Overton, Mays – Homework 12 – Due: Feb 21 2005, 4:00 am – Inst: Turner
1
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The due time is Central
time.
001
(part 1 of 1) 10 points
A curve of radius 59
.
3 m is banked so that
a car traveling with uniform speed 63 km
/
hr
can round the curve without relying on fric
tion to keep it from slipping to its left or right.
The acceleration of gravity is 9
.
8 m
/
s
2
.
2
.
9 Mg
μ
≈
0
θ
What is
θ
?
Correct answer: 27
.
7884
◦
.
Explanation:
Let :
m
= 2900 kg
,
v
= 63 km
/
hr
,
r
= 59
.
3 m
,
and
μ
≈
0
.
Basic Concepts:
Consider the free body
diagram for the car.
The forces acting on
the car are the normal force, the force due to
gravity, and possibly friction.
μ
N
N
N
cos
θ
m g
N
sin
θ
x
y
To keep an object moving in a circle re
quires a force directed toward the center of
the circle; the magnitude of the force is
F
c
=
m a
c
=
m
v
2
r
.
Also remember,
~
F
=
X
i
~
F
i
.
Using the freebody diagram, we have
X
i
F
x
N
sin
θ

μ
N
cos
θ
=
m
v
2
r
(1)
X
i
F
y
N
cos
θ
+
μ
N
sin
θ
=
m g
(2)
(
m g
)
k
=
m g
sin
θ
(3)
m a
k
=
m
v
2
r
cos
θ
(4)
and
,
if
μ
= 0
,
we have
tan
θ
=
v
2
g r
(5)
Solution:
Solution in an Inertial Frame:
Watching from the Point of View of Some
one Standing on the Ground.
The car is performing circular motion with
a constant speed, so its acceleration is just
the centripetal acceleration,
a
c
=
v
2
r
.
The
net force on the car is
F
net
=
m a
c
=
m
v
2
r
The component of this force parallel to the
incline is
X
~
F
k
=
m g
sin
θ
=
F
net
cos
θ
=
m
v
2
r
cos
θ .
In this reference frame, the car is at rest,
which means that the net force on the car
(taking in consideration the centrifugal force)
is zero. Thus the component of the net “real”
force parallel to the incline is equal to the
component of the centrifugal force along that
direction.
Now, the magnitude of the cen
trifugal force is equal to
F
c
=
m
v
2
r
, so
F
k
=
F
net
cos
θ
=
F
c
cos
θ
=
m
v
2
r
cos
θ
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Overton, Mays – Homework 12 – Due: Feb 21 2005, 4:00 am – Inst: Turner
2
F
k
is the component of the weight of the car
parallel to the incline. Thus
m g
sin
θ
=
F
k
=
m
v
2
r
cos
θ
tan
θ
=
v
2
g r
=
(63 km
/
hr)
2
(9
.
8 m
/
s
2
) (59
.
3 m)
×
1000 m
km
¶
2
hr
3600 s
¶
2
= 0
.
526981
θ
= arctan(0
.
526981)
= (0
.
484999 rad)
•
180 deg
π
rad
‚
= 27
.
7884
◦
.
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 Spring '08
 Turner
 Physics, Force, Friction, Work, Correct Answer, Overton, Mays

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