Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 8 - Rotational Kinematics - Problems - Page 214: 29

Answer

The propeller's total displacement is $1.95\times10^4rad$

Work Step by Step

1) Phase 1: acceleration We know $\omega_0=0$, $\alpha=2.9\times10^{-3}rad/s^2$ and $t=2.1\times10^3s$ The propeller's displacement: $$\theta_1=\omega_0t+\frac{1}{2}\alpha t^2=6.39\times10^3rad$$ Its final speed: $$\omega=\omega_0+\alpha t=6.09rad/s$$ 2) Phase 2: constant speed for $\Delta t=1.4\times10^3s$ The propeller's displacement: $$\theta_2=\omega\Delta t=8.53\times10^3rad$$ 3) Phase 3: deceleration We know $\omega_0=6.09rad/s$, $\omega=4rad/s$ and $\alpha=-2.3\times10^{-3}rad/s^2$ The propeller's displacement: $$\theta_3=\frac{\omega^2-\omega_0^2}{2\alpha}=4.58\times10^3rad$$ The propeller's total displacement is $$\sum\theta=\theta_1+\theta_2+\theta_3=1.95\times10^4rad$$
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