Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 19 - Planar Kinetics of a Rigid Body: Impulse and Momentum - Section 19.2 - Principle of Impulse and Momentum - Problems - Page 537: 20


$\omega=18.39 rad/s \circlearrowright$

Work Step by Step

We can determine the required angular velocity as follows: First, we apply the principle of impulse and momentum in normal direction $mv_{Gn1}+\Sigma\int_{t_1}^{t_2} F_n dt=mv_{Gn2}$ $\implies 0+Nt-mgcos30t=0$ $\implies 4N-100(9.81)cos 30(4)=0$ This simplifies to: $N=849.571N$ Now, we apply the impulse momentum principle in tangential direction $mv_{Gt1}+\Sigma\int_{t_1}^{t_2} F_t dt=mv_{Gt2}$ $\implies 0+Tt-mgsin30t+\mu_k Nt=mv_{G2}$ $\implies 4T-100(9.81)sin30(4)+0.1(849.571)(4)=100v_G$ This simplifies to: $T-25v_{G}=405.5429$ [eq(1)] Now $I_G\omega_1+\Sigma\int_{t_1}^{t_2}M_Gdt=I_G\omega_2$ [eq(2)] As $I_G=mk_G^2=100(0.25)^2=6.25Kg\cdot m^2$ and $\omega_1=0$ $\Sigma\int_{t_1=0}^{t_2=4}M_Gdt=\mu_kNtr_A-Ttr_G=0.1(849.57)\times 4\times 0.4-0.2(T)\times 4=135.9312-0.8TN\cdot s$ We plug in the known values in eq(2)to obtain: $0+135.9312-0.8T=6.25\omega$ [eq(3)] and we know that $v_G=0.2\omega$ [eq(4)] We plug in this value from eq(3) into eq(1) to obtain: $T-5\omega=405.5429$ [eq(5)] After solving eq(3) and eq(5), we obtain: $T=313.59N$ and $\omega=18.39 rad/s\circlearrowright$
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