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

Published by Pearson

# Chapter 19 - Planar Kinetics of a Rigid Body: Impulse and Momentum - Section 19.2 - Principle of Impulse and Momentum - Problems - Page 534: 6

#### Answer

$\omega_2=0.01778rad/s$

#### Work Step by Step

We can determine the required angular velocity as follows: We apply the principle of impulse and momentum $I_G\omega_1+\Sigma\int_{t_1}^{t_2}M_G dt=I_G\omega_2~~$[eq(1)] As $I_G=mk_G^2$ $I_G=200(10^3)(15)^2=45\times 10^6Kg\cdot m^2$ and $\omega_1=0$ Similarly, $\Sigma\int_{t_1}^{t_2}M_G=T_A tr_{G/A}-T_B tr_{G/B}$ $\Sigma\int_{t_1}^{t_2}M_G=40\times 10^3\times5\times8-20\times 10^3 \times 5\times 8=8\times 10^5N\cdot m\cdot s$ We plug in the known values in eq(1) to obtain: $0+8\times 10^5=45\times 10^6\omega_2$ This simplifies to: $\omega_2=0.01778rad/s$

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