Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 10 - Rotational Kinematics and Energy - Problems and Conceptual Exercises - Page 329: 104

Answer

$780\ rev$

Work Step by Step

We know that $\Delta \theta_1=\omega_{avg,1}\Delta t_1$ $\implies \Delta \theta_1=0.95(15)=14rev$ $\Delta \theta_2=\omega_{avg,2}\Delta t_2$ $\implies \Delta \theta_2=1.9\times (5.5min\times \frac{60s}{min})=630 rev$ $\Delta \theta_3=\omega_{avg,3}\Delta t_3$ $\implies \Delta \theta_3=0.95\times (2.4min\times \frac{60s}{min})=140 rev$ Now $\Delta \theta=\Delta \theta_1+\Delta \theta_2+\Delta \theta_3$ $\Delta \theta=14+630+140=780\ rev$
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