For each car there is a maximum speed above which it will stay on the track as it goes around the banked curve, but will spin out (and possibly fly off the track) on the flat curve.

Shown above is an electric race track with two cars on it, one in each lane. You use the two pistol-shaped controls to start the cars moving around the track, and to adjust their speeds. The red control operates the car in the outside lane, and the black control operates the car in the inside lane. The box between the controls and the track produces motor sounds when you activate the controls, if you wish. (You can switch these on or off with the yellow knob on the box.) The track has two straightaways, a flat curve and a banked curve. The straightaways are 70.5 cm long, and the radius of the flat curve is 15.1 cm for the inside lane and 20.5 cm for the outside lane. Assuming that the radii around the banked curve are the same as those for the flat curve, the overall length of the loop is 235.0 cm for the inside lane and 269.8 cm for the outside lane. The middle of the banked curve sits at about 34.5 degrees.

As the other demonstrations in this chapter show, in order for an object to move in a circular path, it must experience a sideways force accelerating it toward the center of the circle, whose magnitude is mv²/r, where m is the mass of the object, v is its tangential velocity and r is the radius of the circle in which it moves. (See demonstration 16.03 -- Ball on turntable.) For the toy cars going around the flat curve, as for a car going around a flat curve in a road, friction between the car tires and the road surface provides this centripetal force:

μmg = mv²/r,  or  μg = v²/r

We see that as long as μ ≥ v²/gr, the car stays on the road, but if the car is going fast enough that v²/gr > μ, (v > √(μgr) ), then the car skids outward, and possibly off the road, as it goes around the turn. Note that the friction that acts here is static friction, as long as the wheels of the car are rolling without slipping. Since the centripetal force is inversely proportional to the radius of the curve, the speed at which the car skids off the road is higher for the car in the outside lane than for the car in the inside lane. The car in the outside lane can go faster than can the car in the inside lane, before it skids off the curve. With good enough control of the cars, it may be possible to show this.

If we bank the curve, that is, if we tilt the road surface toward the center of the curve, we can reduce or eliminate the role of friction in keeping the car on the road. If a curve is banked, then the normal force of the road on the car not only balances gravity, but also provides a lateral force on the car toward the center of the curve. These two components are, respectively,

N cos θ = mg  and  N sin θ = mv²/r

From these we get:

tan θ = v²/gr

For the normal force to provide all of the centripetal force, v must equal √(gr tan θ). For the car in the inside lane, this equals √[(980 cm/s²)(15.1 cm) 0.687] = 101 cm/s. For the car in the outside lane, v = √[(980 cm/s²)(20.5 cm) 0.687] = 118 cm/s. Again, for the reason noted above, the speed for which the banked turn keeps the car on the road is higher for the car in the outside lane than for the car in the inside lane. At these speeds, in order for friction to hold the cars around the flat turn, μ would have to equal (101 cm/s)²/[(980 cm/s²)(15.1 cm)] = 0.69 on the inside track (which equals (118 cm/s)²/[(980 cm/s²)(20.5 cm)] = 0.69 on the outside track). For the plastic track and soft plastic or rubber car tires, the coefficient of friction might be similar to this. (For some examples, see here or here.)

If the cars round the banked turn at the speeds calculated above, friction does not act at all in providing the centripetal acceleration; the track provides it exactly via the normal force. If the cars go at greater speeds, the centripetal acceleration now exceeds that provided by the normal force, and friction provides the difference. For either car, as long as the speed is not so great that friction cannot provide the necessary additional centripetal force, the car will make the turn without skidding. It is thus fairly easy to get the cars going around the track, going around both turns, and then to speed them up a bit as they come out of the flat turn, so that they round the banked turn without a problem, but then spin out on the flat turn. With good control, it may be possible to show that the speed at which the car in the inside lane spins out around the flat turn is lower than that at which the car in the outside lane spins out.

The friction that would keep the car from sliding sideways is F = μN. If we take the case described above, in which friction keeps the car from sliding outward, up the slope of the turn, then this friction acts downward, parallel to the road surface. The vertical component of this friction is μN sin θ (in the opposite direction to N cos θ), and the horizontal component is μN cos θ (in the same direction as N sin θ). We have:

mg = N cos θ - μN sin θ  and  mv²/r = N sin θ + μN cos θ

Combining these two equations as above gives:

v_max = √[gr(sin θ + μ cos θ)/(cos θ - μ sin θ)]

which we may also write:

v_max = √[gr(tan θ + μ)/(1 - μ tan θ)]

v is given the subscript “max,” because these equations give the maximum velocity at which the car can travel around the banked turn before it begins to slide outward. We find that with no friction, this gives v_max = √(gr tan θ) as shown above for the banked curve with no friction, and for a flat curve (θ = 0), it gives v_max = √(grμ) as shown above for the flat curve.

If we wish, by squaring the equation above, consolidating the tan θ terms, dividing and simplifying, we can obtain an expression for tan θ in terms of v², g, μ and r. For the range of velocities between that at which the normal force provides the entire centripetal acceleration and that at which the car just begins to slide upward, we have:

tan θ = [(v²/gr) - μ]/[1 + (v²μ/gr)]

If the velocity of the car as it goes around the banked turn is lower than √(gr tan θ), then the horizontal component of the normal force exceeds the centripetal acceleration, and friction must now balance the excess to keep the car from sliding down the incline. Now friction points upward across the incline. Its components are the same as before, except that they point in the opposite direction. The equations become:

mg = N cos θ + μN sin θ  and  mv²/r = N sin θ - μN cos θ

v_min = √[gr(sin θ - μ cos θ)/(cos θ + μ sin θ)]

v_min = √[gr(tan θ - μ)/(1 + μ tan θ)]

v is given the subscript “min,” because these equations give the minimum velocity at which the car can go around the banked turn without sliding down. When you set the cars moving around the track, if you do not give them great enough speed around the banked curve, they may slide out of electrical contact and get stuck. As before, with no friction, this equation becomes v_min = √(gr tan θ). For a flat curve (θ = 0), it gives v_min = √(-grμ). On a flat curve, except for the centripetal force there is no lateral force on the car. As the velocity of the car decreases, so does the centripetal force; as the velocity approaches zero, so does the centripetal force. The car stops, but it stays on the track. There is no minimum velocity, hence the strange-looking result.

As we did before, we can rearrange this to

tan θ = [(v²/gr) + μ]/[1 - (v²μ/gr)]

References:

1) Sears, Francis Weston and Zemansky, Mark W. College Physics (Reading, Massachusetts: Addison-Wesley Publishing Company), pp. 111-112.
2) http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/carbank.html