# Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 199: 23

(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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