## College Physics (4th Edition)

(a) The speed of the bucket is $~3.0~m/s$ (b) The tension in the rope is $8.35~N$ (c) The acceleration of the bucket is $5.63~m/s^2$
(a) Let $M_b$ be the mass of the bucket. Let $M_c$ be the mass of the cylinder. We can use conservation of energy to find the speed at the lowest end of the board: $U_g = KE_{tr}+KE_{rot}$ $M_bgh = \frac{1}{2}M_bv^2+\frac{1}{2}I~\omega^2$ $M_bgh = \frac{1}{2}M_bv^2+\frac{1}{2}(\frac{1}{2}M_cR^2)~(\frac{v}{R})^2$ $M_bgh = \frac{1}{2}M_bv^2+\frac{1}{4}M_cv^2$ $4M_bgh = v^2 (2M_b+M_c)$ $v^2 = \frac{4M_bgh}{2M_b+M_c}$ $v = \sqrt{\frac{4M_bgh}{2M_b+M_c}}$ $v = \sqrt{\frac{(4)(2.0~kg)(9.80~m/s^2)(0.80~m)}{(2)(2.0~kg)+(3.0~kg)}}$ $v = 3.0~m/s$ The speed of the bucket is $~3.0~m/s$ (b) We can consider the energy of the bucket. The tension in the rope does negative work on the bucket as it falls. We can use work and energy to find the tension $F_T$ in the rope: $KE_{tr} = U_g+Work$ $\frac{1}{2}M_bv^2 = M_bgh-F_T~h$ $F_T~h = M_bgh- \frac{1}{2}M_bv^2$ $F_T = M_b~(g- \frac{v^2}{2h})$ $F_T = (2.0~kg)~[9.80~m/s^2- \frac{(3.0~m/s)^2}{(2)(0.80~m)}]$ $F_T = 8.35~N$ The tension in the rope is $8.35~N$ (c) We can find the acceleration of the bucket: $v_f^2 = v_0^2+2ah$ $a = \frac{v_f^2 - v_0^2}{2h}$ $a = \frac{(3.0~m/s)^2 - 0}{(2)(0.80~m)}$ $a = 5.63~m/s^2$ The acceleration of the bucket is $5.63~m/s^2$.