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: 12a

Answer

$9000\ rev/min^2$

Work Step by Step

Given: Initial angular acceleration speed $\omega_i=1200\ rev/s$ Final angular acceleration speed $\omega_f=3000\ rev/s$ Time interval $\Delta t =12\ s$ From the rotational kinetic equations, we have: $\omega_f = \omega_i+\alpha \Delta t$ $3000\ rev/min = 1200\ rev/min +\alpha (12\ s) (\frac{1\ min}{60\ s}) $ $1800\ rev/min =\alpha (12\ s) (\frac{1\ min}{60\ s}) $ $\alpha =9000\ rev/min^2$
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