Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 10 - Rotation - Problems - Page 292: 77a

Answer

$\alpha = -66~\frac{2}{3}~rev/min^2$

Work Step by Step

We can find the angular acceleration: $\omega_f = \omega_0+\alpha~t$ $\alpha = \frac{\omega_f - \omega_0}{t}$ $\alpha = \frac{0 - 33~\frac{1}{3}~rev/min}{0.50~min}$ $\alpha = -66~\frac{2}{3}~rev/min^2$
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