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5 October, 22:22

A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to five times her weight as she goes through the dip. If r = 17.0 m, how fast is the roller coaster traveling at the bottom of the dip

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  1. 5 October, 22:23
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    16.3 m/s

    Explanation:

    Let m = mass of the passenger,

    v = speed of the roller coaster

    Forces on the passenger are

    1. Force by seat = 2.18 mg in upward direction.

    2. Weight mg in downward direction

    Therefore, net upward force = 2.18 mg - mg = 1.18 mg

    Also, net upward force = centripetal force = mv^2/r

    Therefore,

    1.18 mg = mv^2/r

    Dividing by m,

    1.18 g = v^2/r

    Or v^2 = 1.18 gr

    Or v = sqrt (1.18 gr)

    = sqrt (1.18 * 9.81 * 22.9)

    = sqrt (265)

    = 16.3 m/s
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