yang (ey942) – HW09 – TSOI – (58160)
1
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have
27
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001
10.0 points
A circularshaped object of mass 6 kg has an
inner radius of 10 cm and an outer radius of
22 cm.
Three forces (acting perpendicular
to the axis of rotation) of magnitudes 12 N,
26 N, and 14 N act on the object, as shown.
The force of magnitude 26 N acts 26
◦
below
the horizontal.
12 N
14 N
26 N
26
◦
ω
Find the magnitude of the net torque on
the wheel about the axle through the center
of the object.
Correct answer: 3
.
12 N
·
m.
Explanation:
Let :
a
= 10 cm = 0
.
1 m
,
b
= 22 cm = 0
.
22 m
,
F
1
= 12 N
,
F
2
= 26 N
,
F
3
= 14 N
,
and
θ
= 26
◦
.
F
1
F
3
F
2
θ
ω
The total torque is
τ
=
a F
2

b F
1

b F
3
= (0
.
1 m) (26 N)

(0
.
22 m) (12 N + 14 N)
=

3
.
12 N
·
m
,
with a magnitude of
3
.
12 N
·
m
.
002
(part 1 of 2) 10.0 points
A particle is located at the vector position
vectorr
= (2 m)ˆ
ı
+ (3
.
8 m)ˆ
and the force acting on it is
vector
F
= (4
.
7 N)ˆ
ı
+ (4
.
3 N)ˆ
.
What is the magnitude of the torque about
the origin?
Correct answer: 9
.
26 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
= [(2 m) (4
.
3 N)

(3
.
8 m) (4
.
7 N)]
ˆ
k
= [

9
.
26 N m]
ˆ
k .
003
(part 2 of 2) 10.0 points
What
is
the
magnitude
of
the
torque
about the point having coordinates [
a, b
] =
[(0
.
4 m)
,
(8 m)]?
Correct answer: 26
.
62 N m.
Explanation:
Reasoning similarly as we did in the pre
vious section, but with the difference that
relative to the point [(0
.
4 m)
,
(8 m)] the
y

component of the particle is now [
y

(8 m)],
we have
vector
τ
=
{
[
x

a
]
F
y

[
y

b
]
F
x
}
ˆ
k
=
{
[(2 m)

(0
.
4 m)] [4
.
3 N]

[(3
.
8 m)

(8 m)] [4
.
7 N]
}
ˆ
k
=
{
26
.
62 N m
}
ˆ
k .
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yang (ey942) – HW09 – TSOI – (58160)
2
004
10.0 points
A uniform 1 kg rod with length 24 m has
a frictionless pivot at one end.
The rod is
released from rest at an angle of 21
◦
beneath
the horizontal.
12 m
24 m
1 kg
21
◦
What is the angular acceleration of the rod
immediately after it is released? The moment
of inertia of a rod about the center of mass
is
1
12
m L
2
,
where
m
is the mass of the rod
and
L
is the length of the rod. The moment
of inertia of a rod about either end is
1
3
m L
2
,
and the acceleration of gravity is 9
.
8 m
/
s
2
.
Correct answer: 0
.
571818 rad
/
s
2
.
Explanation:
Let :
m
= 1 kg
,
L
= 24 m
,
and
θ
= 21
◦
.
The rod’s moment of inertia about its end
point is
I
=
1
3
m L
2
,
so the angular accelera
tion of the rod is
α
=
τ
I
=
1
2
m g L
cos
θ
1
3
m L
2
=
3
2
g
L
cos
θ
=
3
2
9
.
8 m
/
s
2
24 m
cos 21
◦
=
0
.
571818 rad
/
s
2
.
005
10.0 points
Consider the setup shown, where the inclined
plane has a frictionless surface.
The blocks
have masses
m
2
and
m
1
.
The pulley has
mass
m
3
, and is a uniform disc with radius
R
.
Assume the pulley to be frictionless.
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 Fall '10
 TSIO
 Angular Momentum, Kinetic Energy, Mass, Moment Of Inertia, Rotation

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