## College Physics (4th Edition)

(a) $a = 6.53~m/s^2$ (b) The tension in each cord is $4.24~N$
(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 $4.24~N$