Physics 40S
Motion on a Curve
=> The net force on a car traveling around a curve is the centripetal
force, Fc = m v2 / r, directed toward the center of the curve.
=> For a level curve, the centripetal force will be supplied by the friction
force between the tires and roadway.
=> A banked curve can supply the centripetal force by the normal force
and the weight without relying on friction.
Level Curves
For a level curve, the centripetal force will be supplied by the friction
force between the tires and roadway.
Example What must be the coefficient of friction between the tires and
the level roadway to allow a car to make a curve of radius r = 350 m at a
speed of 80 km/h?
For a level curve, the force of friction is the only horizontal force on a car
and provides the centripetal force. This can be seen from the free-body
diagram:
Physics 40S
The net force must be horizontal--pointing toward the center of the circle-and only the friction force is available to provide it. The normal force and
the weight simply cancel each other.
Calculations are easier if the speed is expressed in m/s so we shall convert
that first.
80 km/hr /3.6 = v = 22.2 m/s
Now we can calculate the necessary centripetal force,
mv 2
Fc 
r
FN  mg
m(22.2) 2
Fc 
FN  9.8m
350
Fc  1.41m
For this flat curve, the centripetal force is supplied by the friction force, Ff,
Ff = FN
F f =  9.8 m
and
Ff = Fc
9.8m = 1.41 m
1.41
 0.141
=
9.8
Of course, any coefficient of friction greater than 0.141 will keep the car
from slipping; this is the minimum value for . Notice that the force of
friction is perpendicular to the velocity. But it still is in the direction
necessary to oppose the motion that would occur without friction. The car
will slide away from the center (to the right in this diagram) if it
encounters a patch of low-friction surface (like ice or oil); therefore the
friction force is directed toward the center.
Notice, too, that we are dealing with static friction. The piece of the tire
in contact with the road is momentarily at rest with respect to the road!
The force of static friction is somewhat greater than the force provided by
kinetic friction, if the tires start to skid.
Physics 40S
Worked example: A banked curve
Question: Civil engineers generally bank curves on roads in such a manner that a car
going around the curve at the recommended speed does not have to rely on friction
between its tires and the road surface in order to round the curve. Suppose that the radius
of curvature of a given curve is r = 60 m, and that the recommended speed is v = 40
km/hr. At what angle  should the curve be banked?
F
Fsin
Fcos
F
Answer: Consider a car of mass m going around the curve. The car's weight, mg, acts
vertically downward. The road surface exerts an upward normal reaction F on the car.
The vertical component of the reaction must balance the downward weight of the car, so
Fcos = mg
The horizontal component of the reaction, Fsin, acts towards the centre of curvature of
mv 2
the road. This component provides the force Fc 
towards the centre of the
r
curvature, which the car experiences as it rounds, the curve. In other words,
mv 2
F sin  
since from the diagram
r
mv 2
 mg tan 
Note that if the car attempts to round the curve at the wrong speed then
r
and the difference has to be made up by a sideways friction force exerted between the
car's tires and the road surface. Unfortunately, this does not always work--especially if
the road surface is wet!