College Physics (4th Edition)

It takes $2.75~seconds$ for the bucket to fall to the bottom of the well.
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 bucket's speed at the bottom of the well: $U_g = KE_{tr}+KE_{rot}$ $M_bgh = \frac{1}{2}M_bv^2+\frac{1}{2}I~\omega^2$ $M_bgh = \frac{1}{2}M_bv^2+\frac{1}{2}(\frac{1}{2}M_cR^2)~(\frac{v}{R})^2$ $M_bgh = \frac{1}{2}M_bv^2+\frac{1}{4}M_cv^2$ $4M_bgh = v^2 (2M_b+M_c)$ $v^2 = \frac{4M_bgh}{2M_b+M_c}$ $v = \sqrt{\frac{4M_bgh}{2M_b+M_c}}$ $v = \sqrt{\frac{(4)(1.10~kg)(9.80~m/s^2)(17.0~m)}{(2)(1.10~kg)+(2.60~kg)}}$ $v = 12.36~m/s$ We can find the average speed of the bucket as it falls: $v_{ave} = \frac{v_0+v_f}{2} = \frac{0+12.36~m/s}{2} = 6.18~m/s$ We can find the time it takes the bucket to fall 17.0 meters: $t = \frac{d}{v_{ave}} = \frac{17.0~m}{6.18~m/s} = 2.75~s$ It takes $2.75~seconds$ for the bucket to fall to the bottom of the well.