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Finding the angle of a car moving on a circular track

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    A circular automobile racetrack is banked at an angle X, to the horizontal, such that no friction between the road and tires are required when a car travels at 30.0 m/s. The radius of the track is 400m. What is the angle?

    The problem i have is setting up the problem. I have drawn a force diagram for the car moving on the incline, but i am very confused relating it to the slope. The only thing i can think of is to take the components of the radius of the track, but in all honesty i am having issues doing this.

    Tips?
     
  2. jcsd
  3. Oct 5, 2011 #2

    Doc Al

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    Staff: Mentor

    What forces act on the car? In what directions?

    Once you get the forces correct, apply Newton's 2nd law. Hint: In what direction is the car accelerating?
     
  4. Oct 5, 2011 #3
    my force diagram has the normal force perpendicular to the incline, the weight perpendicular to the ground, and going in a forward motion with friction in the opposite. The acceleration though is towards the center of the circle I split it into components of X and Y.
    Sum of the forces in the Y direction= 0,
    sum of the forces in the x direction = (mass of the car x radial acceleration which was 2.25 m/s^2)- friction (which here is 0)

    i feel like i am forgetting a force tho
     
  5. Oct 5, 2011 #4

    Doc Al

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    Good, except for the bit about friction--there's no friction here.

    Good. So what are the force components in the y direction?
    Good. So what are the forces in the x direction?

    You need to set up those two equations, putting in the forces. When you take the components, you'll get functions of the angle.
     
  6. Oct 5, 2011 #5
    In the y direction we have Normal force and weight perpendicular to the group (weight times sinX)

    In the X direction we have just radial acceleration of the car, i cannot think of anything else

    I can relate the two because of neuton's second law that the sum of the forces = mass x acceleration. so N -wsinX=2.25. Here tho this feels wrong...
     
  7. Oct 5, 2011 #6

    Doc Al

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    This is a bit mixed up. The weight is vertical and thus is totally in the y-direction. But what's the y-component of the normal force?
    That's the acceleration, but what about the force? Which of the two forces acting on the car has an x-component.

    (Note: I am assuming that x and y refer to horizontal and vertical.)
     
  8. Oct 5, 2011 #7
    the y component of the normal force then would be Normal force x sin(X)

    I'm not quite sure i understand how to get the force from the acceleration without knowing the mass of the vehicle.

    And yes ur note is correct
     
  9. Oct 5, 2011 #8

    Doc Al

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    No, not exactly. You're mixing up your sine and cosine.

    Don't worry about the 'ma' part of the equation. First figure out ƩFx. (You'll find that you won't need the actual mass. Just call it 'm' and see what happens.)
     
  10. Oct 5, 2011 #9
    ƩFx = 2.25 m/s^2 times mass = mass x acceleration. It simplifies to 2.25 = a
    ƩFy = N (the normal force) x cos(X) - mass x gravity = mass times acceleration.

    I plugged in the acceleration i found from ƩFx, but i couldnt simplify it enough to solve it
     
  11. Oct 5, 2011 #10

    Doc Al

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    All you've done here is give the acceleration. You have not evaluated ƩFx. (You need to do something similar to what you did for ƩFy.)

    Good. But what's the acceleration in the y direction?

    Think like this. There are only two forces, the weight and the normal force. List the x and y components of each. To find ƩFx just add up the x-components; To find ƩFy just add up the y-components (you did that one).
     
  12. Oct 5, 2011 #11
    the only force that i know for ƩFx is the acceleration times the mass bc friction is 0

    the acceleration in the y direction should b 0 because there is no movement up or down
     
  13. Oct 5, 2011 #12
    oooo i have the x component of the normal force. so Ncos(X) = 2.25 m/s^2
     
  14. Oct 5, 2011 #13
    woot thank you very much for your help! i've figured it out to the end <3
     
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