homework 19 – FIERRO, JEFFREY – Due: Apr 6 2008, 11:00 pm
1
Question 1, chap 13, sect 4.
part 1 of 1
10 points
A solid steel sphere of density 7
.
85 g
/
cm
3
and mass 0
.
7 kg spins on an axis through its
center with a period of 2
.
4 s.
Given
V
sphere
=
4
3
π R
3
, what is its angular
momentum?
Correct answer: 0
.
000563058 kg m
2
/
s (toler
ance
±
1 %).
Explanation:
The definition of density is
ρ
≡
M
V
=
M
4
3
π R
3
,
Therefore
R
=
bracketleftbigg
3
M
4
π ρ
bracketrightbigg
1
3
=
bracketleftbigg
3 (0
.
7 kg)
4
π
(7850 kg
/
m
3
)
bracketrightbigg
1
3
= 0
.
0277149 m
.
Using
ω
=
2
π
T
=
2
π
(2
.
4 s)
= 2
.
61799 s
−
1
and
I
=
2
5
M R
2
=
2
5
(0
.
7 kg)(0
.
0277149 m)
2
= 0
.
000215072 kg m
2
,
we have
L
≡
I ω
=
4
π M R
2
5
T
=
4
π
(0
.
7 kg)(0
.
0277149 m)
2
5 (2
.
4 s)
=
0
.
000563058 kg m
2
/
s
.
Question 2, chap 13, sect 4.
part 1 of 1
10 points
A force
F
x
= (2 N)ˆ
ı
+ (3
.
5 N)ˆ
is applied
to an object that is pivoted about a fixed axis
aligned along the
z
coordinate axis. The force
is applied at the point
x
= (4
.
1 m)ˆ
ı
+(3
.
9 m)ˆ
.
Find the
z
component of the net torque.
Correct answer: 6
.
55 N m (tolerance
±
1 %).
Explanation:
Basic Concept:
vector
τ
=
vectorr
×
vector
F
Solution:
From the definition of torque
vector
τ
=
vectorr
×
vector
F
= [
x
ˆ
ı
+
y
ˆ
]
×
[
F
x
ˆ
ı
+
F
y
ˆ
]
= [(4
.
1 m)ˆ
ı
+ (3
.
9 m)ˆ
]
×
[(2 N)ˆ
ı
+ (3
.
5 N)ˆ
]
= [(4
.
1 m) (3
.
5 N)
−
(3
.
9 m) (2 N)]
ˆ
k
= (6
.
55 N m)
ˆ
k
In this case, the net torque only has a
z

component.
Question 3, chap 13, sect 4.
part 1 of 2
10 points
A particle is located at the vector position
vectorr
= (1
.
8 m)ˆ
ı
+ (3
.
9 m)ˆ
and the force acting on it is
vector
F
= (5
.
4 N)ˆ
ı
+ (1
.
6 N)ˆ
.
What is the magnitude of the torque about
the origin?
Correct answer:
18
.
18
N m (tolerance
±
1
%).
Explanation:
Basic Concept:
vector
τ
=
vectorr
×
vector
F
Solution:
Since neither position of the par
ticle, nor the force acting on the particle have
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homework 19 – FIERRO, JEFFREY – Due: Apr 6 2008, 11:00 pm
2
the
z
components, the torque acting on the
particle has only
z
component:
vector
τ
= [
x F
y
−
y F
x
]
ˆ
k
= [(1
.
8 m) (1
.
6 N)
−
(3
.
9 m) (5
.
4 N)]
ˆ
k
= [
−
18
.
18 N m]
ˆ
k .
Question 4, chap 13, sect 4.
part 2 of 2
10 points
What
is
the
magnitude
of
the
torque
about the point having coordinates [
a, b
] =
[(4 m)
,
(6 m)]?
Correct answer: 7
.
82 N m (tolerance
±
1 %).
Explanation:
Reasoning similarly as we did in the pre
vious section, but with the difference that
relative to the point [(4 m)
,
(6 m)] the
y

component of the particle is now [
y
−
(6 m)],
we have
vector
τ
=
{
[
x
−
a
]
F
y
−
[
y
−
b
]
F
x
}
ˆ
k
=
{
[(1
.
8 m)
−
(4 m)] [1
.
6 N]
−
[(3
.
9 m)
−
(6 m)] [5
.
4 N]
}
ˆ
k
=
{
7
.
82 N m
}
ˆ
k .
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 Spring '08
 Turner
 Force, Friction, Kinetic Energy, Mass, Work, Velocity, Correct Answer

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