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Conservation of angular momentum and energy

  1. Jun 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi. Just look at the picture, I will explain it:
    index.php?action=dlattach;topic=58603.0;attach=10538.gif
    It says that a bullet impacts the rod as you can see (in the center of mass forming a 30 degree angle with the horizontal and keeps inside the rod) with a speed of: 25m/s. The bullet has a mass 0.1Kg, the rod 0.9Kg and the length of the rod is 0.45m.

    I'm asked to find the maximum angular displacement that has the rod respect to the vertical. (the bullet keeps in the rod).


    2. Relevant equations



    3. The attempt at a solution

    What I did first was to realize that the angular momentum is conserved just when the bullet impacts the rod, after that what is conserved is the energy so I wrote:

    [tex] \displaystyle {{L}_{i}}=r\wedge mv[/tex]

    [tex] \displaystyle {{L}_{i}}=\frac{L}{2}mv\sin 120{}^\text{o}[/tex]

    And the final angular momentum (when the bullet gets inside the rod):
    [tex] \displaystyle {{L}_{f}}=I\omega [/tex]
    [tex] \displaystyle {{L}_{f}}=\left( \frac{M{{L}^{2}}}{3}+m\frac{{{L}^{2}}}{4} \right)\omega [/tex]

    So,

    [tex] \displaystyle \frac{L}{2}mv\sin 120{}^\text{o}=\left( \frac{M{{L}^{2}}}{3}+m\frac{{{L}^{2}}}{4} \right)\omega [/tex]

    And soving for w I get that the initial angular speed is 8.13 rad/s. Now I use conservation of energy:

    [tex] \displaystyle \frac{1}{2}I{{\omega }^{2}}=\left( m+M \right)gh[/tex]

    where h is the maximum height of the center of mass. Solving for h I get h=0,22m.
    But now, how can I get the angle? The correct answer is 79,5 degrees.

    Thanks!
     
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  3. Jun 28, 2012 #2

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    Everything looks correct to me so far. :smile:
    I ended up getting a different answer (I double checked too). You might want to try recalculating the numbers.
    I think that's the right idea. But if I'm right about your ω being a little off, this number is a little off too.
    Draw a diagram with the rod at an angle. Label h on the diagram. (Actually, it might be more intuitive to label 2h; that's what I did anyway.) Also label L with the rod in its original position, and in its final position. You should have a right triangle now. You can use that right triangle to solve for the angle (the maximum angular displacement) in terms of h and L.
     
  4. Jun 28, 2012 #3
    Oh! Yeah! Hohoho! 8 radians per second! Now I realize LOL!! Yeah, I probably got my calculator in radian mode instead of degrees (I keep changing it). Now I get 0.30 rad/s which is reasonable.

    Alright, yes, the diagram is the first thing I do always. I have that triangle but I don't understand why you say 2h when it's actually 1h. I think you are taking 2h from the floor, arent you? Anyway, I cannot realize how to find out the angular displacement just having h- Thank you!

    ** h is the height the center of mass displaced. Im not taking it from the floor
     
  5. Jun 28, 2012 #4
    This is what I have:

    attachment.php?attachmentid=48730&stc=1&d=1340933012.gif
     

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  6. Jun 28, 2012 #5

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    Wait, hold on. My answer for ω was definitely bigger than 0.3 rad/s (it was more than twice that, actually).

    Try that one more time. And yes, make sure your calculator is in degrees. :smile:
    Yes, that is what I did. If the center of the rod rises up by h, then the end of the rod rises up by 2h. But you can do it your way if you like, you will just need to draw a smaller triangle.
    Okay, we'll try it a different way. Label L/2 on the diagram in two places: One at the rod's original position, and again when the rod is at maximum angular displacement (L/2 is at an angle at the latter position). Now label h on the y-axis.

    You should see a right triangle now with hypotenuse of L/2. You should now be able to determine what the adjacent side of the triangle is in terms of L/2 and h.

    [tex] cos \ \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotensuse}} [/tex]
     
  7. Jun 28, 2012 #6

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    I see your diagram now.

    Rather than label h somewhere off in the middle, label h on the y-axis, with the "bottom" of h at L/2. Then connect a straight line between the top of h and the point at where the rod's L/2 is at its final position.
     
  8. Jun 28, 2012 #7
    Ahhhh now I see the right triangle! I thought it was not a right angle but now I see it.

    BTW now I get 7.4 rad/s LOL!!

    Thank you very much!
     
  9. Jun 28, 2012 #8
    Wait, the right triangle you say is the one which is connected by O (point of rotation) initial L/2 and final L/2? Now I see two right angles. The one I told you and the other which is formed by connecting O, L/2-h initial and L/2 final
     
  10. Jun 28, 2012 #9

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    There may be more than one similar triangle, if one works at it. :smile:

    But you've found at least one triangle that's good enough. In one triangle (where the angle in question is at the point O), the adjacent side is (L/2 - h) and the hypotenuse is L/2. With that, you should be able to solve for the maximum angular displacement (i.e. the angle up at O).

    And yes, the initial angular velocity of 7.4 rad/s sounds just right (to two significant figures, that is). :approve:
     
  11. Jun 28, 2012 #10
    I got it! Thank you very much!

    BTW 7.4 rad/s doesn't mean it completes a spin, does it? It cannot spin because of the gravity, that's why it reaches its maximum value..
     
  12. Jun 28, 2012 #11

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    It doesn't complete a whole spin. 7.4 rad/sec is just the initial rate of rotation, not the amount of rotation.

    The total amount of rotation is your final answer, 79.5o (which converts to around 1.39 rad). That's not even a quarter circle.

    You can tell that that is the case by looking at your diagram (and numbers), since h is less than L/2.

    (By the way, I used a lot of precision in my calculations, and I came up with an answer that is just a tiny, tiny bit different than 79.5o when rounding. I think the official answer might have a very tiny rounding/precision mistake. But it's really close though, so it's not a big deal)
     
  13. Jun 29, 2012 #12
    Yeah, me too, I got the exact value of h which is:

    [tex] \displaystyle h=\frac{3125}{17004}m[/tex]

    And the angle is: 79.4438678º . It would round to 79.4º but it's the same!

    Thank you!!
     
  14. Jun 29, 2012 #13

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    That's precisely the same answer I got. :smile:
     
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