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Question on Motion of a car round a banked track

  1. Sep 15, 2007 #1
    Question on Motion of a car round a banked track [SOLVED]

    1. The problem statement, all variables and given/known data

    A racing car of 1000kg moves round a banked track at a constant speed of 108km/h. Assuming the total reaction at the wheel is normal to the track, and the horizontal radius is 100m, calculate the angle of inclination of the track to the horizontal and the reaction at the wheels.

    2. Relevant equations

    tan(theta) = v^2 / rg where r is the radius and g is 10m/s

    v = rw, w is the angular speed, r is the radius

    F = mrw^2 = mv^2 / r
    , F is the force towards the centre of the track, r is the radius, w is the angular speed

    3. The attempt at a solution

    Currently in a Junior College, not sure what that equates to in any part of the world but it's sort of a Pre-University education. So that should give you some idea on my knowledge.

    I tried drawing a vector diagram as attached. But I can't figure out whether the car is at the extreme of the track so that its horizontal distance is 100m from the track.

    However, I'm also unsure if the above equations with radius means horizontal radius. That said, I assumed it to be and I used the above equations to try solving but I couldn't get an answer. Spent over 2 hours and I'm stumped... Anyone care to help? Any help is appreciated!

    P.S. the Answer is 42 degrees and 13450N.
     

    Attached Files:

    Last edited: Sep 15, 2007
  2. jcsd
  3. Sep 15, 2007 #2
    The 100m means that the radius of the centrifugal force is 100m. The fact that the reaction at the wheel is normal to the surface means that the frictional force along the track and the centrifugal force along the track are equal. This should give you the angle.
     
  4. Sep 15, 2007 #3

    learningphysics

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    Draw the freebody diagram of the car... It is on an incline at an angle theta... what are the forces acting on it...

    EDIT: nevermind... I'll have a look at the picture you've attached when it is approved...
     
  5. Sep 15, 2007 #4

    learningphysics

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    The tan(theta) equation you have posted gives you the angle... can you show your calculations?
     
  6. Sep 15, 2007 #5
    Okay guys, thanks for your input. I forgot that by converting 108km/h gives me 30m/s. I was under the impression that it was 108 x 1000 x 3600, when it should have been
    (108 x 1000) / 3600 = 30m/s.

    So by using tan (theta) = v^2 / rg,

    (theta) = tan^-1 (900 / 100 x 10) = 41.9 ~ 42 degrees.

    I've yet to find the reaction force but I'm sure I can handle that.

    Thanks for the fast replies! Greatly appreciated.
     
  7. Sep 15, 2007 #6

    learningphysics

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    Although the formula works... you should be able to derive the formula yourself... using the freebody diagram... what are the vertical forces? what are the horizontal forces? what is the vertical acceleration? what is the horizontal acceleration?
     
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