Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems - Page 111: 21

Answer

See the angular position-versus-time graph below,

Work Step by Step

We know that the angular velocity is given by $$\omega=\dfrac{\theta}{t}$$ and to find the angular position of an object from its angular velocity versus time graph, we need to find the area under the curve since $$\theta=\omega t$$ Thus, from $t=0$ s to $t=4$ s, the area under the curve is given by $$A_1=\theta_1=\omega_1 \Delta t_1=20\cdot 4=\bf 80\;\rm rad$$ And the area under the curve from $t=4$ s to $t=6$ s is given by $$A_2=\theta_2=\omega_2 \Delta t_2=2\cdot 2=\bf 0\;\rm rad$$ Finally, the area under the curve from $t=6$ s to $t=8$ s is given by $$A_3=\theta_3=\omega_3\Delta t_3=-10\cdot 2=\bf -20\;\rm rad$$ Now we can draw the angular position-versus-time graph, as we see below.
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