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Am I missing something for this to be calculated?

  1. Oct 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi, (firsly apologies for posting in wrong area originally)

    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.


    2. Relevant equations

    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.)


    3. The attempt at a solution
    I am happy with gaining a value for the radius from the standard car and this will also then be used for the racing car, but I am struggling with what the down force effect will do. I researched other similar questions and generally there would be a mass involved to assist in calculating the change of the down force, but there was no mass given in this example and so the friction coeffiecient is what I would expect to see as the factor which would be doubled. However, the question did state that the friction co-effiecient was constant (which I understand to mean it will be the same for both vehicles), so this has thrown me. Any pointers on how I am going wrong here would really help!!
     
  2. jcsd
  3. Oct 14, 2014 #2

    SteamKing

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    Rubber is rubber, and asphalt is asphalt, so it is entirely plausible that the coefficient of friction will be the same for the regular car tires as the racing car tires. The only factor which changes is the total amount of down force generated by the two vehicles. The racing car is able to generate this extra force due to the spoilers and wings, which only generate a downward force when the race car is moving. Presumably, the regular car does not have these aerodynamic enhancements to help keep it on the track.
     
  4. Oct 14, 2014 #3
    If the weight of the car is mg, what is the normal force of the road exerted on the car? If the downward force doubles, what is the normal force of the road exerted on the car?

    Chet
     
  5. Oct 14, 2014 #4

    phinds

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    Actually, although you apparently are supposed to ignore the fact for the purposes of this problem, race car tires are wider than regular tires and so have more traction.
     
  6. Oct 14, 2014 #5
    Correct Steam King, the standard car in this example doesn't have the extra downforce enhancements and it is possible the coefficient of friction will be the same between the tyres and tarmac on both vehicles. However, with the given formula, I could only see the coefficient of friction as being the variable here which is what could be affected by the downforce doubling?

    Chet,
    Forgive me here, I only have a very basic understanding and I'm trying to get my head around this. If the weight of the car is mg, the force of the road would be equal to that of mg would it not? If the downforce is doubled, I can see that this would have the effect of raising the level of coefficient between tyre and road surface, so this is why I believe the CofF would be the variable. The only fly in the ointment is it said the CofF was a constant. I take that to mean it would be the same for both vehicles, in which case we would not double this? Would perhaps, the value of the acceleration due to gravity be doubled in this instance?
     
  7. Oct 14, 2014 #6
    That's a fair point and yes I would expect this would have an effect on increasing the coefficient of friction between tyre and tarmac but the question did state the coefficient was constant, which is why this has baffled me!!
     
  8. Oct 14, 2014 #7

    phinds

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    Well, it's very common in learning situations to make simplifying assumptions such as that.
     
  9. Oct 14, 2014 #8

    SteamKing

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    The coefficient of friction is a ratio between only the friction force and the normal force. For Coulomb type friction, the area of contact is irrelevant.

    http://en.wikipedia.org/wiki/Friction

    Coulomb type friction is only one model of friction, however.
     
  10. Oct 14, 2014 #9

    Mark44

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    Asphalt is asphalt, but there are different blends of rubber to make them "stickier", a capability that comes at a cost of reduced lifetime.
    And different rubber compounds, as noted above.
     
  11. Oct 14, 2014 #10
    I agree there can be more variables to consider such as tyre rubber compound effect on tarmac, whether the circular course is banked etc. Using the information provided it appears there is no option, in lack of evidence otherwise, than to assume the CofF is equal between the two cars. Am I right in saying that if you are told something is a "constant" that this will remain constant throughout the calculation of the problem? This is what I understand a constant to mean.

    If this is the case, then I see there can be no variables given the formula which has been provided to use. I would therefore conclude that if the Cof F is constant and the radius of the circle is the same, gravity remains the same as they are at the same height and that the resulting final velocity for the racing car would be the same as for the standard car. There simply is no place in the formula provided in which to factor in the doubled rate of downforce.

    This seems both strange and unlikely but sticking to the facts and figures of the formula - this would be the answer?
     
  12. Oct 14, 2014 #11
    This is the danger of using a formula without understanding the limitations of the formula. If you are looking at the coefficient of friction as the cause of your dilemma, then you are looking in the wrong place. If the downward force as a result of the aerodynamics is double the weight of the car (2mg), this has the same effect as if gravity g were doubled to 2g.

    Chet
     
    Last edited: Oct 14, 2014
  13. Oct 14, 2014 #12

    SteamKing

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    Yes, this is all true, but for the purposes of working a problem in Intro Physics, these topics are too esoteric to consider. Perhaps the problem could have been worded better, but you pays yer money and you takes yer chance.
     
  14. Oct 15, 2014 #13
    Ok, thank you for your replies.

    Chet, I appreciate your help here. My initial calculations to this question were exactly as you have described as the gravitational element was the only factor which, logically, was a downward producing force in this formula. Then when I mentioned the problem to someone else, it was suggested that the downforce did not affect the weight of the car, it was an effect which pushed the car into closer contact to the surface which, in turn, aids the grip coefficient of friction. This then set me thinking about the frictionional grip and I guess threw me off track. I should have stuck with my original instinct! Many thanks for your help, I'm afraid I've got a lot to read up on again!
     
  15. Oct 15, 2014 #14

    haruspex

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    Thing is, the tyres and the road can't tell the difference between increased g, aerodynamic forces, or (except for the consequent change to horizontal forces) increased mass of car.
     
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