practice 12 – – Due: Apr 17 2007, noon
1
Question 1, chap 3, sect 4.
part 1 of 2
7 points
Given:
Two vectors
vector
A
=
A
x
ˆ
ı
+
A
y
ˆ
and
vector
B
=
B
x
ˆ
ı
+
B
y
ˆ
,
where
A
x
=
−
2,
A
y
= 1,
B
x
= 2, and
B
y
= 1.
Find the
z
component of
vector
A
×
vector
B
.
Correct answer:
−
4
(tolerance
±
1 %).
Explanation:
It follows from the definition of cross prod
uct that
(
vector
A
×
vector
B
)
x
=
A
y
B
z
−
A
z
B
y
= 0
(
vector
A
×
vector
B
)
y
=
A
z
B
x
−
A
x
B
z
= 0
(
vector
A
×
vector
B
)
z
=
A
x
B
y
−
A
y
B
x
= (
−
3) (3)
−
(4)(2)
=
−
4
.
Question 2, chap 3, sect 4.
part 2 of 2
7 points
Find the angle between
vector
A
and
vector
B.
Correct answer: 126
.
87
◦
(tolerance
±
1 %).
Explanation:
vector
A
·
vector
B
=
A
x
B
x
+
A
y
B
y
= (
−
2) (2) + (1) (1)
=
−
3
,
A
=
radicalBig
A
2
x
+
A
2
y
=
radicalBig
(
−
2)
2
+ (1)
2
= 2
.
23607
,
and
B
=
radicalBig
B
2
x
+
B
2
y
=
radicalBig
(2)
2
+ (1)
2
= 2
.
23607
.
Therefore, using
vector
A
·
vector
B
=

A
 
B

cos
θ ,
and solving for
θ
, gives
θ
= cos
−
1
bracketleftBigg
vector
A
·
vector
B
A B
bracketrightBigg
= cos
−
1
bracketleftbigg
(
−
3)
(2
.
23607) (2
.
23607)
bracketrightbigg
= 2
.
2143 rad
= 126
.
87
◦
.
Question 3, chap 13, sect 4.
part 1 of 1
7 points
A particle is located at the vector position
vectorr
= (2
.
3 m)ˆ
ı
+ (4
.
5 m)ˆ
and the force acting on it is
vector
F
= (3 N)ˆ
ı
+ (1
.
7 N)ˆ
.
What is the magnitude of the torque about
the origin?
Correct answer: 9
.
59 N m (tolerance
±
1 %).
Explanation:
Basic Concept:
vector
τ
=
vectorr
×
vector
F
Solution:
Since neither the position of the
particle, nor the force acting on the particle
have a
z
component, the torque acting on the
particle has only a
z
component
vector
τ
= [
x F
y
−
y F
x
]
ˆ
k
= [(2
.
3 m) (1
.
7 N)
−
(4
.
5 m) (3 N)]
ˆ
k
=
[
−
9
.
59 N m]
ˆ
k
.
Question 4, chap 13, sect 3.
part 1 of 1
10 points
A space station shaped like a giant wheel
has a radius of 111 m and a moment of inertia
of 4
.
9
×
10
8
kg
·
m
2
(when it is unmanned). A
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practice 12 – – Due: Apr 17 2007, noon
2
crew of 150 live on the rim, and the station
is rotating so that the crew experience an
apparent acceleration of 1
g
. When 100 people
move to the center of the station, the angular
speed changes.
The acceleration of gravity is 9
.
8 m
/
s
2
.
What apparent acceleration is experienced
by those remaining at the rim?
Assume:
The average mass of for each in
habitant is 67
.
6 kg.
Correct answer: 13
.
1112 m
/
s
2
(tolerance
±
1
%).
Explanation:
Given :
r
= 111 m
,
I
= 4
.
9
×
10
8
kg
·
m
2
,
and
m
= 150
×
67
.
6 kg
.
For an apparent acceleration of g,
summationdisplay
F
c
=
m a
c
=
m g
v
2
r
=
g
r ω
2
0
=
g
ω
0
=
radicalbigg
g
r
Initially,
L
0
=
I ω
0
+
(
150
m r
2
)
ω
0
and after 100 people move to the center,
L
f
=
I ω
+ 50
m r
2
ω
Angular momentum is conserved, so
L
f
=
L
i
(
I
+ 150
m r
2
)
ω
0
=
(
I
+ 50
m r
2
)
ω
ω
=
I
+ 150
m r
2
I
+ 50
m r
2
radicalbigg
g
r
and the apparent gravity is
g
n
=
r ω
2
=
parenleftbigg
I
+ 150
m r
2
I
+ 50
m r
2
parenrightbigg
2
g
Since
I
+ 150
m r
2
= 4
.
9
×
10
8
kg
·
m
2
+ 150(67
.
6 kg)(111 m)
2
= 6
.
14935
×
10
8
m
/
s
2
and
I
+ 50
m r
2
= 4
.
9
×
10
8
kg
·
m
2
+ 50(67
.
6 kg)(111 m)
2
= 5
.
31645
×
10
8
m
/
s
2
,
then
g
n
=
parenleftbigg
6
.
14935
×
10
8
m
/
s
2
5
.
31645
×
10
8
m
/
s
2
parenrightbigg
2
(
9
.
8 m
/
s
2
)
=
13
.
1112 m
/
s
2
.
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 Spring '08
 Turner
 Angular Momentum, Force, Friction, Mass, Rotation

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