Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems: 23

Answer

(a) $v = 3.8~m/s$ $a_c = 0.96~m/s^2$ (b) The ratio of the apparent weight to the true weight is 0.90 (c) The ratio of the apparent weight to the true weight is 1.1

Work Step by Step

(a) We find the speed first; $v = \frac{distance}{time}$ $v = \frac{2\pi~r}{t}$ $v = \frac{(2\pi)(15~m)}{25~s}$ $v = 3.8~m/s$ We then find the centripetal acceleration; $a_c = \frac{v^2}{r}$ $a_c = \frac{(3.8~m/s)^2}{15~m}$ $a_c = 0.96~m/s^2$ (b) Let $F_N$ be the normal force of the seat pushing up on the person. Note that $F_N$ is equal to the person's apparent weight. $\sum F = \frac{Mv^2}{r}$ $Mg-F_N = \frac{Mv^2}{r}$ $F_N = Mg-\frac{Mv^2}{r}$ $F_N = (M)(9.80~m/s^2)-\frac{M(3.8~m/s^2)^2}{15~m}$ $F_N = (M)(9.80~m/s^2-0.96~m/s^2)$ $F_N = (M)(8.84~m/s^2)$ We can find the ratio of $F_N$ to $mg$: $ratio = \frac{F_N}{mg}$ $ratio = \frac{(M)(8.84~m/s^2)}{(M)(9.80~m/s^2)}$ $ratio = 0.90$ The ratio of the apparent weight to the true weight is 0.90 (c) We can find an expression for the normal force at the bottom. $\sum F = \frac{Mv^2}{r}$ $F_N-Mg = \frac{Mv^2}{r}$ $F_N = Mg+\frac{Mv^2}{r}$ $F_N = (M)(9.80~m/s^2)+\frac{M(3.8~m/s^2)^2}{15~m}$ $F_N = (M)(9.80~m/s^2+0.96~m/s^2)$ $F_N = (M)(10.76~m/s^2)$ We can find the ratio of $F_N$ to $mg$: $ratio = \frac{F_N}{mg}$ $ratio = \frac{(M)(10.76~m/s^2)}{(M)(9.80~m/s^2)}$ $ratio = 1.1$ The ratio of the apparent weight to the true weight is 1.1
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