Answer
(a) $a = 6.53~m/s^2$
(b) The tension in each cord is $4.24~N$
Work Step by Step
(a) We can find an expression for the tension $F_T$ in each cord:
$\tau = I~\alpha$
$(R)~(2F_T) = (\frac{1}{2}MR^2)~(\frac{a}{R})$
$F_T = \frac{1}{4}Ma$
We can find the acceleration of the cylinder:
$\sum F = Ma$
$Mg-2F_T = Ma$
$Mg-2(\frac{1}{4}Ma) = Ma$
$Mg = \frac{3}{2}Ma$
$a = \frac{2g}{3}$
$a = \frac{(2)(9.80~m/s^2)}{3}$
$a = 6.53~m/s^2$
(b) We can find the tension $F_T$ in each cord:
$F_T = \frac{Ma}{4}$
$F_T = \frac{(2.6~kg)(6.53~m/s^2)}{4}$
$F_T = 4.24~N$
The tension in each cord is $4.24~N$
