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71. We denote the cat with subscript 1 and the ring with subscript 2. The cat has a mass
m
1
=
M/
4, while the mass of the ring is
m
2
=
M
= 8.00 kg. The moment of inertia of the
ring is
22
12
()
/
2
Im
RR
=+
(Table 102), and
I
1
=
m
1
r
2
for the cat, where
r
is the
perpendicular distance from the axis of rotation.
Initially the angular momentum of the system consisting of the cat (at
r
=
R
2
) and the ring
is
2
2
2
21
111
2 2
1 0 2
2
1
2
0
1 2
0
2
11
1
1
.
ii
i i
mR
Lm
v
rI
mRR
m
R
ωω
ω
ªº
§·
=
+
+
=
+
+
«»
¨¸
©¹
¬¼
After the cat has crawled to the inner edge at
1
rR
=
the final angular momentum of the
system is
2
2
1
1
2
1
1.
ff
f
f
R
m
R
R
m
R
+
=
+
+
Then from
L
f
= L
i
we obtain
2
2
2
2
2
2
01
2
1
2
1 2(0.25 1)
(2.0)
1.273
(
14
)
1
2
f
R
R
++
==
=
.
Thus,
0
1.273
f
=
. Using
0
=8.00 rad/s, we have
f
=10.2 rad/s. By substituting
I
=
L
/
into
2
/2
KI
=
, we obtain
KL
=
. Since
L
i
= L
f
, the kinetic energy ratio
becomes
0
1.273.
f
f
i
=
which implies
0.273
f
KK K
K
Δ= − =
. The cat does positive work while walking toward
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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, Inertia, Mass

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