This **preview** has intentionally **blurred** sections. Sign up to view the full version.

chaney (glc568) – Rotations and Gravity – murthy – (21118)
1
This print-out should have 47 questions.
Multiple-choice questions may continue on
the next column or page – fnd all choices
beFore answering.
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 6
◦
From the vertical when the car has a speed oF
33
.
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: 1
.
03002 m
/
s
2
.
Explanation:
±or the plumb bob, along the vertical direc-
tion we have
T
cos
θ
=
mg
along the horizontal direction we have
T
sin
θ
=
mv
2
r
±rom 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
=
r
g r
tan
θ .
The de²ection angle is
θ
1
= 6
◦
, so
a
c
=
g
tan
θ
1
=
(
9
.
8 m
/
s
2
)
tan6
◦
=
1
.
03002 m
/
s
2
.
002
(part 2 oF 3) 10.0 points
What is the radius oF the curve?
Correct answer: 1076
.
57 m.
Explanation:
The radius is
r
=
v
2
g
tan
θ
1
=
(33
.
3 m
/
s)
2
(9
.
8 m
/
s
2
) tan6
◦
=
1076
.
57 m
.
003
(part 3 oF 3) 10.0 points
What is the speed oF the car iF the plumb
bob’s de²ection is 12
.
2
◦
while rounding the
same curve?
Correct answer: 47
.
7606 m
/
s.
Explanation:
Here the de²ection angle is
θ
2
= 12
.
2
◦
so
v
2
=
r
g r
tan
θ
2
=
R
(9
.
8 m
/
s
2
) (1076
.
57 m) tan12
.
2
◦
=
47
.
7606 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
= 18
◦
so that the cars can take
this curve at higher speeds.
Consider a car oF mass 982 kg whose
tires have static Friction coe³cient
μ
= 0
.
59
against the pavement.
The acceleration oF gravity is 9
.
8 m
/
s
2
.

chaney (glc568) – Rotations and Gravity – murthy – (21118)
2
top view
R
= 44 m
18
◦
rear
view
μ
= 0
.
59
How fast can the car take this curve without
skidding to the outside of the curve?
Correct answer: 22
.
0925 m
/
s.
Explanation:
Let :
R
= 44 m
,
θ
= 18
◦
,
m
= 982 kg
,
and
μ
= 0
.
59
.
Basic Concepts:
(1) To keep an object
moving in a circle requires a
net
force of
magnitude
F
c
=
ma
c
=
m
v
2
r
directed toward the center of the circle. (2)
Static friction law:
|F| ≤
μ
N
.
Solution:
Consider the free body diagram
for the car. Looking from the rear of the car,
we have:
N
F
mg
a
c
Let the
x
axis go
parallel the highway sur-
face
, to the
left
and 18
◦
below the horizontal,
and let the
y
axis be perpendicular to the sur-
face, 18
◦
leftward from vertically up. The
car’s centripetal acceleration is directed hor-
izontally to the left, so in our coordinates it
has components
a
x
= +
v
2
R
cos

This is the end of the preview. Sign up to
access the rest of the document.