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Lm
v
r
I
mR
ii
i
i
=+
=
+
111
2 2
1 0
2
20
2
1
2
ωω
ω
.
After the cockroach has completed its walk, its position (relative to the axis) is
rR
f
1
2
=
so the final angular momentum of the system is
R
ff
f
=
F
H
G
I
K
J
+
1
2
2
2
2
1
2
.
Then from
L
f
= L
i
we obtain
f
1
4
1
2
1
2
1
2
1
2
2
2
+
F
H
G
I
K
J
F
H
G
I
K
J
.
Thus,
22
12
2
1
00
0
0
2
1
21
(
/
)
2
1.33
.
42
1
/
4
(
/
)
2
1
/
4
2
f
m m
§·
§
·
++
+
==
=
=
¨¸
¨
¸
+
©¹
©
¹
With
0
= 0.260 rad/s, we have
f
=0.347 rad/s.
(b) We substitute
I
=
L
/
into
KI
=
1
2
2
and obtain
KL
=
1
2
. Since we have
L
i
= L
f
,
the kinetic energy ratio becomes
/2
1.33.
f
L
K
=
(c) The cockroach does positive work while walking toward the center of the disk,
increasing the total kinetic energy of the system.
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Angular Momentum, Mass, Momentum

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