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Car moving around a curve

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data

    A curve with a 200-m radius on a level road is banked at the correct angle for a speed of 60 km/h, i.e. a car traveling at this speed would remain on the road, even if the surface were frictionless.
    a) If a car travels around this curve at a speed of 90 km/h, what is the minimum coefficient of static friction between the tires and the road that will prevent skidding?
    b) If the surface were wet, would the speeding car in part (a) remain on the road?
    c) If it was raining, and the car had to slow down to a crawl (for example, to safely pass a stranded vehicle), would it slide down the inclined roadway?

    2. Relevant equations
    Friction = force applied/(mass*gravity)
    F=ma

    3. The attempt at a solution
    I made a post earlier about another problem... I'm staring at these problems... and I have no idea how to solve them. I took notes on things relevant to these, but I have no idea how to apply them. In parentheses, mindless ranting [It might seem like I'm out just trying to get the easy answer to these.. but I really just need some kind of hint or something. I have no idea how to apply the relevant equations to the problem. Once again i would appreciate some kind of explanation as to how these kinds of problems work, so I can attempt to solve these kinds of problems. I'm better at math than physics, because I really am bad at application of equations.] Mainly, I don't understand how you can drive along a curve without any friction. I could understand that if there were something on the sides guiding you on the curve, but I think you would drift out of the curve if there were no friction.
     
  2. jcsd
  3. Feb 6, 2012 #2

    NascentOxygen

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    Staff: Mentor

    To cause a body to travel in a circle, you need a force acting towards the centre of that circular path. This force can (i) come from friction with the roadway, or (ii) from a rope anchored to a pole at the centre of the circle, or (iii) by banking the roadway as you see in a velodrome. When the roadway is banked, a component of the weight of the body tends to cause the body to slide towards the centre of the circular path. In a velodrome, there's an optimal speed where the cyclist will travel in a level circle; slower than this it will tend to slide down towards the centre; much faster than optimal and the bicycle would tend to rise up the slope. Friction of tyres with the track allows the cyclist to make corrections and travel a little faster or slower than optimal for the slope of that track.
     
  4. Feb 6, 2012 #3
    Hi,

    An interesting problem!

    When going through a curve of radius [itex]r[/itex] with speed [itex]v_1[/itex], you have a centripetal acceleration of [itex]v_1^2/r[/itex]. Hence, according to Newton's law, you need a centripetal force of [itex]m v_1^2/r[/itex], where [itex]m[/itex] is the mass of the car (this will cancel out later).

    Now, suppose that the road exerts a normal force [itex]F_n[/itex] on the car. Since there is no vertical acceleration, the vertical component of the normal force balances the weight of the car: [itex]F_n \cos(\theta) = m g [/itex], where [itex]\theta[/itex] is the angle of the banking slope and [itex]g[/itex] is the acceleration of gravity. The horizontal component [itex]F_n \sin(\theta)[/itex] of the normal force provides the centripetal force, and is equal to [itex]m v_1^2/r[/itex] where [itex]v_1[/itex] is the "normal" speed of 60 km/h. Incidentally, it follows that the banking slope [itex]\theta[/itex] is given by [tex]\tan(\theta)=\frac{v_1^2}{r g}[/tex]
    so that, with [itex]v_1[/itex]= 60 km/h = 16.67 m/s, [itex]r[/itex]= 200 m, and [itex]g[/itex]=9.8 m/s^2, I obtain a slope of 8.1 degrees, or 14%, which sounds plausible.

    Continuing to question (a), suppose the car is going [itex]v_2[/itex] = 90 km/h instead of 60 km/h. This requires [itex]m(v_2^2-v_1^2)/r[/itex] extra centripetal force, which must be provided by the horizontal component of the friction force [itex]F_f[/itex] on the tires, which is equal to [itex]F_f\cos(\theta)[/itex]. The friction force is given by [itex]F_f = \mu F_n[/itex], where [itex]\mu[/itex] is the coefficient of static friction. Using my previously obtained result [itex]F_n = m g / \cos(\theta)[/itex], I find [itex]F_f cos(\theta) = \mu m g[/itex]. Equating this to the required extra centripetal force and solving for [itex]\mu[/itex], I obtain [tex]\mu=\frac{v_2^2-v_1^2}{r g}.[/tex] With [itex]v_2[/itex] = 90 km/h = 25 m/s and the parameters above, I obtain [itex]\mu[/itex] = 0.18.

    As for question (b), my answer would be that this depends on the value of the static friction coefficient when it is wet, [itex]\mu_{wet}[/itex]. If [itex]\mu_{wet}<0.18[/itex] the car will skid, otherwise not.

    As for (c), the requirement here is that the horizontal component of the normal force, [itex]F_n \cos(\theta)[/itex], is smaller than the horizontal component of the friction force, [itex]F_f \sin(\theta)[/itex]. Filling in [itex]F_f = \mu_{wet} F_n[/itex] and some manipulation yields the requirement [tex] \tan(\theta) < \mu_{wet}[/tex] which in view of the parameters given means that we must have [itex]\mu_{wet}>0.14[/itex].

    I hope this helps!

    Cheers,
    Kurt
     
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