Answer
$L=3.92~slug\cdot ft/s$
Work Step by Step
We can determine the required linear momentum as follows:
$L=mv_G=\frac{W}{g}v_G=\frac{10}{32.2}v_G=0.3106v_G~~~$[eq(1)]
We know that
$T=\frac{1}{2}mv_G^2+\frac{1}{2}I_G\omega^2~~~$[eq(2)]
As $I_G=\frac{W}{g}k_G^2=\frac{10}{32.2}(0.6)^2=0.1118slug\cdot ft$
and $v_G=1.2\omega$
$\implies \omega=\frac{v_G}{1.2}$
We plug in the known values in eq(2) to obtain:
$31=\frac{1}{2}(\frac{10}{32.2})v_G^2+\frac{1}{2}(0.1118)(\frac{v_G}{1.2})^2$
This simplifies to:
$v_G=12.6377ft/s$
Now, We plug in the known values in eq(1) to obtain:
$L=0.3106(12.6377)=3.92~slug\cdot ft/s$
