Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 72

Answer

(a) t = 6.32 s (b) The motor turns through an angle of 84.3 radians.

Work Step by Step

(a) At the moment when the motor changes direction, the angular velocity will be zero. We can find the time when the angular velocity is zero as; $\omega(t) = (20-\frac{1}{2}t^2)~rad/s = 0$ $t^2 = 40$ $t = \sqrt{40}$ $t = 6.32~s$ (b) The angle that the motor turns is equal to the area under the omega versus time graph. We can integrate to find this area; $\Delta \theta = \int_{0}^{\sqrt{40}}\omega(t)~dt$ $\Delta \theta = \int_{0}^{\sqrt{40}}(20-\frac{1}{2}t^2)~dt$ $\Delta \theta = (20t-\frac{1}{6}t^3)\vert_{0}^{\sqrt{40}}$ $\Delta \theta = (20)(\sqrt{40})-\frac{1}{6}(\sqrt{40})^3$ $\Delta \theta = 84.3~rad$ The motor turns through an angle of 84.3 radians.
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