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    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    A kid swings from a rope 8.5 m long on a stationary pole on a playground. His body mass is 70 kg and he is moving 6.0 m/s when he passes through the lowest part of his path.
    What is the tension in the rope at the lowest part of his path?
    When he reaches the highest point, what angle does the rope make with the vertical?

    © BrainMass Inc. brainmass.com September 24, 2022, 5:54 pm ad1c9bdddf
    https://brainmass.com/physics/circular-motion/circular-motion-and-tension-in-a-rope-38319

    SOLUTION This solution is FREE courtesy of BrainMass!

    At the lowest part the forces acting are
    Gravitational force downwards
    Tension in the rope upwards
    The net force = Tension - mg provides the centripetal force = m v^2/r for the kid to move in a circular path

    m= 70 kg
    g= 9.81 m/s^2
    v= 6.0 m/s
    r= 8.5 m

    T-mg = mv^2/r
    or
    T=mg + mv^2/r = 983 N =70x9.81+70x6^2/8.5

    tension in the rope at the lowest part of his path= 983 N

    To calculate the maximum height (h)to which the kid reaches
    we use the equation v^2= 2 gh

    v= 6.0 m/s
    g= 9.81 m/s^2

    Therefore h= 1.83 m =6^2/ (2 x9.81)

    Cos theta = (r-h)/r (theta is the angle with the vertical)
    r= 8.5 m
    h= 1.83 m

    Substituting the values:
    Cos theta = (8.5-1.83)/8.5= 0.7847

    or theta = 38 degrees

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com September 24, 2022, 5:54 pm ad1c9bdddf>
    https://brainmass.com/physics/circular-motion/circular-motion-and-tension-in-a-rope-38319

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