Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - Problems: 13

Answer

People should feel an effect of 0.90 g when the station rotates at a rate of 1700 rev/day.

Work Step by Step

$\frac{v^2}{r} = a$ $v^2 = ar$ $v = \sqrt{ar}$ $v = \sqrt{(0.90)(9.8 ~m/s^2)(550 ~m)}$ $v = 69.65 ~m/s$ We can use the velocity to find the number of revolutions per second. Let $z$ be the number of revolutions each second. So, $z\times 2\pi r = v$ $z = \frac{v}{2 \pi r}$ $z =\frac{69.65 ~m/s}{(2)(\pi)(550 ~m)}$ $z = 0.020 ~rev/s$ We can convert the rate of rotation to revolutions per day: $z = (0.020 ~rev/s)(\frac{24\times 3600 ~s}{1 ~day})$ $z = 1700 ~rev/day$ People should feel an effect of 0.90 g when the station rotates at a rate of 1700 rev/day.
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