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.