fisica1-youn-e-freedman-exercicios-resolvidos

10.66: The accelerations of blocks A and B will have the same magnitude a. Since the
a
cord does not slip, the angular acceleration of the pulley will be α = . Denoting the
tensions in the cord as T A
and T B
, the equations of motion are
m g − T
T
B
A
T
− m g = m a
A
B
− T
A
B
= m
=
A
B
a
I
a, 2
R
where the last equation is obtained by dividing τ = Iα
by R and substituting for α in
terms of a.
Adding the three equations eliminates both tensions, with the result that
mA
− mB
a = g
2
mA
+ mB
+ I / R
Then,
a mA
− mB
α = = g
.
R mAR
+ mB
R + I / R
The tensions are then found from
R
T
T
A
B
2mAmB
+ mA
I R
= mA
( g − a)
= g
2
m + m + I R
2mBmA
+ mB
I R
= mB
( g + a)
= g
2
m + m + I R
A
A
B
B
2
2
.
As a check, it can be shown that ( T − T ) R Iα.
A B
=