oldhomewk 27 – YOO, HEE – Due: Apr 1 2008, 4:00 am
1
Question 1, chap 13, sect 4.
part 1 of 3
10 points
A 3 kg bicycle wheel rotating at a
2509 rev
/
min angular velocity has its shaft
supported on one side, as shown in the Fg
ure. The wheel is a hoop of radius 0
.
6 m, and
its shaft is horizontal. The distance from the
center of the wheel to the pivot point is 0
.
6 m.
When viewing from the left (from the posi
tive
x
axes), one sees that the wheel is rotat
ing in a clockwise manner.
Assume:
All of the mass of the system is
located at the rim of the bicycle wheel. The
acceleration of gravity is 9
.
8 m
/
s
2
.
y
z
x
b
mg
ω
R
The magnitude of the angular momentum
of the wheel is given by
1.
b
v
L
b
=
1
4
mR
2
ω
2
2.
b
v
L
b
=
1
4
mR
2
ω
3.
b
v
L
b
=
1
2
mR
2
ω
4.
b
v
L
b
=
mR
2
ω
2
5.
b
v
L
b
=
1
2
mR
2
ω
2
6.
b
v
L
b
=
mR
2
ω
correct
Explanation:
Basic Concepts:
v
τ
=
d
v
L
dt
Solution:
The magnitude of the angular
momentum
I
of the wheel is
L
=
I ω
=
mR
2
ω ,
and is along the negative
x
axis.
Question 2, chap 13, sect 4.
part 2 of 3
10 points
Let :
m
= 3 kg
,
ω
= 2509 rev
/
min
b
= 0
.
6 m
,
and
R
= 0
.
6 m
.
±ind the change in the precession angle
after a 2 s time interval.
Correct answer: 7
.
12358
◦
(tolerance
±
1 %).
Explanation:
Let :
ω
= 2509 rev
/
min
=
2
π
(2509 rev
/
min)
(60 s
/
min)
= 262
.
742 rad
/
s
.
Solution:
The magnitude of the angular
momentum
I
of the wheel is
L
=
I ω
=
mR
2
ω ,
and is along the negative
x
axis.
±rom the Fgure in Part 2, we get Δ
φ
=
Δ
L
L
.
Using the relation, Δ
L
=
τ
Δ
t,
where
τ
is the
torque,
v
τ
=
v
b
×
mvg .
The precession angle Δ
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 Spring '08
 Turner
 Physics, Angular Momentum, Kinetic Energy, Momentum, Moment Of Inertia, Correct Answer

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