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001
10.0 points
An amusement park ride consists of a large
vertical cylinder that spins about its axis
fast enough that any person inside is held
up against the wall when the floor drops away
(see figure). The coefficient of static friction
between the person and the wall is
μ
and the
radius of the cylinder is
R
.
ω
R
What is the minimum tangential velocity
needed to keep the person from slipping down
ward?
1.
v
=
1
μ
radicalbig
g R
2.
v
=
μ
radicalbig
2
g R
3.
v
= 2
radicalbig
g R
4.
v
=
radicalbig
2
μ g R
5.
v
=
1
2
radicalbig
g R
6.
v
=
μ
radicalbig
g R
7.
v
=
radicalBigg
g R
μ
correct
8.
v
= 2
μ
radicalbig
g R
9.
v
=
μ
radicalbig
2
π g R
10.
v
=
radicalbig
g R
Explanation:
Basic Concepts:
Centripetal force:
F
=
m v
2
r
Frictional force:
f
s
≤
μ
N
=
f
max
s
Solution:
The maximum frictional force
due to friction is
f
max
=
μ
N
,
where
N
is the
inward directed normal force of the wall of
the cylinder on the person.
To support the
person vertically, this maximal friction force
f
max
s
must be larger than the force of gravity
m g
so that the actual force, which is less
than
μ
N
, can take on the value
m g
in the
positive vertical direction.
Now, the normal
force supplies the centripetal acceleration
v
2
R
on the person, so from Newton’s second law,
N
=
m v
2
R
.
Since
f
max
s
=
μ
N
=
μ m v
2
R
≥
m g ,
the minimum speed required to keep the per
son supported is at the limit of this inequality,
which is
μ m v
2
min
R
=
m g,
or
v
min
=
parenleftbigg
g R
μ
parenrightbigg
1
2
.
002
(part 1 of 4) 10.0 points
The following figure shows a Ferris wheel that
rotates 2 times each minute and has a diame
ter of 19 m.
The acceleration of gravity is 9
.
8 m
/
s
2
.
R
ω
oldmidterm2 01 – Turner – (58185)
1
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hinojosa (jlh3938) – oldmidterm2 01 – Turner – (58185)
2
What is the centripetal acceleration of a
rider?
Correct answer: 0
.
416717 m
/
s
2
.
Explanation:
Let :
r
=
D
2
= 9
.
5 m
.
The period of the Ferris wheel is
T
=
one minute
n
=
60 s
2
= 30 s
and the speed of the wheel is
v
=
2
π r
T
=
2
π
(9
.
5 m)
30 s
= 1
.
98968 m
/
s
,
so the centripetal acceleration is
a
=
v
2
r
=
(1
.
98968 m
/
s)
2
9
.
5 m
=
0
.
416717 m
/
s
2
.
003
(part 2 of 4) 10.0 points
What force does the seat exert on a 33 kg
rider at the lowest point of the ride?
Correct answer: 337
.
152 N.
Explanation:
The force exerted by the seat balances the
gravity and provides the centripetal force, so
F
l
=
m
(
g
+
a
)
= (33 kg) (9
.
8 m
/
s
2
+ 0
.
416717 m
/
s
2
)
=
337
.
152 N
.
004
(part 3 of 4) 10.0 points
What force does the seat exert on a 33 kg
rider at the highest point of the ride?
Correct answer: 309
.
648 N.
Explanation:
The gravity is partly balanced by the force
exerted by the seat and this resultant provides
the centripetal force, so
F
l
=
m
(
g
−
a
)
= (33 kg) [(9
.
8 m
/
s
2
)
−
(0
.
416717 m
/
s
2
)]
=
309
.
648 N
.
005
(part 4 of 4) 10.0 points
What force (magnitude) does the seat exert
on a rider when the rider is halfway between
top and bottom?
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 Spring '08
 Turner
 Force, Friction, Mass, Correct Answer, kg, Hinojosa

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