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Circular Motion - Banked Curve

  1. Nov 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Designing an on ramp for the 401 the engineer wants cars to be able to make the turn with a radius of 50 m while travelling 40km/hr in conditions with no friction. What angle must he bank the curve at to make this possible?


    2. Relevant equations

    FR = (m)(aR)
    FR = (m)(v2)/(r)


    3. The attempt at a solution

    I first started by drawing a diagram and noting the values I have for the question which are the velocity (40 km/h) and the radius (50 m). The value I we're looking for in the question is the angle the bank makes.

    I first know FN(Sin θ) = (m)(v2)/R

    which I can then isolate FN through which becomes

    FN = (m*g)/(cos θ)


    Since in the actual equation has no movement in the y axis it is:
    FNsinθ = m * v2/r

    Then I sub in the FN I got from the previous equation and sub it into this one getting:

    (m*g/cos θ)(sin θ) = m * v2/r

    which then simplifies into:

    tan θ = v2/r*g

    I then put in the original values I had into the equation and got 3.26 then I put 3.26 into the tan-1 thing in the calculator and got 73o degrees.

    Is my answer and thought process correct?
     
  2. jcsd
  3. Nov 27, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Your thought process is perfect but your answer is not. (Convert the speed to standard units. :wink:)
     
  4. Nov 28, 2013 #3
    Converted 40 km/h to 11.111 m/s.

    Now I input the following into the formula:

    tan θ = (11.1 m/s)2/(50)(9.8)

    Which equals 0.2514489...

    I then input that into the tan-1 function into the calculator and get:

    14.11o degrees

    Is this correct?
     
  5. Nov 28, 2013 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good!
     
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