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--- course_planning:todor [2022/11/27 23:28] – created pwirving
+++ course_planning:todor [2022/12/06 21:20] (current) – pwirving
@@ Line 1: / Line 1: @@
-====== Project 12 Problem: Part A: You spin me right round ======
+====== Project 14 Solution: Part A: Showdown at boar tiger corral ======
-For some reason, the area in which you live in keeps on going through periods of blackouts where you have no power. You decide that you need to build an alternative generator using a merry-go-round. The merry-go-round ride has one bar moving from the centre of the ride to a point at the edge of the ride. In building your generator you have to dig up the perimeter of merrygoround in order to accomodate all the piping needed to develop it and this has resulted in you being unable to run alongside the merrygoround pushing it. Which means you are going to have to create a collision with the bar somehow in order to make the merry-go-round rotate to create energy. You need to figure out what velocity the mass you have chosen to collide with the bar needs to be in order to achieve your goal generating 6500 Joules in order to create enough electricity to keep your phone charged. You also have available to you seven 5kg weights with some straps. The uniform density disk that comprises the ride is 300kg and has a radius of 2.5m. Your team which is working on the roundabout consists of 4 members.
+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 $$v_{\rm cm}=\frac{mv}{m+M}.$$
-{{course_planning:screen_shot_2022-11-21_at_12.18.02_pm.png}}
+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  $v_{\rm end}=\frac{mL^{2}v}{4I+mL^{2}}$   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:  $\vec{v}_{\rm tot}=\vec{v}_{\rm cm}+\vec{v}_{\rm end}$
-====== Project 12 Solution: Part A: You spin me right round ======
+<WRAP tip>
+==Tutor Questions:==
-The idea here is to have the group of people run and land on the platform.
+  * **Question:**  What system have you chosen?
+  * **Expected Answer:**  The cannon ball and the block.
-For review, angular momentum outlines and formulas can be found in the [[183_notes:ang_momentum| Angular Momentum Notes]]
+  * **Question:**  Why does conservation of linear momentum work here?
+  * **Expected Answer:**  Since the cannon ball and the block are in the system, we have no external forces to cause a change in momentum.
-The initial momentum of the group of people having total mass  $\tilde{m}$  running with a velocity  $v$  is given by  $p=\tilde{m}v$ , where  $\tilde{m}=q\mathcal{M}+sm$  is the total mass of the group ( $q$  is the number of people in the group and  $s$  is the number of bags that are being carried along).  Accordingly, the initial angular momentum as measured from the axis of the platform is given by  $L_{i}=r\tilde{m}v,$  where  $r$  is the distance from the axle of the platform.
+  * **Question:**  Where is the axis of rotation of the cannon ball and block system?
+  * **Expected Answer:**  We assume that the center of mass of the cannon ball and block system is not significantly different from that of the block alone.  Thus, the cannon ball and block system rotates about the geometric center of the block.
-After landing on the platform (we assume they all land at the maximum possible radius  $r$ ), we know that it will rotate with angular velocity  $\omega$ .  Since angular momentum is conserved, we must have \begin{eqnarray*}L_{i}&=&L_{f}\\
+  * **Tutor Question:**  Why does conservation of angular momentum work here?
-r\tilde{m}v&=&I\omega.\end{eqnarray*}
+  * **Expected Answer:**  Again, considering the cannon ball and block in the system, we have no external torque to cause a change in angular momentum.
+  * **Tutor Question:**  Does energy conservation work here?
+  * **Expected Answer:**  No, we can't use energy conservation because of the inelastic collision of the cannon ball and block.
+  * **Tutor Question:**  What motion will the end of a block trace out (draw it on the white board) as it moves towards the boar tigers?
+  * **Expected Answer:**  ...
+  * **Tutor Question:**  How does the speed of the center of mass and the speed of the end of the block depend on where you shoot the cannon ball at the block?  What would happen if you shoot the cannon ball at the center of the block?
+  * **Expected Answer:**  The further out you shoot the cannon ball, the faster the rotation.  If you shoot the cannon ball at the center of the block, there will be no torque and so no rotation.  The center of mass speed is independent of the location of the collision.
+  * **Tutor Question:**  How does the speed of each point on the block due to rotation compare to its neighbor?
+  * **Expected Answer:**  Each point on the block is moving faster (going through a larger distance in the same amount of time) as you move away from the axis of rotation.
+  * **Tutor Question:**  What is the maximum velocity with which the boar tigers can be struck?
+  * **Expected Answer:**  This occurs when the length axis is perpendicular to the velocity of the center of mass, where we have the simple sum (at least on one side of the block) $v_{\rm max}=v_{\rm cm}+v_{\rm end}$.
+{{course_planning:qr_code_for_post_questions_boar_tiger_coral.png}}
+QR code for conceptual questions.
+</WRAP>
-Now, to get rid of  $\omega$ , we must recognize the relationship between energy (formula for rotation kinetic energy can be review in [[183_notes:rot_ke|Rotational Kinetic Energy notes]]) and angular velocity:  $\frac{1}{2}I\omega^{2}=E_{\rm min}\qquad\longrightarrow\qquad \omega=\sqrt{\frac{2E_{\rm min}}{I}}.$
-Combining everything together, we have  $\begin{eqnarray*}r\tilde{m}v&=&I\omega\\r\tilde{m}v&=&I\sqrt{2E_{\rm min}/I}\\r\tilde{m}v&=&\sqrt{2E_{\rm min}I}\end{eqnarray*}$   Finally, this can be solved for either the mass  $\tilde{m}$  that must be used, or the velocity  $v$  that must be acquired by the group.