University Physics with Modern Physics (14th Edition)

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

Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 298: 9.64

Answer

(a) $v = 75.1~m/s$ (b) $a_{rad} = 54200~m/s^2$

Work Step by Step

(a) The angular velocity of the motor is 3450 rev/min. Since the shaft has half the diameter, then the angular velocity is double, that is, 6900 rev/min. $\omega = (6900~rev/min)(\frac{1~min}{60~s})(2\pi~rad/rev)$ $\omega = 722.57~rad/s$ We can find the tangential speed on the rim of the blade. $v = \omega ~r = (722.57~rad/s)(0.104~m)$ $v = 75.1~m/s$ (b) $a_{rad} = \frac{v^2}{r} = \frac{(75.1~m/s)^2}{0.104~m}$ $a_{rad} = 54200~m/s^2$
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