Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 12 - Kinematics of a Particle - Section 12.8 - Curvilinear Motion: Cylindrical Components - Problems - Page 83: 186

Answer

$a=8.66$ m/s$^2$

Work Step by Step

$\dot{\theta}=5$ $\ddot{\theta}=0$ $r=100(2-\cos \theta)$ $\dot{r}=100\dot{\theta} \sin \theta = 500\sin \theta$ $\ddot{r}=500 \dot{\theta} \cos \theta = 2500 \cos \theta$ At $\theta=120$ $a_r=\ddot{r}-r\dot{\theta}^2=2500\cos \theta - 100(2-\cos\theta)(5)^2=5000(\cos 120 - 1)=-7500 mm/s^2$ $a_s=r\ddot{\theta}+2\dot{r}\dot{\theta}=0+2(500\sin \theta)(5)=5000\sin 120 = 4330.1$ mm/s$^2$ $a=\sqrt{(-7500)^2+(4330.1)^2}=8660.3$ mm/s$^2 = 8.66$ m/s$^2$
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