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

Published by Pearson

# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems - Page 114: 68

#### 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. $\omega(t) = (20-\frac{1}{2}t^2)~rad/s = 0$ $t^2 = 40$ $t = \sqrt{40}$ $t = 6.32~s$ (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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