## Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson

# Chapter 16 - Planar Kinematics of a Rigid Body - Section 16.4 - Absolute Motion Analysis - Problems - Page 342: 40

#### Answer

$\omega=28.9~rad/s \circlearrowright$ $\alpha=470~rad/s^2\circlearrowleft$

#### Work Step by Step

The required angular velocity and angular acceleration can be determined as follows: We know that $\frac{dx}{dt}=V=-0.2sin\theta\frac{d\theta}{dt}$ $\omega=\frac{d\theta}{dt}$ $\implies V=-0.2\omega sin\theta$ We plug in the known values to obtain: $-5=-0.2\omega sin 60$ This simplifies to: $\omega=28.9~rad/s \circlearrowright$ As $V=-0.2\omega sin\theta$ Differentiating with respect to $t$, we obtain: $\frac{dV}{dt}=a=-0.2sin\theta\frac{d\omega}{dt}-0.2\omega cos\theta \frac{d\theta}{dt}$ This simplifies to: $a=-0.2\alpha sin\theta-0.2\omega^2 cos\theta$ We plug in the known values to obtain: $-2=-0.2\alpha sin60-0.2(28.8675)^2cos60$ $\implies \alpha=470~rad/s^2\circlearrowleft$

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