Answer:
4.203 m/s
Explanation:
First, we solve for the translational kinetic energy ( [tex]K_{\text {trans }}[/tex] ) in the definition of the total kinetic energy [tex]\left(K_{\text {tot }}\right)[/tex]. We also have an equation for the definition of [tex]K_{\text {trans }}[/tex] (third line). From these two equations we can solve for the speed of the center of mass [tex]\left(v_{\mathrm{CM}}\right)[/tex]
[tex]\begin{aligned}
K_{\text {tot }} &=K_{\text {trans }}+K_{\text {rot }} \\
\Longrightarrow K_{\text {trans }} &=K_{\text {tot }}-K_{\text {rot }} \\
K_{\text {trans }} &=\frac{1}{2} m_{\text {tot }} v_{\mathrm{CM}}^{2} \\
\Rightarrow \frac{1}{2} m_{\text {tot }} v_{\mathrm{CM}}^{2} &=K_{\text {tot }}-K_{\text {rot }} \\
\Rightarrow v_{\mathrm{CM}} &=\sqrt{\frac{2\left(K_{\text {tot }}-K_{\text {rot }}\right)}{m_{\text {tot }}}} \\
&=\sqrt{\frac{2(398 \mathrm{~J}-80 \mathrm{~J})}{36 \mathrm{~kg}}}=4.203 \frac{\mathrm{m}}{\mathrm{s}}
\end{aligned}[/tex]
