is (2Rg(1-costhetha))^1/2 <--- square root) then everything should cancle and you get costheta = .75

    cheers!
     
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