Answer
a) $a=0.69\frac{m}{s^2}$
b) $\%_{error}=2.1\%$
Work Step by Step
a) positive:
$F_{TA}$ and $F_{GB}$
clockwise torque
negative:
$F_{TB}$ and $F_{GA}$
anti clockwise torque
Newton's Second Law
$(65kg)a=F_{TA}-F_{GA}$
$F_{TA}=(65kg)a+(65kg)g$
$(75kg)a=F_{GB}-F_{TB}$
$F_{TB}=-(75kg)a+(75kg)a$
$\sum \tau=I \alpha$
$\sum \tau=\frac{MR^2}{2}\times\frac{a}{R}=R(F_{TB}-F_{TA})$
$\frac{Ma}{2}=-(75kg)a+(75kg)a-(65kg)a-(65kg)g$
$a=\frac{(75kg)g-(65kg)g}{\frac{6.0kg}{2}+65kg+75kg}=0.69\frac{m}{s^2}$
b) When Moment of Inertia ignored
$a_{Ii}=\frac{(75kg)g-(65kg)g}{65kg+75kg}=0.7\frac{m}{s^2}$
$\%_{error}=\frac{a_{Ii}-a}{a_{Ii}}\times100\%=\frac{0.69\frac{m}{s^2}-0.7\frac{m}{s^2}}{0.69\frac{m}{s^2}}\times100\%=2.1\%$
