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Please help me solve a vehicle cornering problem?

  1. Oct 14, 2014 #1
    Hi,

    I'm new to the forum and this is my first post....so go easy!!
    I have a physics problem to solve and if possible I would love a bit of help here.

    This is the question:
    A standard car is driven around in a measured circle, increasing its speed as it goes and is able to reach 85mph before it loses traction and slides away. A racing car, with spoilers and wings fitted, produces twice the amount of downforce as the standard car. It is driven around the same circle and in the same circumstances. How fast can the racing car drive around the circle until it too breaks away and loses traction.
    The formula given to calculate the answer is:

    v =[Square root of] u g r (where v = final velocity, u = co-efficient of grip and is a constant, g = gravity, and r = radius of circle.)

    Given that gravity would normally be a constant at land level (I believe) and the coefficient of grip is a constant, I am struggling to understand how the doubled downforce would fit into this equation. Hence, how would I calculate the critical speed of the racing car?

    Any help here would be much appreciated.

    AJH.
     
  2. jcsd
  3. Oct 14, 2014 #2

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Hi and welcome to PF.
    If this is a homework / coursework question it really should be in the appropriate forum (see rules at the top of the General Physics Forum.
    Do you have any ideas about this, before PF steps in to help? We do so hate to spoon feed people. Looking at that formula, do you know how it has been arrived at? In arriving at it, the mass of the car (leading to its weight) has been cancelled out. That formula, as it stands, assumes that just the mass of the car is involved. What should be different in the formula when applied to the racing car? (Going back to the basic formula and derivation could give you a clue about that.)
    See this link
    and this link
    See if you can figure it out, with the help of the links.
     
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