## University Physics with Modern Physics (14th Edition)

(a) We can find the mass of the wheel rim. $M_r = (2\pi~R)(25.0~g/cm)$ $M_r = (2\pi)(21.0~cm)(25.0~g/cm)$ $M_r = 3.299~kg$ We can find the mass of one spoke. $M_s = (21.0~cm)(25.0~g/cm)$ $M_s = 0.525~kg$ We can find the moment of inertia of the wheel. $I = M_rR^2+6\times \frac{1}{3}M_sR^2$ $I = (M_r+2M_s)~R^2$ $I = [(3.299~kg)+(2)(0.525~kg)](0.210~m)^2$ $I = 0.1918~kg~m^2$ We can use conservation of energy to find the speed at the bottom of the hill. $K= PE$ $\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2=Mgh$ $\frac{1}{2}Mv^2+\frac{1}{2}I(\frac{v}{R})^2=Mgh$ $MR^2v^2+Iv^2=2MghR^2$ $v^2=\frac{2MghR^2}{MR^2+I}$ $v=\sqrt{\frac{2MghR^2}{MR^2+I}}$ $v=\sqrt{\frac{(2)(3.824~kg)(9.80~m/s^2)(58.0~m)(0.210~m)^2}{(3.824~kg)(0.210~m)^2+(0.1918~kg~m^2)}}$ $v = 23.1~m/s$ We can find the angular speed. $\omega = \frac{v}{R} = \frac{23.1~m/s}{0.210~m}$ $\omega = 110~rad/s$ (b) We can find the mass of the wheel rim. $M_r = (2\pi~R)(50.0~g/cm)$ $M_r = (2\pi)(42.0~cm)(50.0~g/cm)$ $M_r = 13.19~kg$ We can find the mass of one spoke. $M_s = (42.0~cm)(50.0~g/cm)$ $M_s = 2.10~kg$ We can find the moment of inertia of the wheel. $I = M_rR^2+6\times \frac{1}{3}M_sR^2$ $I = (M_r+2M_s)~R^2$ $I = [(13.19~kg)+(2)(2.10~kg)](0.420~m)^2$ $I = 3.068~kg~m^2$ We can use conservation of energy to find the speed at the bottom of the hill. $K= PE$ $\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2=Mgh$ $\frac{1}{2}Mv^2+\frac{1}{2}I(\frac{v}{R})^2=Mgh$ $MR^2v^2+Iv^2=2MghR^2$ $v^2=\frac{2MghR^2}{MR^2+I}$ $v=\sqrt{\frac{2MghR^2}{MR^2+I}}$ $v=\sqrt{\frac{(2)(15.29~kg)(9.80~m/s^2)(58.0~m)(0.420~m)^2}{(15.29~kg)(0.420~m)^2+(3.068~kg~m^2)}}$ $v = 23.1~m/s$ We can find the angular speed. $\omega = \frac{v}{R} = \frac{23.1~m/s}{0.420~m}$ $\omega = 55~rad/s$ The translational speed at the bottom does not change but the angular speed is half the value found in part (a).