Example 5-11 A Turn on a Banked Road
Unlike the level off-ramp in Example 5-10, many freeway off-ramps, as well as corners at racetracks, are banked.
As an example, the corners at the Indianapolis Motor Speedway are banked by 9.20°. The radius of each corner is
256 m. What is the fastest speed at which a race car could take one of these corners if the coefficient of static friction is
ms = 0.550, as in Example 5-10?
Set Up
In contrast to the situation in Example 5-10,
the net centripetal force on the car comes
from two forces exerted by the banked road:
the horizontal component of the normal force
and the horizontal component of the static
friction force. Because we want the car to
move as fast as possible and hence have the
maximum possible centripetal acceleration, it
has to have the maximum available centripetal
force exerted on it. In this extreme case, the
static friction force will be at its maximum
value. Our goal is to find the value of v in this
situation.
Solve
As in Example 5-10, we write Newton’s
second law for the car in component form,
with the static friction force set equal to its
maximum value fs, max. Again, the horizontal (x)
acceleration equals the centripetal acceleration;
there is zero vertical (y) acceleration. The
difference is that now the normal force and
friction force have both x and y components.
y
Newton’s second law equation for
the car:
s
a Fext on car
s + w
s car + sf s
= n
scar
= mcar a
x
toward
center
of circle
Centripetal acceleration:
2
acar = acent =
v
r
n
acar
fs
O
(5-10)
wcar
Magnitude of the static friction force:
fs, max = msn
(5-1b)
Newton’s second law in component
form for the car, assuming that
the static friction force has its
maximum value:
x: n sin u + fs, max cos u = mcar acent
y: n cos u + (2wcar)
+ (2fs, max sin u) = 0
y
nx = n sin O
x
fs, max,x = fs,max cosO
fs,max,y = –fs,max sinO
Substitute the expressions for the
maximum force of static friction,
the centripetal acceleration, and
the weight of the car:
fs, max = msn
v2
r
wcar = mcarg
acent =
Then the Newton’s law equations become
v2
x: n sin u + ms n cos u = mcar
r
y: n cos u 2 mcar g 2 ms n sin u = 0
Solve the y component equation for the normal
force n.
Substitute this expression for n into the
x component equation and solve for the
speed v.
Rearrange the y component equation to find n:
n cos u 2 msn sin u = mcar g
n(cos u 2 ms sin u) = mcar g
mcar g
n =
cos u - ms sin u
Rearrange the x component equation:
v2
r
v2
n1sin u + ms cos u2 = mcar
r
n sin u + ms n cos u = mcar
n
fs,max
O
ny = n cosO
O
wcar,y = –wcar
Substitute the expression for n and solve for v:
mcar g
1sin u + ms cos u2
cos u - ms sin u
sin u + ms cos u
v2
v2
ga
b =
r
r
cos u - ms sin u
sin
u
+
m
cos
u
s
v 2 = gr a
b
cos u - ms sin u
= mcar
v =
Use the values of r, u, and ms given for the
Indianapolis Motor Speedway.
B
gr a
sin u + ms cos u
b
cos u - ms sin u
Using r = 256 m, u = 9.20°, and ms = 0.550,
sin u = sin 9.20° = 0.160
cos u = cos 9.20° = 0.987
v =
C
19.80 m>s 2 2 1256 m2 a
= 44.3 m>s
= 144.3 m>s2 a
Reflect
This speed is quite a bit less than that reached
by race cars on the straightaway (200 to
300 km>h), so it’s necessary to slow down
before turning this corner.
We can check our result by looking at two
special cases: (a) If the curve is not banked,
then u = 0, sin u = 0, and cos u = 1. In this
case, our result reduces to what we found for
the flat road in Example 5-10. (b) If the curve
is banked but there is no friction, then ms = 0
(because then the maximum force of friction
is fs, max = msn = 0). In this situation, our result
simplifies to what we found for the airplane
making a level turn in Example 5-9. (The
s on
airplane isn’t on a road, but the lift force L
a turning airplane plays exactly the same role
s on a car on a frictionless
as the normal force n
banked curve.)
You can show that in case (a) the car’s
maximum speed would be only 37.1 m>s,
while in case (b) it would be only 20.2 m>s.
A banked curve with friction allows the highest
speeds!
0.160 + 10.5502 10.9872
0.987 - 10.5502 10.1602
1 km
3600 s
ba
b = 159 km>h
1000 m
1h
= 1159 km>h2 a
1 mi
b = 99.1 mi>h
1.61 km
y
(a) If the curve is not banked
(so u = 0):
gr a
C
0 + ms 112
1 - ms 102
= 2ms gr
n
x
sin 0 + ms cos 0
v = gr a
b
C
cos 0 - ms sin 0
=
fs
b
acar
wcar
This equation is equivalent to what we
found in Example 5-10 for a flat road:
r =
v2
, so v 2 = ms gr and v = 2ms gr
ms g
(b) If the curve is banked but there is
no friction (so ms = 0):
v =
gr a
C
sin u + 102 cos u
cos u - 102 sin u
sin u
= gr a
b
C
cos u
= 2gr tan u
b
This equation is equivalent to what we
found in Example 5-9 for an airplane
making a banked turn:
v2
, so v 2 = gr tan u and
r =
g tan u
v = 2gr tan u
b
n
y
x
acar
O
wcar