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 324: 26

Answer

(a) $17453.29rad/s^2$ (b) $38484.5rad$

Work Step by Step

(a) We can find the required angular acceleration as follows: $\alpha=\frac{\omega-\omega_{\circ}}{t}$ We plug in the known values to obtain: $\alpha=\frac{350000rpm-0rpm}{2.1s}$ $\alpha=17453.29rad/s^2$ (b) We can find the required number of revolutions as follows: $\theta=\frac{\omega^2-\omega_{\circ}^2}{2\alpha}$ We plug in the known values to obtain: $\theta=\frac{(350000rpm)^2-(0rpm)^2}{2(17453.29rad/s^2)}$ $\theta=\frac{(350000rpm(\frac{2\pi rad}{60s}))^2}{2(17453.29rad/s^2)}$ $\theta=38484.5rad$
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