## 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: 41

#### Answer

$\omega=8.7027rad/s$ $\alpha=-50.4968rad/s^2$

#### Work Step by Step

We can determine the required angular velocity and angular acceleration as follows: As $cos\theta=\frac{-y}{0.3}$ $\implies y=-0.3cos\theta$ We know that $v_y=\frac{dy}{dt}$ $\implies v_y=\frac{dy}{d\theta}\times \frac{d\theta}{dt}$ $\implies v_y=0.3sin\theta(\frac{d\theta}{dt})~~~$[eq(1)] We plug in the known values to obtain: $2=0.3sin50(\frac{d\theta}{dt})$ This can be rearranged as: $\frac{d\theta}{dt}=\frac{2}{0.3sin50}$ $\implies \omega=8.7027rad/s$ After differentiating eq(1), we obtain: $a_y=0.3cos\theta(\frac{d\theta}{dt})^2+0.3sin\theta\frac{d^2\theta}{dt^2}$ We plug in the known values to obtain: $3=(0.3cos 50)\omega^2+(0.3sin50)\alpha$ As $\omega=8.7027rad/s$ $\implies \alpha=\frac{3-(0.3cos50\times (8.7027)^2)}{0.3sin50}$ This simplifies to: $\alpha=-50.4968rad/s^2$

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