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

$t=1.04s$
We can determine the required time as follows: First, we apply the principle of impulse and momentum to figure (1) $I_B\omega_{B_1}+\Sigma \int_{t_1}^{t_2}M_Bdt=I_B\omega_{B_2}$ [(eq1)] We know that $r_B=\frac{9in}{12in/ft}=0.75ft$ $\implies I_B=\frac{1}{2}m_Br_B^2=\frac{1}{2}(\frac{50}{32.2})(0.75)^2=0.4367 slug\cdot ft^2$ and $\omega_1=0 rad/s$ $\omega_{B_2}=60 rad/s$ Similarly $\Sigma\int_{t_1}^{t_2}M_Bdt=0.75\int_{0}^{t}T_1dt-0.75\int_{0}^{t}T_2dt$ We plug in the known values in eq(1) to obtain: $0+0.75\int_{0}^{t} (T_1-T_2)dt=0.4367\times 60$ This simplifies to: $\int_{0}^{t} (T_1-T_2)dt=34.9379$ [(eq2)] Now, we apply the principle of impulse and momentum to figure(1) $I_A\omega_1+\Sigma\int_{t_1}^{t_2}M_Adt=I_A\omega_{A_2}$ [(eq3)] As $r_A=\frac{6in}{12in/ft}=0.5ft$ and $I_A=m_Ak_A^2=(\frac{30}{32.2})(\frac{4in}{12in/ft})^2=0.1035slug\cdot ft^2$ Similarly $\omega_1=0$ and $\omega_2=\frac{r_B}{r_A}\omega_B=(\frac{0.75}{0.5})(60)=90 rad/s$ $\Sigma\int_{t_1}^{t_2}M_Adt=0.5\int_{0}^{t}T_2dt-0.5\int_{0}^{t}T_1dt+\int_{0}^{t} 50tdt=25t^2+0.5\int_{0}^{t}(T_2-T_1)dt$ We plug in the known values in eq(3) to obtain: $0+25t^2+0.5\int_{0}^{t} (T_2-T-1)dt=0.1035\times 90$ [eq(4)] We plug in the known values from eq(2) into eq(4) to obtain: $25t^2+0.5(-34.9379)=9.315$ This simplifies to: $t=1.04s$