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

$\omega_2=36.3 rad/s$
We can determine the required angular velocity as follows: First, we apply the principle of impulse and momentum in the x-direction $mv_{x_1}+\Sigma\int_{t_1}^{t_2}F_xdt=mv_{x_2}$ We plug in the known values to obtain: $0+Nt-(\frac{5}{13})F_{OB}t=0$ $6N=6(\frac{5}{13})F_{OB}$ This simplifies to: $N=\frac{5}{13}F_{OB}~~~$[eq(1)] Now, we apply the impulse and momentum principle in the y-direction $mv_{y_1}+\Sigma\int_{t_1}^{t_2}F_y dt=mv_{y_2}$ $\implies 0+\mu_k Nt+(\frac{12}{13})F_{OB}t+Pt-mgt=0$ We plug in the known values to obtain: $0.2N+6(\frac{12}{13})F_{OB}+40-40\times 9.81=0$ This simplifies to: $F_{OB}=352.4N$ We plug in this value in eq(1) to obtain: $N=134.5385N$ We know that $I_{\circ}\omega_1+\Sigma\int_{t_1}^{t_2}M_{\circ}dt=I_{\circ}\omega_2~~~$[eq(2)] As $I_{\circ}=mk_{\circ}=40(0.08)^2=0.256Kg\cdot m^2$ Now, from eq(2) $0+Ptr-\mu_kNrt=0.256\omega_2$ We plug in the known values to obtain: $40(0.12)(6)-0.2(135.54)(0.12)(6)=0.256\omega_2$ This simplifies to: $\omega_2=36.3 rad/s$