Bautista, Aldo – Homework 10 – Due: Nov 8 2005, 4:00 am – Inst: Maxim Tsoi
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The due time is Central
time.
001
(part 1 of 1) 10 points
A uniform solid sphere
I
sphere
=
2
5
m r
2
¶
of
mass 1
.
8 kg and radius
r
= 0
.
209 m, is placed
on the inside surface of a hemispherical bowl
of radius
R
= 1
.
6 m. The sphere is released
from rest at an angle
θ
= 46
.
1
◦
from the
vertical and rolls without slipping (see the
figure).
The acceleration of gravity is 9
.
8 m
/
s
2
.
θ
R
r
What is the angular velocity of the sphere
when it reaches the bottom of the bowl?
Correct answer: 11
.
6914 rad
/
s.
Explanation:
Basic Concepts:
Energy conservation
(potential + kinetic = constant):
U
i
+
K
i
=
U
f
+
K
f
.
Use energy conservation for this problem. The
initial energy is the potential energy of the
center of mass of the sphere above its lowest
point.
The distance between the center of
mass of the sphere at its highest (rest) and its
lowest points is
h
= (
R

r
)

(
R

r
) cos
θ
= (
R

r
)(1

cos
θ
)
= (1
.
6 m

0
.
209 m)(1

cos 46
.
1
◦◦
)
= 0
.
426478 m
so that
U
i
=
m g h
= (1
.
8 kg) (9
.
8 m
/
s
2
) (0
.
426478 m)
= 7
.
52308 J
,
U
f
= 0 J
,
and
Δ
U
=
U
i

U
f
= 7
.
52308 J
.
where
m
is the mass of the sphere.
The
moment of inertia of the sphere is
I
sphere
=
2
5
m r
2
and
ω
=
v
r
. The energy of the sphere
at the bottom of the bowl is all kinetic. This
final kinetic energy:
K
f
=
1
2
m v
2
f
+
1
2
I ω
2
f
=
1
2
+
2
10
¶
m v
2
f
=
7
10
m v
2
f
.
Energy conservation gives
7
10
m v
2
f
=
m g h
v
f
=
r
10
7
g h
=
r
10
7
(9
.
8 m
/
s
2
) (0
.
426478 m)
= 2
.
4435 m
/
s
.
ω
f
=
v
f
r
=
2
.
4435 m
/
s
0
.
209 m
= 11
.
6914 rad
/
s
.
002
(part 1 of 3) 10 points
A
3
kg
bicycle
wheel
rotating
at
a
2676 rev
/
min angular velocity has its shaft
supported on one side, as shown in the fig
ure. The wheel is a hoop of radius 0
.
5 m, and
its shaft is horizontal. The distance from the
center of the wheel to the pivot point is 0
.
5 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
.
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Bautista, Aldo – Homework 10 – Due: Nov 8 2005, 4:00 am – Inst: Maxim Tsoi
2
y
z
x
b
m g
ω
R
The magnitude of the angular momentum
of the wheel is given by
1.
k
~
L
k
=
1
2
m R
2
ω
2.
k
~
L
k
=
1
4
m R
2
ω
3.
k
~
L
k
=
1
4
m R
2
ω
2
4.
k
~
L
k
=
m R
2
ω
correct
5.
k
~
L
k
=
1
2
m R
2
ω
2
6.
k
~
L
k
=
m R
2
ω
2
Explanation:
Basic Concepts:
~
τ
=
d
~
L
dt
Solution:
The magnitude of the angular
momentum
I
of the wheel is
L
=
I ω
=
m R
2
ω ,
and is along the negative
x
axis.
003
(part 2 of 3) 10 points
Let :
m
= 3 kg
,
ω
= 2676 rev
/
min
b
= 0
.
5 m
,
and
R
= 0
.
5 m
.
Find the change in the precession angle
after a 1
.
3 s time interval.
Correct answer: 5
.
20964
◦
.
Explanation:
Let :
ω
= 2676 rev
/
min
=
2
π
(2676 rev
/
min)
(60 s
/
min)
= 280
.
23 rad
/
s
.
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 Fall '08
 Turner
 Force, Work, Correct Answer, Bautista, Maxim Tsoi

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