Chapter 5 - Problems - Page 188: 62

(a) At the top, the person's apparent weight is: $~~mg - m~\omega^2~R$ (b) At the bottom, the person's apparent weight is: $~~mg + m~\omega^2~R$

Work Step by Step

(a) The radial acceleration is: $~a_r = \omega^2~R$ At the top, we can find the normal force $F_N$ exerted on the person by the scale on the floor of the Ferris wheel. Note that the normal force is equal to the person's apparent weight. $\sum F = m~a_r$ $mg-F_N = m~a_r$ $F_N = mg - m~\omega^2~R$ At the top, the person's apparent weight is: $~~mg - m~\omega^2~R$ (b) At the bottom, we can find the normal force $F_N$ exerted on the person by the scale on the floor of the Ferris wheel. Note that the normal force is equal to the person's apparent weight. $\sum F = m~a_r$ $F_N - mg = m~a_r$ $F_N = mg + m~\omega^2~R$ At the bottom, the person's apparent weight is: $~~mg + m~\omega^2~R$.

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