karna (pk4534) – HW 12 – coker – (58245)
1
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001
(part 1 of 2) 10.0 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
= (3
.
5 N)ˆ
ı
+ (3
.
4 N)ˆ
.
What is the magnitude of the torque about
the origin?
Correct answer: 7
.
53 N m.
Explanation:
Basic Concept:
vector
τ
=
vectorr
×
vector
F
Solution:
Since neither position of the par
ticle, nor the force acting on the particle have
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) (3
.
4 N)
−
(3
.
9 m) (3
.
5 N)]
ˆ
k
= [
−
7
.
53 N m]
ˆ
k .
002
(part 2 of 2) 10.0 points
What
is
the
magnitude
of
the
torque
about the point having coordinates [
a, b
] =
[(0
.
8 m)
,
(6 m)]?
Correct answer: 10
.
75 N m.
Explanation:
Reasoning similarly as we did in the pre
vious section, but with the difference that
relative to the point [(0
.
8 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)
−
(0
.
8 m)] [3
.
4 N]
−
[(3
.
9 m)
−
(6 m)] [3
.
5 N]
}
ˆ
k
=
{
10
.
75 N m
}
ˆ
k .
003
10.0 points
A uniform horizontal rod of mass 1
.
5 kg and
length 0
.
98 m is free to pivot about one end
as shown. The moment of inertia of the rod
about an axis perpendicular to the rod and
through the center of mass is given by
I
=
m ℓ
2
12
.
0
.
98 m
F
76
◦
pivot
1
.
5 kg
If a 5
.
2 N force at an angle of 76
◦
to the hor
izontal acts on the rod as shown, what is the
magnitude of the resulting angular accelera
tion about the pivot point? The acceleration
of gravity is 9
.
8 m
/
s
2
.
Correct answer: 4
.
70298 rad
/
s
2
.
Explanation:
Let :
ℓ
= 0
.
98 m
,
m
= 1
.
5 kg
,
θ
= 76
◦
,
and
F
= 5
.
2 N
.
By the parallel axis theorem, the moment
of inertia of a stick pivoted at the end is
I
=
I
cm
+
m d
2
=
1
12
m ℓ
2
+
m
parenleftbigg
ℓ
2
parenrightbigg
2
=
1
3
m ℓ
2
.
The sum of the torques (counterclockwise ro
tation defined as positive) is
summationdisplay
τ
=
F ℓ
sin
θ
−
m g
ℓ
2
=
I α
F ℓ
sin
θ
−
m g l
ℓ
2
=
parenleftbigg
1
3
m ℓ
2
parenrightbigg
α
6
F
sin
θ
−
3
m g
= 2
m ℓ α
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karna (pk4534) – HW 12 – coker – (58245)
2
vectorα
=
6
F
sin
θ
−
3
m g
2
m ℓ
Since
6
F
sin
θ
−
3
m g
= 6 (5
.
2 N) sin 76
◦
−
3 (1
.
5 kg) (9
.
8 m
/
s
2
)
=
−
13
.
8268 N
,
then
bardbl
vectorα
bardbl
=
vextendsingle
vextendsingle
vextendsingle
−
13
.
8268 N
vextendsingle
vextendsingle
vextendsingle
2 (1
.
5 kg) (0
.
98 m)
=
4
.
70298 rad
/
s
2
.
004
10.0 points
A onepiece cylinder has a core section pro
truding from the larger drum and is free to ro
tate around its central axis. A rope wrapped
around the drum of radius 1
.
22 m exerts a
force of 3
.
4 N to the right on the cylinder.
A rope wrapped around the core of radius
0
.
61 m exerts a force of 6
.
6 N downward on
the cylinder.
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 Spring '08
 Turner
 Physics, Angular Momentum, Kinetic Energy, Moment Of Inertia, kg

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