University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises: 10.73

Answer

(a) The translational speed is 23.1 m/s and the angular speed is 110 rad/s (b) The translational speed is 23.1 m/s and the angular speed is 55 rad/s The translational speed at the bottom does not change but the angular speed is half the value found in part (a).

Work Step by Step

(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).
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