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Lab issue - Please check my projectile launcher calculations

  1. Nov 6, 2013 #1
    Lab issue -- Please check my projectile launcher calculations

    I am making a lab for a class I teach and my results aren't working out and I need another pair of eyes to see if my method is wrong (I apologize if this should go in homework, but it isn't a homework question so I thought it should go here).

    I have a projectile launcher. I used a force probe to find it took 7.5 N to push the launcher plug 3.5 cm or 0.035 m which gives me a spring constant of k=7.5/0.035=214. Using W=1/2kx^2 I found the KE of the projectile leaving the launcher to be KE=W=1/2(214)(0.035)^2=0.131 j. Setting KE=U I found the maximum height from 0.131=mgh where the projectile is 17 g or 0.017 kg so h=0.131/((0.017)(9.8))=0.79. When I fired the projectile my height was off by almost a factor of 2 so I think I am missing a factor of 2 somewhere. Any help would be appreciated.
     
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  3. Nov 6, 2013 #2

    russ_watters

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    Is the spring fully extended in the launcher? You said the launcher has a "plug" on it - did you consider its mass? I don't see anything wrong in your calc.
     
  4. Nov 6, 2013 #3

    Simon Bridge

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    Since you have assumed that all the energy in the spring goes to height (gpe), and the energy in the spring is the work done to compress it, then Fd=mgh right? (where h is measured from the position of the projectile on the compressed spring ... did you apply a constant force all the way?)

    h=(7.5x0.035)/(9.8x0.017)=1.5756m ... almost twice your calculation.
    i.e. the result for a constant force applied the entire movement.

    ... Did your launcher fire 2x higher than calculated, or half as high?

    Check that the spring obeys hooks law?
     
    Last edited: Nov 6, 2013
  5. Nov 7, 2013 #4
    I had not thought of the mass of the plug. My actual height was 0.43 so it was a little more than half of my predicted height. The plug would have to have a mass of 0.173 kg to make that much difference (U(stored in spring)-U(gravity at max height from launcher)=U(gravity change in launcher), 1/2(214)(0.035)^2- (0.017)(9.8)(0.43)=M(9.8)(0.035). The entire launcher only weighs 0.192 kg so I don't think the magnetic plug is anywhere near that. The force can't be constant because the spring force increases as you push the projectile in. I calculated k for x=0.068 and got 208 so it appears to follow hookes law.

    I just had a thought. The launcher is magnetized to keep the ball from falling out. Could that be what is slowing the ball down? Thanks for the responses!
     
  6. Nov 7, 2013 #5

    Simon Bridge

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    OK - so,
    • for a spring that obeys hooks law (did you check)
    • firing the projectile directly upwards
    • neglecting -
      • air resistance
      • mass of the plug
      • rotation of the projectile (was it a ball?)
      • friction in the launcher
      • pull from the magnet
      • recoil of the launcher

    ##\frac{1}{2}Fd = mgh \Rightarrow h=79\text{cm}##​

    You measured: ##h=43\text{cm}##

    Your rig sounds like a commercial one I've used before - it has an aluminium tube mounted on a protractor+plumb-line thing that you can bolt to a bench. The magnet is very weak so has almost no effect compared to other factors. iirc recoil of the launcher and air resistance were the biggest effects and we saw some effect from the rotation of the ball we used as a projectile.

    The launcher had several settings that could be compared.
    You can also compare time-of-flight for firing horizontally with just dropping the projectile from the same height.
    (We used a pressure plate and a photogate.)
    Also compare predicted range with actual range for horizontal firing - iirc: air resistance was not significant for the vertical movement but was for the horizontal ... higher speeds. It was also significant for firing vertically upwards - but I don't remember how significant it was.
     
    Last edited: Nov 7, 2013
  7. Nov 11, 2013 #6
    Thanks for all of the help. I think the launcher just doesn't covert the potential spring energy into kinetic energy very efficiently. Thanks again, especially Simon.
     
  8. Nov 11, 2013 #7

    Simon Bridge

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    How did you determine that?
    Did you account for the assumption you made in the test runs?
    Could you test it by attempting to make the launcher more efficient.

    Oh well - so long as you are happy with the results.
     
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