Alvarado, Patrick – Homework 6 – Due: Oct 19 2006, 10:00 am – Inst: Andrei Sirenko
1
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printout
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have
19
questions.
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before answering.
The due time is Central
time.
001
(part 1 of 2) 10 points
A cylinder with moment of inertia 23
.
6 kg m
2
rotates with angular velocity 5
.
5 rad
/
s on a
frictionless vertical axle.
A second cylinder,
with moment of inertia 40
.
4 kg m
2
, initially
not rotating, drops onto the first cylinder and
remains in contact.
Since the surfaces are
rough, the two eventually reach the same an
gular velocity.
I
2
I
1
ω
0
Before
ω
After
Calculate the final angular velocity.
Correct answer: 2
.
02813 rad
/
s.
Explanation:
From conservation of angular momentum
(
I
1
+
I
2
)
ω
=
I
1
ω
0
,
or
ω
=
I
1
I
1
+
I
2
ω
0
=
(23
.
6 kg m
2
)
(23
.
6 kg m
2
) + (40
.
4 kg m
2
)
(5
.
5 rad
/
s)
=
2
.
02813 rad
/
s
.
002
(part 2 of 2) 10 points
Show that energy is lost in this situation by
calculating the ratio of the final to the initial
kinetic energy.
Correct answer: 0
.
36875 .
Explanation:
K
i
=
1
2
I
1
ω
0
2
and
K
f
=
1
2
(
I
1
+
I
2
)
ω
2
,
Therefore
K
f
=
1
2
(
I
1
+
I
2
)
ω
2
=
1
2
(
I
1
+
I
2
)
I
1
I
1
+
I
2
¶
2
ω
2
0
=
I
1
I
1
+
I
2
¶
1
2
I
1
ω
2
0
=
I
1
I
1
+
I
2
¶
K
i
.
So
K
f
K
i
=
I
1
I
1
+
I
2
=
(23
.
6 kg m
2
)
(23
.
6 kg m
2
) + (40
.
4 kg m
2
)
=
0
.
36875
<
1
.
keywords:
003
(part 1 of 1) 10 points
Two disks of identical mass but different radii
(
r
and 2
r
) are spinning on frictionless bear
ings at the same angular speed
ω
0
but in op
posite directions. The two disks are brought
slowly together. The resulting frictional force
between the surfaces eventually brings them
to a common angular velocity.
2
r
ω
0
r
ω
0
What is the magnitude of that final angular
velocity in terms of
ω
0
?
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Alvarado, Patrick – Homework 6 – Due: Oct 19 2006, 10:00 am – Inst: Andrei Sirenko
2
1.
ω
f
=
4
5
ω
0
2.
ω
f
=
2
5
ω
0
3.
ω
f
=
1
4
ω
0
4.
ω
f
=
1
3
ω
0
5.
ω
f
=
1
5
ω
0
6.
ω
f
=
3
4
ω
0
7.
ω
f
=
1
2
ω
0
8.
ω
f
=
3
5
ω
0
correct
9.
ω
f
=
2
3
ω
0
Explanation:
Note:
Since the disks are spinning in oppo
site directions, let
ω
1
=
ω
0
and
ω
2
=

ω
0
.
The inertia of the larger disk is
I
1
=
1
2
m
(2
r
)
2
= 2
m r
2
,
and of the smaller disk
I
2
=
1
2
m r
2
.
Using conservation of angular momentum,
I
i
ω
i
=
I
f
ω
f
I
1
ω
0

I
2
ω
0
= (
I
1
+
I
2
)
ω
f
ω
f
=
I
1

I
2
I
1
+
I
2
ω
0
=
2
m r
2

1
2
m r
2
2
m r
2
+
1
2
m r
2
ω
0
=
3
5
ω
0
.
keywords:
004
(part 1 of 3) 10 points
Consider an Earthlike planet hit by an aster
oid.
The planet has mass
M
p
= 5
.
54
×
10
23
kg
and radius
R
p
= 5
.
89
×
10
6
m, and you may
approximate it as a solid ball of uniform
density.
It rotates on its axis once every
T
= 15 hr.
The asteroid has mass
M
a
=
3
.
82
×
10
17
kg and speed
v
a
= 23200 m
/
s
(relative to the planet’s center); its velocity
vector points
θ
= 73
◦
below the Eastward hor
izontal. The impact happens at an equatorial
location.
The picture below shows the view
from above the planet’s North pole:
v
m
θ
ω
R
First, calculate the planet’s angular mo
mentum (relative to its spin axis)
before
the
impact.
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 Spring '08
 OPYT
 Angular Momentum, Kinetic Energy, Mass, Work, Rotation

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