TAM 212 Worksheet 9: Cornering and banked turns
The aim of this worksheet is to understand how vehicles drive around curves, how slipping and rolling limit
the maximum speed, and how banking the road can increase the m
...
TAM 212 Worksheet 9: Cornering and banked turns
The aim of this worksheet is to understand how vehicles drive around curves, how slipping and rolling limit
the maximum speed, and how banking the road can increase the maximum speed.
The photos above show two different views of a Setra S 411 HD bus cornering at high speed on the MercedesBenz test track at Untert¨urkheim, on the Neckar River just outside of Stuttgart, Germany.
Below is a top view of the track, showing the bus and its velocity and acceleration vectors. In this worksheet
we will always assume a circular track of fixed radius ρ and a bus traveling at constant speed v.
ρ
~v
~a
1Cornering on a flat road
First we consider the bus driving around a curve on a flat horizontal road, as shown in the front-view free
body diagram below. We will model the bus as a rigid body that contacts the ground at two points (the
wheel centers) with normal forces N1 and N2, and we assume that the friction force F applies equally to each
wheel. Gravity g acts through the center of mass C.
C
mg
~a
N1
N2
F=2 F=2
h
‘ ‘
N1
N2
F=2 F=2
^{
|^
1. What is the acceleration ~a as a function of the bus speed v and radius of curvature ρ of the road?
Solution: ~a = −v2
ρ
^{
22. If the bus is not rolling then it has zero angular velocity about the k^ direction, so the total moment about
the k^ axis will be zero. The kinetics will thus satisfy P F~ = m~a and P M~ C = 0. Use these two equations to
find the forces F , N1, and N2 in terms of all other variables.
Solution: The rigid body equations are:
X F~ = m~a
−F^{ + N1|^+ N2|^− mg|^ = m~a = −mv2
ρ
^{
X M~ C = 0
−Fhk^ − N1‘k^ + N2‘k^ = 0:
Equating components gives the equations:
−F = −mv2
ρ
N1 + N2 − mg = 0
−Fh − N1‘ + N2‘ = 0
and solving these gives:
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