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 539: 27



Work Step by Step

The required speed can be determined as follows: We apply the principle of impulse and momentum as follows $0+\Sigma\int_{t_1}^{t_2}M_{\circ}dt=I_{\circ}\omega+m_Av_A r_A+m_Bv_Br_B$ [eq(1)] As $I{\circ}=m_pk_{\circ}^2=15(0.11)^2=0.1815Kg\cdot m^2$ Similarly $v_A=r_A\omega=0.2\omega$ and $v_B=r_B\omega=0.075\omega$ We know that $\int_{0}^{3} M_{\circ}dt=m_A gr_At-m_Bgr_Bt$ $\int_{0}^{3} M_{\circ}dt=40(9.81)(0.2)(3)-85(9.81)(0.075)(3)=47.8238N\cdot m\cdot s$ We plug in the known values in eq(1) to obtain: $0+47.8238=0.1815\omega+40(0.2)(0.2\omega)+85(0.075)(0.075\omega)$ This simplifies to: $\omega=21.1645rad/s$ Now $v_B=0.075(21.1645)=1.59~m/s$
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