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

$v_B=24.0876m/s$
We can determine the required speed as follows: We know that $I_O \omega_1+ \Sigma \int_{t_1}^{t_2} M_O dt=-m_Av_B(0.2)-I_O \omega_2~~~$[eq(1)] As $I_O=mk_O^2=(15)(0.11)^2=0.1815Kg\cdot m^2$ and $\omega_1=0$ $Sigma \int_{t_1}^{t_2}M_{O}dt=m_Agtr_{O/A}-Ftr_{O/B}$ $Sigma \int_{t_1}^{t_2}M_{O}dt=40(9.81)(3)(0.2)-2000(3)(0.075)=-214.56N\cdot m\cdot s$ Similarly $v_B=r_{O/B}\omega_2=0.2\omega_2$ $\implies \omega_2=5v_B$ We plug in the known values in eq(1) to obtain: $0-214.56=-40\times 0.2v_B-0.1815(5v_B)$ This simplifies to: $v_B=24.0876m/s$