khounvivongsy (sk27799) – Homework 8 – Weathers – (17104)
1
This
printout
should
have
12
questions.
Multiplechoice questions may continue on
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before answering.
001
(part 1 of 2) 10.0 points
An amusement park ride consists of a rotating
circular platform 11
.
3 m in diameter from
which 10 kg seats are suspended at the end
of 3
.
65 m massless chains. When the system
rotates, the chains make an angle of 38
.
1
◦
with
the vertical.
The acceleration of gravity is 9
.
8 m
/
s
2
.
θ
l
d
What is the speed of each seat?
Correct answer: 7
.
79242 m
/
s.
Explanation:
In the vertical direction we have
T
cos
θ
=
m g ,
where
T
is the tension in the chain.
In the
horizontal direction we have
T
sin
θ
=
m v
2
r
.
Since
r
=
ℓ
sin
θ
+
d
2
= (3
.
65 m) sin 38
.
1
◦
+
11
.
3 m
2
= 7
.
90218 m
,
we have
v
=
radicalbig
g r
tan
θ
=
radicalBig
(9
.
8 m
/
s
2
) (7
.
90218 m) tan 38
.
1
◦
= 7
.
79242 m
/
s
.
002
(part 2 of 2) 10.0 points
If a child of mass 46
.
4 kg sits in a seat, what is
the tension in the chain (for the same angle)?
Correct answer: 702
.
371 N.
Explanation:
M
= 46
.
4 kg
From the first part we have
T
cos
θ
= (
m
+
M
)
g
T
=
(
m
+
M
)
g
cos
θ
=
(10 kg + 46
.
4 kg) (9
.
8 m
/
s
2
)
cos 38
.
1
◦
= 702
.
371 N
.
003
10.0 points
A curve of radius 53
.
6 m is banked so that
a car traveling with uniform speed 48 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
.
1
.
9 Mg
μ
≈
0
θ
What is
θ
?
Correct answer: 18
.
6981
◦
.
Explanation:
Let :
m
= 1900 kg
,
v
= 48 km
/
hr
,
r
= 53
.
6 m
,
and
μ
≈
0
.
Basic Concepts:
Consider the free body
diagram for the car.
The forces acting on
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khounvivongsy (sk27799) – Homework 8 – Weathers – (17104)
2
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,
vector
F
=
summationdisplay
i
vector
F
i
.
Using the freebody diagram, we have
summationdisplay
i
F
x
N
sin
θ

μ
N
cos
θ
=
m
v
2
r
(1)
summationdisplay
i
F
y
N
cos
θ
+
μ
N
sin
θ
=
m g
(2)
(
m g
)
bardbl
=
m g
sin
θ
(3)
m a
bardbl
=
m
v
2
r
cos
θ
(4)
and
,
if
μ
= 0
,
we have
tan
θ
=
v
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 Spring '09
 weathers
 Physics, Force, Friction, Correct Answer, tan θ

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