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 112: 32

Answer

(a) $v = 3.9~m/s$ (b) The merry-go-round makes 2.5 revolutions.

Work Step by Step

(a) We can find the angular velocity as: $\omega = \frac{2\pi~rad}{4.0~s}$ $\omega = 1.57~rad/s$ We can find the speed of a child on the rim as: $v = \omega ~r$ $v = (1.57~rad/s)(2.5~m)$ $v = 3.9~m/s$ (b) We can find the angle $\theta$ that the merry-go-round rotates through as it slows down: $\theta = (\frac{\omega_0+0}{2})(20~s)$ $\theta = (\frac{1.57~rad/s+0}{2})(20~s)$ $\theta = 15.7~rad$ We can convert $\theta$ to revolutions; $\theta = (15.7~rad)(\frac{1~rev}{2\pi~rad})$ $\theta = 2.5~rev$ The merry-go-round makes 2.5 revolutions.

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