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Help with design project

  1. Mar 29, 2007 #1
    1. The problem statement, all variables and given/known data
    I need to design, and build a scale model of something that can throw a 10 lbf pumpkin 300 feet.

    No variables right now are for certain other than that I know I need a 98.3 ft/s launch velocity at 45 degrees. Right now I've pretty much just been playing with symbols trying to get an equation that gives values that seem right

    For the design my group has chosen http://members.lycos.nl/onager/OnagerPic.jpg

    2. Relevant equations
    [tex]W_{sp} = \frac{1}{2}k(x_i^2-x_f^2)[/tex]
    [tex]W_{net}=\Delta KE[/tex]

    3. The attempt at a solution

    The kinematic analysis isn't hard; I won't list the steps, but I know that for a distance of 300ft and 45 degree launch angle that it will require an initial velocity of 98.3 ft/s. Different parts of the arm will be moving at different speeds, but in the end the entire arm/projectile should be moving at the same angular velocity W = (98.3 ft/s)/(6 ft/rad) W = 16.4 rad/s (for an arm length of 5 ft and end part of 1 ft).

    A kinetic analysis is what has got me stuck. The machine gets its energy from the ropes that the arm run through. The ropes act like a spring but I can't think of how to measure an experimental K value. In a normal spring you can just stretch it out, measure how much force, the distance from its neutral position and do K = F/x. In this design the spring isn't pulled it is rotated. I have never dealt with anything like this before.

    What I was thinking was measure how much force it takes to rotate the arm 360 degrees; However that force is dependent on how far away from the center you hold the arm, but the amount of torque would be constant. So if you spun the arm around 360 degrees and measured that it took 300 lbf to hold the arm at the very end (say the arm is 5 feet) then the K value may be considered as

    K = (5ft)*(300lbf)/(360 degrees) or
    K = 4.16 ft*lbf/degree

    Then I thought adjusting the formula from
    Wsp = .5*k*x^2 to
    Wsp = .5*k*(theta)^2

    I know right away that this assumption is wrong because of how the units work out (also such a large number).
    Wsp = .5*(4.16ft*lbf/degree)*(360degrees)^2.
    Wsp = 269,568 ft*lbf*degree

    I know that this type of rope setup was common in ancient balista/catapult design. I've found many sources on how to build them, but nothing on how to theoretically calculate the work that they output. Any advice/sources on how to calculate this would be helpful.

    Aside from needing to know how to calculate the work that the spring/rope does, I need to calculate the total amount of work that is required. I think I have this right. I know that Wnet = change in kinetic energy. I know the change in kinetic energy of the projectile, but I'm skeptical on whether I know how to calculate the change in kinetic energy of the arm. This is what I have done

    Total change in kinetic energy = .5*m*v^2 + .5*M*V^2 where m/v is in regards to the arm and M/V is in regards to the projectile.

    Ke = .5*(w/g)*[(angular velocity)*(.5*length of arm)]^2 + .5*(W/g)*[(angular velocity)*(total length)]^2

    I know this is probably confusing to read but basically it is just Ke = 1/2 mv^2 + 1/2 mv^2. Where the first term is mass of the arm * velocity at the centroid, and the second term is mass of projectile * velocity at the end. That equation should give me the total work that needs to be done by the spring (excluding work done by gravity/friction for now). Is calculating the total change in kinetic energy this way correct?
  2. jcsd
  3. Mar 29, 2007 #2
    I think you might be making the spring constant too hard. While rotating it is not exactly the same as simple linear stretch, for your purposes, it is entirely possible that F=-kN where n is the number of loops. But that will take building it and testing. Can you just buy a short length of rope and measure vert. stretch with people of various wts hanging from it? If the rope is too stout, maybe you could find one of the same material and use a smaller gauge. Just blue sky ideas here.
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