alexander (jra2623) – homework 12 – Turner – (92510)
1
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001
(part 1 of 3) 10.0 points
A student builds and calibrates an accelerom
eter, which she uses to determine the speed of
her car around a certain highway curve. The
accelerometer is a plumb bob with a protrac
tor that she attaches to the roof of her car.
A friend riding in the car with her observes
that the plumb bob hangs at an angle of 20
◦
from the vertical when the car has a speed of
27
.
3 m
/
s.
The acceleration of gravity is 9
.
8 m
/
s
2
.
At this instant, what is the centripetal ac
celeration of the car rounding the curve?
Correct answer: 3
.
56691 m
/
s
2
.
Explanation:
For the plumb bob, along the vertical direc
tion we have
T
cos
θ
=
m g
along the horizontal direction we have
T
sin
θ
=
m v
2
r
From these two equations the centripetal ac
celeration is
a
c
=
v
2
r
=
g
tan
θ ,
the radius of the curve is
r
=
v
2
g
tan
θ
,
and the speed of the car
v
=
radicalbig
g r
tan
θ .
The deflection angle is
θ
1
= 20
◦
, so
a
c
=
g
tan
θ
1
=
(
9
.
8 m
/
s
2
)
tan 20
◦
=
3
.
56691 m
/
s
2
.
002
(part 2 of 3) 10.0 points
What is the radius of the curve?
Correct answer: 208
.
946 m.
Explanation:
The radius is
r
=
v
2
g
tan
θ
1
=
(27
.
3 m
/
s)
2
(9
.
8 m
/
s
2
) tan 20
◦
=
208
.
946 m
.
003
(part 3 of 3) 10.0 points
What is the speed of the car if the plumb
bob’s deflection is 8
.
5
◦
while rounding the
same curve?
Correct answer: 17
.
4936 m
/
s.
Explanation:
Here the deflection angle is
θ
2
= 8
.
5
◦
so
v
2
=
radicalbig
g r
tan
θ
2
=
radicalBig
(9
.
8 m
/
s
2
) (208
.
946 m) tan 8
.
5
◦
=
17
.
4936 m
/
s
.
004
10.0 points
A highway curves to the left with radius of
curvature
R
= 44 m. The highway’s surface
is banked at
θ
= 16
◦
so that the cars can take
this curve at higher speeds.
Consider a car of mass 1900 kg whose
tires have static friction coefficient
μ
= 0
.
67
against the pavement.
The acceleration of gravity is 9
.
8 m
/
s
2
.
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alexander (jra2623) – homework 12 – Turner – (92510)
2
top view
R
= 44 m
16
◦
rear
view
μ
= 0
.
67
How fast can the car take this curve without
skidding to the outside of the curve?
Correct answer: 22
.
5977 m
/
s.
Explanation:
Let :
R
= 44 m
,
θ
= 16
◦
,
m
= 1900 kg
,
and
μ
= 0
.
67
.
Basic Concepts:
(1) To keep an object
moving in a circle requires a
net
force of
magnitude
F
c
=
m a
c
=
m
v
2
r
directed toward the center of the circle.
(2)
Static friction law:
F ≤
μ
N
.
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 Summer '08
 Kaplunovsky
 Force, Friction, Work, Correct Answer, Alexander, v2

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