What tension must a 40.4 cm length of string support in order to whirl an attached 1,000.0 g stone in a circular path at 2.53 m/s?

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  • Anonymous
    6 months ago

    It depends if the circle is in a horizontal or a vertical plane, which you haven't stated.

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  • 6 months ago

    here is a very similar Q/A, just change a few numbers and recalculate

    v = 2.59 m/s

    r = 41.8 cm

    there are two forces acting on the stone, both of which cause tension.

    assuming this circular path is horizontal.

    1. Centripetal force f = mV²/r = mrω²

    ω is angular velocity in radians/sec

    1 radian/sec = 9.55 rev/min

    m is mass in kg

    r is radius of circle in meters

    V is the tangental velocity in m/s = ωr

    f is in Newtons

    F = (1 kg)(2.59 m/s)² / 0.418 m = 16.05 N

    2. force of gravity F = mg = 1•9.8 = 9.8 N

    Those two are orthogonal, therefore the sum of the two is

    F = √(16.05² + 9.8²) = 18.80 N

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  • oubaas
    Lv 7
    6 months ago

    T = m*V^2/r = 1.00*2.53^2/0.404 = 15.8 N

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