Figure 1 shows the relevant forces at work on a car in a banked turn.

• w is the weight of the car (mg) and its components are shown.
• Since the car is neither sinking through the roadway nor acclerating vertically, the gravitational force (weight) and its components are balanced by F_n
• The roadway pushes up on the car with the normal force F_n.
  Since F_n has one component that points in the same direction as the centripetal acceleration a_c (see figure 2 below), this component can contribute to the centripetal force F_c. In fact, what if we set up a situation where ALL of the centripetal force is supplied by this component of the normal force?
  • Remember that a_c is dependent upon the radius of the circle r and the velocity of the car v as
  • Study the diagram and notice that F_n is going to vary only with the weight of the car and angle₁.
  • Radius and velocity do not contribute to F_n.
  • So, for any car of weight w and roadway banked at ₁, the maximum value of F_n (which we will calculate in a minute) is fixed, and so is its component which supplies F_c.Another way of looking at this is that F_n is the equilibrant of the component of weight perpendicular to the roadway, and the magnitude of that component of weight depends upon (mg) and the angle of bank, and thus so does F_n.

Think about this.

• If we use ONLY the component of F_n which points along the radius of the circle to provide F_c,
• F_c = ma_c tells us there is a maximum value of a_c for that component and
• since there is a maximum velocity that can result from a_c and therefore from any particular magnitude of F_n

What then, is the maximum velocity a car can travel at and still have enough centripetal acceleration to negotiate the curve without the aid of friction when F_n supplies F_c ?

Check it out...

• Figure 2 shows the normal force F_n in green and its components in blue. These components are equal in magnitude to their counterparts, the components of the weight, shown in figure 1 above.
• Let's calculate the magnitudes of those components F_c and -mg
  
• By similar triangles, angle in figure 2 is the same as the angle of bank ₁ in up in figure 1. This is should look a little like the layout for the frictio-ramp lab you did in the previous chapter.
• So the value of F_n 's horizontal component which points along the radius r, and provides the centripetal force F_c, is found by
  

  and

  1.

• The vertical component of F_n is equal in magnitude and opposite in direction to the weight (mg) of the car...
• note again that the vertical forces in this situation are balanced, and that vertical component in figure 2 is calculated as

  2.

Newton's Second law tells us again that centripetal force F_c = ma_c, and since then the centripetal force F_c is found by

3.

• Now, the tangent of is , or

  4.

  note that the mass again cancels out of this expression ( I love it when that happens!) such that

5.

solving this for v, we find that

6.

• v is the maximum velocity at which a car can negotiate a curve of bank angle and depend only on the normal force to provide the centripetal force.
• This means that a highway engineer can calculate the angle at which to bank a curve such that a car can negotiate it at a particular speed and not depend upon friction AT ALL.
  • You try: find the velocity at which a car can negotiate a curve banked at 6 degrees when the radius of curvature is 100 meters.
  • You try: find the bank angle needed to permit cars to travel safely at 88 km/hr around a curve who's radius is 100 meters.
• In other words, no matter how slippery the surface is, if you drive at or below this maximum velocity you should get around the corner ok. Coefficient of friction is now OUT OF THE PICTURE! But.. if there is some friction anyway, good. It adds an extra margin of safety.
• Hmmmm... ever see those speed signs on highway curves? Ever wonder why the speeds are set at a particular value?

back to chapter 7 notes, main page.

10/98 bunning