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Magnitude and direction of the force of the seat on you?

  1. Dec 14, 2008 #1
    1. The problem statement, all variables and given/known data
    One fine spring day you decide to skip class and go to the amusement park to ride the loopthe-
    loop roller coaster. At the bottom of the loop, you’re moving with a speed of 20 m/s in a vertical circle 20 m
    in diameter. What’s the magnitude and direction of the force of the seat on you? Assume your mass is 50 kg.


    2. Relevant equations

    Vt=omega x radius
    Fr=mass x omega2 x radius

    3. The attempt at a solution

    Vt=omega x radius
    20m/s=omega x 10m
    omega=10 rad/s

    Fr=mass x omega2 x radius
    Fr=50kg x 10rad/s x 10m
    =5,000 N

    I'm not sure what I'm doing at all on this problem please help me out.

    I drew the FBD and the only forces are the Normal which I assume is the seats force on the rider which I further moved towards thinking that would be the Force towards the center thus the equation
    Fr=mass x omega2 x radius, and the other force is (of course) g.

    Please help me out I am in a bit of confusion.

    Thank you for reading!

    -Charan
     
    Last edited: Dec 14, 2008
  2. jcsd
  3. Dec 14, 2008 #2
    Re: Loop-the-loop

    F(centripetal)=V^2/R because you are moving in a circle there is a centripetal force, always. At the bottom it is opposite of weight and the normal you feel is w-Fc at the top it is creating a normal if I remember correctly. so It is Fc+W=N
     
  4. Dec 15, 2008 #3
    Re: Loop-the-loop

    Oh I see thank you I also just found it in my book

    at the bottom N=mg+mv2/r
    and at the top N=mv2/r - mg

    so in this case it is asking for the Normal force at the bottom and that is the first equation.

    so:

    N=(50kg)(g) + (50kg)(20m/s)2/10m
    N=2,490N
     
  5. Dec 15, 2008 #4
    Re: Loop-the-loop

    Charan,

    Just so you know how to get that equation and not memorize it, here are some steps you might take to help you understand why that is the case.

    First draw a free-body and analyze the forces acting on you due to the seat. After drawing the diagram you will see that you have a Fn going up and a Fw going down. So if you use newtons second law you get.

    Efy=ma(c)

    Fn-Fw=m(V^2/r) -----> Fn=Fw+m(V^2/r)

    It will probably stick better if you understand why those equations are true
     
  6. Dec 15, 2008 #5

    PhanthomJay

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    Science Advisor
    Homework Helper
    Gold Member

    Re: Loop-the-loop

    That's all very well, but please don't blindly find formulas in the book and try to apply them without understanding them.
     
  7. Dec 16, 2008 #6
    Re: Loop-the-loop

    When you are at the bottom, then Fseat is upward and your weight Mg is downward.
    Therefore Fr = Fseat - Mg
    Also Fr = Mv^2/r
    Therefore Mv^2/r = Fseat - Mg
    Fseat = Mv^2/r + Mg
    Solve the above to calculate Fseat.

    For centripetal force, you have used M*omega^2*r. That is also correct, but in this problem, v is given. So it is easier to use Mv^2/r
     
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