52. We denote the cockroach with subscript 1 and the disk with subscript 2.
(a) Initially the angular momentum of the system consisting of the cockroach and the disk is
L
i
= m
1
v
1i
r
1i
+ I
2
ω
2i
= m
1
ω
0
R
2
+
1
2
m
2
ω
0
R
2
.
After the cockroach has completed its walk, its position (relative to the axis) is r
1f
= R/2 so the
final angular momentum of the system is
L
f
= m
1
ω
f
R
2
2
+
1
2
m
2
ω
f
R
2
.
Then from L
f
= L
i
we obtain
ω
f
1
4
m
1
R
2
+
1
2
m
2
R
= ω
0
m
1
R
2
+
1
2
m
2
R
2
.
Thus,
ω
f
− ω
0
=
ω
0
m
1
R
2
+ m
2
R
2
/2
m
1
R
2
/4 + m
2
R
2
/2
− ω
0
=
ω
0
m + 10m/2
m/4 + 10m/2
− 1
=
ω
0
(1.14
− 1)
which yields ∆ω = 0.14ω
0
. For later use, we note that ω
f
/ω
i
= 1.14.
(b) We substitute I = L/ω into K =
1
2
Iω
2
and obtain K =
1
2
Lω. Since we have L
i
= L
f
, the the
kinetic energy ratio becomes
K
K
0
=
1
2
L
f
ω
f
1
2
L
i
ω
i
=
ω
f
ω
i
= 1.14 .
(c) The cockroach does positive work while walking toward the center of the disk, increasing the total
kinetic energy of the system.