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

$L=3.92~slug\cdot ft/s$
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: $L=0.3106(12.6377)=3.92~slug\cdot ft/s$
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