85. (First problem in Cluster 2)

The last line of the problem indicates our choice of positive directions: up for m

2

, down for m

1

and

counterclockwise for the two-pulley device. This allows us to write R

2

α = a

2

and R

1

α = a

1

with all

terms positive. We apply Newton’s second law to the elements of this system:

T

2

− m

2

g

=

m

2

a

2

= m

2

R

2

α

m

1

g

− T

1

=

m

1

a

1

= m

1

R

1

α

T

1

R

1

− T

2

R

2

=

Iα

Multiplying the first equation by R

2

, the second by R

1

and adding the equations leads to

α =

m

1

gR

1

− m

2

gR

2

I + m

1

R

2

1

+ m

2

R

2

2

.

(a) Therefore, again using R

1

α = a

1

, we obtain

a

1

=

m

1

gR

2

1

− m

2

gR

1

R

2

I + m

1

R

2

1

+ m

2

R

2

2

.

(b) Once more, we use R

2

α = a

2

and find

a

2

=

m

1

gR

1

R

2

− m

2

gR

2

2

I + m

1

R

2

1

+ m

2

R

2

2

.