# A Banked Turn With Friction

##    Conceptual:  v < videal v > videal
Suppose we consider a particular car going around a particular banked turn. The centripetal force needed to turn the car (mv2/r) depends on the speed of the car (since the mass of the car and the radius of the turn are fixed) - more speed requires more centripetal force, less speed requires less centripetal force. The centripetal force available to turn the car (the horizontal component of the normal force = if you follow the mathematics) is fixed (since the mass of the car and the bank angle are fixed). So, it makes sense that we found one particular speed at which the centripetal force needed to turn the car equals the centripetal force supplied by the road. This is the "ideal" speed, videal, at which the car the car will negotiate the turn - even if it is covered with perfectly-smooth ice. Any other speed, v, will require a friction force between the car's tires and the pavement to keep the car from sliding up or down the embankment:

1. v > videal (right diagram): If the speed of the car, v, is greater than the ideal speed for the turn, videal, the horizontal component of the normal force will be less than the required centripetal force, and the car will "want to" slide up the incline, away from the center of the turn. The friction force will oppose this motion and will act to pull the car down the incline, in the general direction of the center of the turn.
2. v < videal (left diagram): If the speed of the car, v, is less than the ideal (no friction) speed for the turn, videal. In this case, the horizontal component of the normal force will be greater than the required centripetal force and the car will "want to" slide down the incline toward the center of the turn. If there is a friction force present between the car's tires and the road it will oppose this relative motion and pull the car up the incline.

## Mathematical:

### Case 1: v > videal: A free-body diagram for the car is shown at left. Both the normal force, N (blue components) and the friction force, f (red components) have been resolved into horizontal and vertical components. Notice that there are now 3 vectors in the vertical direction (there were 2 vectors in the no-friction case), and: Using the approximation , where is the coefficient of friction, gives: If the coefficient of friction is zero, this reduces to the same normal force as we derived for no-friction, which is reassuring. If the coefficient of friction is not zero, notice that the normal force will be larger than it was in the no-friction case. Now, in the horizontal direction: Here, the term is friction's contribution to the centripetal force. Also, notice again that if mu = 0, this result reduces to the same Fnet as the no-friction case. Now, since the net force provides the centripetal force to turn the car: So, what can be done with this?

### Example 4:

Suppose you want to negotiate a curve with a radius of 50 meters and a bank angle of 15o (See the Example 1). If the coefficient of friction between your tires and the pavement is 0.50, what is the maximum speed that you can safely use?

#### Solution:

Continuing the derivation above, we can get: First, note that if the coefficient of friction were zero, the expression given above for v would reduce to the same expression we derived for the no-friction case. This is always a good, quick check. Next, notice that this velocity is about twice the no-friction velocity from Example 1.

### Example 5:

In Example 3, I noted that NASCAR race cars actually go through the turns at Talladega Motor Speedway at about 200 mi/hr. If that is the case, what coefficient of friction exists between the car's tires and the pavement?

#### Solution:

First, (It is not always "good form" to generate intermediate results in a calculation, of course, but we have a pretty "hairy" calculation coming up, so I think I can forgive myself for getting the units straightened out at this point. Notice that I've kept an extra significant digit in the result, though, just for safety's sake.) Now, if we continue working from the last equation above Example 4, we can cross-multiply and solve for mu: Is this correct? Well, it is a reasonable answer, and notice that the units work out correctly, which is always a good, quick check.

### Case 2: v < videal: A free-body diagram for the car is shown at left. Both the normal force, N (blue components) and the friction force, f (red components) have been resolved into horizontal and vertical components. Notice that the friction force acts up the incline, to keep the car from sliding toward the center of the turn. This derivation is very similar to the previous case, and is left as an "exercise for the reader".    last update February 6, 2006 by JL Stanbrough