Answer
$t=1.04s$
Work Step by Step
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:
$t=1.04s$
