Answer
$t=0.6125s$
Work Step by Step
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:
$t=0.6125s$
