## College Physics (4th Edition)

(a) The speed of the yo-yo when it reaches the distance of 1.00 m is $1.48~m/s$ (b) It takes $1.35~seconds$ for the yo-yo to fall 1.00 meter.
(a) Let $R_o$ be the outer radius of the yo-yo. Let $R_a$ be the axle radius of the yo-yo. We can use conservation of energy to find the yo-yo's speed after falling 1.0 meter: $U_g = KE_{tr}+KE_{rot}$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{2}MR_o^2)~(\frac{v}{R_a})^2$ $Mgh = \frac{1}{2}Mv^2+\frac{1}{4}M(\frac{R_0^2}{R_a^2}~)v^2$ $4gh = v^2 ~(2+\frac{R_0^2}{R_a^2})$ $v^2 = \frac{4gh}{2+\frac{R_0^2}{R_a^2}}$ $v = \sqrt{\frac{4gh}{2+\frac{R_0^2}{R_a^2}}}$ $v = \sqrt{\frac{(4)(9.80~m/s^2)(1.0~m)}{2+\frac{(0.020~m)^2}{(0.0050~m)^2}}}$ $v = 1.48~m/s$ The speed of the yo-yo when it reaches the distance of 1.00 m is $1.48~m/s$ (b) We can find the average speed of the yo-yo as it falls: $v_{ave} = \frac{v_0+v_f}{2} = \frac{0+1.48~m/s}{2} = 0.74~m/s$ We can find the time it takes the yo-yo to fall 1.00 meter: $t = \frac{d}{v_{ave}} = \frac{1.00~m}{0.74~m/s} = 1.35~s$ It takes $1.35~seconds$ for the yo-yo to fall 1.00 meter.