Physics (10th Edition)

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

Chapter 8 - Rotational Kinematics - Problems - Page 214: 33



Work Step by Step

At the instant the child spots the horse, the distance between the child and the horse is 1/4 a turn, equaling $\Delta\theta=\frac{1}{4}\times2\pi rad=\frac{\pi}{2}rad$ Take the shortest time it takes for the child to catch up with the horse to be $t$. During this time, - The child has a displacement of $\theta_c=0.25t$ - The merry-go-round has a displacement of $\theta_m=\omega_0t+\frac{1}{2}\alpha t^2=0+\frac{1}{2}(0.01)t^2=(5\times10^{-3})t^2$ Since we know the distance between them, we see that $$\Delta\theta=\theta_c-\theta_m=\frac{\pi}{2}rad$$ $$(5\times10^{-3})t^2-0.25t+\frac{\pi}{2}=0$$ $$t_{min}=7.37s$$
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