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Centripetal Acceleration on a car

  1. Jun 19, 2011 #1
    Hey guys,

    Ok so for my final year project I've researched about traction and slip, etc, and I came across the traction circle, where each tyre has a specified 'budget' of friction that should never be exceeded during cornering. Now the longitudinal acceleration exerted will be different for each wheel, but my question was does the lateral acceleration exerted by each wheel have any correlation so the acceleration experienced by the centre of gravity? I was planning on using an accelerometer at the COG of the model car I had to be able to give an indication of the traction budget for the car, until I realised that the traction must be specified at each wheel (which would be tricky to do).

    Any ideas much appreciated.
  2. jcsd
  3. Jun 19, 2011 #2


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    Hey Raag90! :smile:
    If an object has instantaneous radius of curvature r, and speed v, then the https://www.physicsforums.com/library.php?do=view_item&itemid=27"(lateral acceleration) is v2/r.

    We can also write this as ω2r, where ω is the angular velocity, v/r … this has the advantage that ω is the same for every part of the car (which v is not).

    Obviously, the centre of the car has a smaller r than the outside tyres, and a larger r than the inside tyres, but the average value of ω2r will be the same as ω2r for the centre of mass. :wink:
    Last edited by a moderator: Apr 26, 2017
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