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

$t=0.6125s$
First we apply the principle of angular impulse and momentum to figure (1) $I_{\circ}\omega_1+\Sigma \int_{t_1}^{t_2} M_{\circ} dt=I_{\circ} omega_2~~~$[eq(1)] We know that $I_{\circ}=m_A k_{\circ}^2=(30)(0.125)^2=0.4688Kg\cdot m^2$ and $\omega_1=0$ $\omega_2=20 rad/s$ Similarly, $\int_{t_1}^{t_2} M_{\circ}= Frt=(0.15)Ft$ We plug in the known values in eq(1) to obtain: $0+0.15Ft=0.4688\times 20$ $Ft=62.5067N\cdot s$ Now, we apply the principle of angular impulse and momentum to figure (2) $m_Av_{B_1}+\Sigma \int_{t_1}^{t_2} M_{\circ} dt=m_Bv_{B_2}~~~$[eq(2)] Now, $v_{B_1}=r_B\omega_1=0$ and $v_{B_2}=r_B\omega_2=(0.15)(20)=3m/s$ Similarly, $\int_{t_1}^{t_2} M_{\circ} Pt-Ft=200t-62.5067$ We plug in the known values in eq(2) to obtain: $0+200t-62.5067=20(3)$ This simplifies to: $t=0.6125s$