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Math Help - verticle circle radius

  1. #1
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    verticle circle radius

    I believe this is another Verticle circular motion problem, they still confuse me, why is it i must add the net forces for the answer, and do i use an equation (m * v^2) / r more than once?

    please help, i'm not sure what is the logic behind the equation.

    heres the problem:

    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 twice her weight as she goes through the dip.

    If r = 16.5 m, how fast is the roller coaster traveling at the bottom of the dip?

    in m/s
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  2. #2
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    Quote Originally Posted by rcmango View Post
    I believe this is another Verticle circular motion problem, they still confuse me, why is it i must add the net forces for the answer, and do i use an equation (m * v^2) / r more than once?

    please help, i'm not sure what is the logic behind the equation.

    heres the problem:

    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 twice her weight as she goes through the dip.

    If r = 16.5 m, how fast is the roller coaster traveling at the bottom of the dip?

    in m/s
    One would suspect either that you don't read my posts or you don't tell me if you are confused by the answers/hints that I give you.

    As always, start with a Free-Body Diagram. I will choose a +y direction upward. There is a weight on the coaster, w, straight down. There is an upward (presumably a normal) force equal to 2w acting straight up. That's it for the vertical direction.

    So what is the net force on any object moving in a circular path? Right! It's the centripetal force. Thus
    \sum F = -w + 2w = F_c

    I will leave you with the rest.

    -Dan
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  3. #3
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    Okay then i reached the answer using this equation: v = sqrt(r * g)

    however, i don't believe i am supposed to solve this problem using that.

    also, i am confused with the formula you refer to: F = -w +2w = Fc

    can you break this down in an explanation of each variable, or what it really means.

    Thankyou.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    Okay then i reached the answer using this equation: v = sqrt(r * g)

    however, i don't believe i am supposed to solve this problem using that.

    also, i am confused with the formula you refer to: F = -w +2w = Fc

    can you break this down in an explanation of each variable, or what it really means.

    Thankyou.
    Where did the v = \sqrt{rg} come from?

    We are looking at the car at the bottom of the circular dip. The net force on any object moving in a circular path is equal to the centripetal force. The Free-Body Diagram for the girl at this point contains two forces: the weight w acting straight down, and the normal force N acting straight up. So Newton's 2nd says:
    \sum F = -w + N = F_c
    (where I have chosen the positive direction as upward.)

    The problem states that the car is acting on the girl with a force magnitude of twice her weight. Thus N = 2w. Thus
    \sum F = -w + 2w = \frac{mv^2}{r}

    w = \frac{mv^2}{r}

    mg = \frac{mv^2}{r}

    v^2 = rg

    v = \sqrt{rg}
    (Now you can use this equation since you've just derived it.)

    -Dan
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