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Physics

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A loop-the-loop ride at an amusement park has a radius of 5 m. At the highest point, the rider moves at 8 m/s. At the lowest point the rider moves at 13 m/s. Find what you would feel like you weigh at those points. (This force represents the force of the seat against your rear end; its F(app) and it takes the place of Tension for the ball on the string.)

I don't understand. If I replace Tension in the following equation, I will have 2 unknowns:

(highest point): T+mg=(mv^2)/R
F(app)+mg=(mv^2)/R
(lowest point): T-mg=(mv^2)/R
F(app)-mg=(mv^2)/R

  • Physics - ,

    i agree with your result, but tension is not involved.

    At the top of the loop, the seat force F PLUS the weight provides the centripetal acceleration, and

    F = MV^2/R -Mg

    At the bottom of the loop,

    F = M V^2/R + M g

  • Physics - ,

    Forgive my ignorance, but how do I find the mass when I do not know the seat force?

  • Physics - ,

    OK, so they don't tell you the mass. All you can say in response is BY WHAT FACTOR the force exceeds your weight at the bottom or is less at the top of the arc.

    At the top of the arc, V^2/R = 12.8 m/s^2
    F = M (12.8 - 9.8) = M* 3 m/s^2.

    This is 3/9.8 or 30.6% of the weight of the person

    At the bottom of the arc, V^2/R = 33.8 m/s^2, and the seat force is 4.44 times the weight.

  • Physics - ,

    I followed up to this point:

    F = M (12.8 - 9.8) = M* 3 m/s^2.

    I do not understand why we must divide by g (9.8).

  • Physics - ,

    You are comparing actual forces exerted by the seat to the weight of the person, which is Mg. That invoves taking a ratio, Force/(M*g).

    Since they don't tell you the person's mass, all you can calculate is force ratios

  • Physics - ,

    Thanks drwls!

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