Answer
$v_B=1.59~m/s$
Work Step by Step
The required speed can be determined as follows:
We apply the principle of impulse and momentum as follows
$0+\Sigma\int_{t_1}^{t_2}M_{\circ}dt=I_{\circ}\omega+m_Av_A r_A+m_Bv_Br_B$ [eq(1)]
As $I{\circ}=m_pk_{\circ}^2=15(0.11)^2=0.1815Kg\cdot m^2$
Similarly $v_A=r_A\omega=0.2\omega$
and $v_B=r_B\omega=0.075\omega$
We know that
$\int_{0}^{3} M_{\circ}dt=m_A gr_At-m_Bgr_Bt$
$\int_{0}^{3} M_{\circ}dt=40(9.81)(0.2)(3)-85(9.81)(0.075)(3)=47.8238N\cdot m\cdot s$
We plug in the known values in eq(1) to obtain:
$0+47.8238=0.1815\omega+40(0.2)(0.2\omega)+85(0.075)(0.075\omega)$
This simplifies to:
$\omega=21.1645rad/s$
Now $v_B=0.075(21.1645)=1.59~m/s$
