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Newton's 2nd law in circular motion

  1. Oct 19, 2007 #1
    1. The problem statement, all variables and given/known data

    Daytona International Speedway in Florida. Both of its courses feature four story, 31.0 degree banked curves with max radius of 316 if a car negotiates the curve too slowly it tends to slip down the incline of the turn whearas if it going too fast it may begin to slide up the incline. find the centripetal acceleration on the curve so won't slip up or down and calculate the speed (neglect friction

    2. Relevant equations


    3. The attempt at a solution

    I have the solution it is 5.89 m/s^2 and 43.1 m/s, but i don't understand why the normal force uses cos. It says use the y-component of Newton's second law to solve for the normal force n....so why am I supposed to use cos instead of sin and how can I know for future problems that are similar?

    ma= F = n + mg
    ncostheta -mg = 0
    n= mg/ costheta
    F = mg tan theta
    centripetal acceleration = g*tan*theta

    then v^2/r = 43.1
    Last edited: Oct 19, 2007
  2. jcsd
  3. Oct 19, 2007 #2
    I am having trouble understanding why cos is the y-component of New's second law in regards to solving for the normal force of a problem regarding circular motion (car on a banked racetrack) and in linear motion the y-component of Newton's second law was sin. How do I know whether the y component pertains to cos or sin for a circular problem or if it is always cos then why?!?!
  4. Oct 19, 2007 #3


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    You have to look at which angle you have labelled. Do you have a specific question?
  5. Oct 19, 2007 #4


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    Thanks to whoever combined these.

    If you draw a diagram it should become apparent. The curve is banked at 30 degrees, so calculating the y component of the weight you will need to use cosine.

    Try drawing the diagram, then if you are still stuck ask away.
  6. Oct 19, 2007 #5
    I can't seem to upload the paint picture I drew, but the components as they are drawn in in the book are a vertical line with n cos theta at the top arrow and -mg straight down from it and n sin theta horizontal and pointing to the left and the angle of the road is to the left of the picture. If I was to put these components on an axis it seems like the components of the normal force should be switched with sin corresponding to the y direction. (In linear motion sin was always in the y direction why does it seem different, what am I missing or what I'm I not seeing?) Thanks for your help!!
  7. Oct 19, 2007 #6
    has the problem, i don't understand the component diagram to the right of the car, it seems like if you line the axis up with the gravity component on the vertical axis then the normal force you should be using is n sin theta...before without the circular motion it was always mg sin of theta for the y-direction and ramps why do we now use cos for the y-direction doesn't sin always correspond to the y direction?
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