Husain, Zeena – Homework 10 – Due: Nov 4 2003, 4:00 am – Inst: H L Berk
1
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The due time is Central time.
001
(part 1 of 3) 10 points
Given:
g
= 9
.
8 m
/
s
2
.
A bicycle wheel of mass
m
rotating at an
angular velocity
ω
has its shaft supported
on one side, as shown in the figure.
When
viewing from the left, one sees that the wheel
is rotating in a counterclockwise manner. The
distance from the center of the wheel to the
pivot point is
b
.
We assume the wheel is a
hoop of radius
R
, and the shaft is horizontal.
b
τ
W=mg
ϖ
The magnitude of the angular momentum
of the wheel is given by
1.
1
4
m R
2
ω
2.
1
2
m R
2
ω
3.
m R
2
ω
2
4.
m R
2
ω
correct
5.
1
2
m R
2
ω
2
6.
1
4
m R
2
ω
2
Explanation:
Solution: Basic Concepts:
~
τ
=
d
~
L
dt
Top view
L
τ
L
+
∆
L
∆
L
∆φ
The magnitude of the angular momentum of
the wheel,
L
, is
L
=
I ω
=
m R
2
ω,
since the moment of inertia of the wheel,
I
, is
m R
2
.
002
(part 2 of 3) 10 points
Given:
the mass 3 kg, the angular velocity
10 rad
/
s, the axil length
b
= 0
.
5 m, and
the radius of the wheel
R
= 0
.
47 m. Find the
precession angle in the time interval
t
= 1
.
8 s.
Correct answer: 228
.
768
◦
.
Explanation:
From the figure below, we get ∆
φ
=
∆
L
L
.
Using the relation, ∆
L
=
τ
∆
t
, where
τ
is the
magnitude of the torque,
mg
·
b
, we get
∆
φ
=
∆
L
L
=
τ
∆
t
L
=
m g b
∆
t
m R
2
ω
=
g b
∆
t
R
2
ω
=
(9
.
8 m
/
s
2
)(0
.
5 m)(1
.
8 s)
(0
.
47 m)
2
(10 rad
/
s)
= 3
.
99276 rad
= 228
.
768
◦
.
003
(part 3 of 3) 10 points
The direction of precession as viewed from the
top is:
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Husain, Zeena – Homework 10 – Due: Nov 4 2003, 4:00 am – Inst: H L Berk
2
1.
along the direction of rotation of the
wheel
2.
clockwise
3.
opposite to the direction of rotation of the
wheel
4.
counterclockwise
correct
Explanation:
From the figure, we can see the direction of
precession is counterclockwise.
004
(part 1 of 1) 10 points
Given:
g
= 9
.
8 m
/
s
2
.
Two weights attached to a uniform beam
of mass 26 kg are supported in a horizontal
position by a pin and cable as shown in the
figure.
41 kg
20 kg
3
.
3 m
7 m
28
◦
26 kg
What is the tension in the cable which sup
ports the beam?
Correct answer: 1
.
09233 kN.
Explanation:
Let :
m
= 26 kg
,
M
1
= 41 kg
,
M
2
= 20 kg
,
L
1
= 3
.
3 m
,
L
2
= 7 m
,
and
θ
= 28
◦
.
Basic Concepts:
X
~
F
= 0
X
~
τ
= 0
.
Solution:
The sum of the torques about the
pivot is
T L
2
sin
θ

mg
L
2
2

M
1
gL
1

M
2
gL
2
= 0
Solving for the tension
T
T
=
m g
L
2
2
+
M
1
g L
1
+
M
2
g L
2
L
2
sin
θ
=
(26 kg)(9
.
8 m
/
s
2
)
7 m
2
¶
(7 m) sin
θ
+
(41 kg)(9
.
8 m
/
s
2
)(3
.
3 m)
(7 m) sin
θ
+
(20 kg)(9
.
8 m
/
s
2
)(7 m)
(7 m) sin
θ
= 1
.
09233 kN
.
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 Fall '08
 Turner
 Force, Work, Sin, Cos, FY, Husain

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