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--- course_planning:todor [2022/12/06 21:13] – pwirving
+++ course_planning:todor [2022/12/06 21:20] (current) – pwirving
@@ Line 1: / Line 1: @@
 ====== Project 14 Solution: Part A: Showdown at boar tiger corral ======
-In order to determine the speed of the center of mass, we have to use momentum conservation.  Accordingly, we must have  $\begin{eqnarray*}p_{i}&=&p_{f}\\mv&=&(M)v_{\rm cm},\end{eqnarray*}$  where  $m$  is the mass of the cannon ball,  $v$  is the velocity of the cannon ball,  $M$  is the mass of the block.  Thus, we can solve for the center of mass speed $$v_{\rm cm}=\frac{mv}{m+M)]}.$$
+In order to determine the speed of the center of mass, we have to use momentum conservation.  Accordingly, we must have \begin{eqnarray*}p_{i}&=&p_{f}\\mv&=&(m+M)v_{\rm cm},\end{eqnarray*} where  $m$  is the mass of the cannon ball,  $v$  is the velocity of the cannon ball,  $M$  is the mass of the block.  Thus, we can solve for the center of mass speed  $v_{\rm cm}=\frac{mv}{m+M}.$
-Now, in order to determine the speed of the end of the block with respect to the center of mass motion as it spins, we have to use angular momentum conservation.  If we assume that the cannon ball strikes the rectangular block at its end, we must have  $\begin{eqnarray*}L_{i}&=&L_{f}\\mvr&=&\tilde{I}\omega\\mv(L/2)&=&\big[m(L/2)^{2}+I\big]\frac{v_{\rm end}}{L/2},\end{eqnarray*}$  where $$I=\frac{1}{12}\rho\big[V_{\rm out}\big(L^{2}+W^{2}\big)-V_{\rm in}\big((L-t)^{2}+(W-t)^{2}\big)\big]\approx 37454.3\,{\rm kg\cdot m^{2}}$$ is the [[http://www.health.uottawa.ca/biomech/courses/apa4311/solids.pdf|moment of inertia]] of a block about its geometric center along the height axis with length $L$ and width $W$ (they should Google this), and we have used the fact that $v_{\rm end}=\omega(L/2)$.
+Now, in order to determine the speed of the end of the block with respect to the center of mass motion as it spins, we have to use angular momentum conservation.  If we assume that the cannon ball strikes the rectangular block at its end, we must have  $\begin{eqnarray*}L_{i}&=&L_{f}\\mvr&=&\tilde{I}\omega\\mv(L/2)&=&\big[m(L/2)^{2}+I\big]\frac{v_{\rm end}}{L/2},\end{eqnarray*}$  where $$I=\frac{1}{12}*M\big(L^{2}+W^{2})$$ is the [[http://www.health.uottawa.ca/biomech/courses/apa4311/solids.pdf|moment of inertia]] of a block about its geometric center along the height axis with length $L$ and width $W$ (they should Google this), and we have used the fact that $v_{\rm end}=\omega(L/2)$.
-Thus, we can solve for the speed of the end of the block with respect to the center of mass $$v_{\rm end}=\frac{mL^{2}v}{4I+mL^{2}}\approx 51.67\,{\rm m/s}. $Now, assuming that the contraption will strike the boar tigers at its corner, they will see a different velocity depending on their location with respect to the center of mass:$ \vec{v}_{\rm tot}=\vec{v}_{\rm cm}+\vec{v}_{\rm end}$$
+Thus, we can solve for the speed of the end of the block with respect to the center of mass  $v_{\rm end}=\frac{mL^{2}v}{4I+mL^{2}}$   Now, assuming that the contraption will strike the boar tigers at its corner, they will see a different velocity depending on their location with respect to the center of mass:  $\vec{v}_{\rm tot}=\vec{v}_{\rm cm}+\vec{v}_{\rm end}$
 <WRAP tip>