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See Saw Catapult Physics

  1. Apr 21, 2005 #1
    I posted this on the k-12 forum but i haven't been getting any responses.

    At first glance the physics of a seesaw seem quite simple, however i've been stumped on a problem. For part of my Physics project, I am incorporating a seesaw catapult where one weight will be dropped on one side launching the hacky sack on the other side. Simple right? Of course.... but the hacky sack needs to land .76 m to the left, on a stool .76 m high.

    I figured my launch angle to be approx.... 80 degrees ( I can always change this) which would yeild me a 5.45 m/s vertical velocity and a .8 (approx) m/s horizontal velocity. The Tanget velocity would therefore need to be 5.55 m/s (approx). Now how can I work backwards to solve for how heavy and at what height the object would have to be dropped from in order to launch it at that velocity? Lets say the arm with the hacky sack is 4 times larger then the other arm, so we have 4 and 1. I'm guessing I'll have to use Torque, and force of course.

    I gave it some thought tho, and in order for the hacky to even move the mass falling must have enough force to bring the seesaw to equilibrium. So once I calculate that force required, then anything more than that will begin to bring the acceleration and velocity of the hacky up from zero. Also I must find a way to factor in the weight of the arms, it will make a difference. Any help guys?! Sorry if it was confusing
     
  2. jcsd
  3. Apr 21, 2005 #2
    Forgive me if I missunderstand but ( to the left ?? ) left of what , surely a seesaw
    can only launch in one direction (along it's length ) .
    But aside from that some things appear missing --- the height of the seesaw matters
    because that resricts it's movement also the actual launch point is dependant on whether the article thrown is held in place ( other wise it may pre launch ) even going backwards . The precise launch point has to be controlled and determined other wise you have no clue where the projectile will go .
    The mass of the seesaw is extremely iportant as it has to avoid being broken in two by the dropping mass . ( you did not say what the mass was or from how high it could be dropped --- perhaps only lifted and placed on one end ) .
    This is a fascinating subject Milch there are many sites on the web devoted to Catapults and they deal with quite complicated physics .
    Best of luck with your project Ray.
     
  4. Apr 21, 2005 #3
    I've really done some extensive research on catapults and seesaw's and I have had no luck finding what I need. Catapults are a little different because they only have 1 arm, which makes it so much easier. If you check on the k-12 forum, I've posted some more information on my thread titled "Seesaw Physics" I don'thave the mass or the height or anything because i haven't built it yet. I'm trying to figure out how I should build it. HOW do I find the velocity of the hacky sack if launched perpendicular to its arm knowing the mass ofthe object falling, the height of its fall, the lengths of the arms, the mass of the hacky, the mass of each arm, and to make things a little easier, lets assume the arms are level, IE, there is a mass supporting the arm up with the hack sack. I'm looking for an equation for the velocity of the hacky.... so i can work backwards and figure out what I would want.
     
  5. Apr 21, 2005 #4
    Tricky stuff

    The type of equations needed are energy (potential and Kinetic )
    But there are serious problems --- the dropping mass does not transfer all it's energy if for instance it hits the ground , I did a catapult design and simulation
    based on the following
    a) I assumed the start angle was about - 45 degrees ( i.e backward sloping )
    b) arms say 4 : 1 ratio at some real physical length this tells you how high the fulcrum has to be (the one end on the ground ) and also how high the mass end is .
    c) I then attached the driving mass via a rope to the catapult end running over a pulley so that the mass almost drops vertically and comes to a stop
    when arms come into line with the pulley ( say +45 degrees )
    This means all energy of the drop is used.
    d) I had to assume some real arms with some mass (they could be tapered )
    but the trick is to convert this mass into an equivalent at the projectile location (i.e. what mass at the catapult end would be the same as the real arms rotating around about the fulcrum ) .
    we now have M.d = 1/2 . ( m1 + m2 ) . V^2
    where M is the dropped mass and d the drop distance
    m1 is the projectile and m2 the arm equivalent
    v is the projectile speed at release in the direction of where the arms stop.
    In the most ideal case if m2 were zero this scheme would be 100 % efficient.
    In other words it is the arm design which is critical to the whole thing .
    It should as light but strong as possible .
    just for fun let m2 = 0 , let M = 100 Kgms and d = 1 meter and let m1 = 0.1 kgm
    V then is ~ 44 m/sec --- assume release is at 45 degrees so that the horizontal and vertical speeds are ~ 31 m/sec .
    the rise time is t = v/g ( g being gravity at ~ 9 m/s/s ) or ~ 3.5 secs
    so total flight time will be say twice this (roughly ) or 7 secs
    so the throw distance would be ~ 217 meters . :rofl:
    Ray. As said have fun.
     
  6. Apr 21, 2005 #5
    Thanks that really helps! I'll look into it some more!
     
  7. Apr 22, 2005 #6
    Wait btw, there is no where to factor in the angle in that equation for velocity, does teh angle make a difference in the tanget velocity?

    Also how would I convert the mass of the arm into a mass at the end of the arm?
     
    Last edited: Apr 22, 2005
  8. Apr 22, 2005 #7
    The equation only relates potential with Kinetic energy it gives the velocity at the point at which accelleration stops and the projectile leaves . That velo is in the direction that the projectle has at that time.
    In My illustration this occured at +45 dgrees because the rope ceases to turn the arm when the arm points to the pulley. In theory this is a variable lever as the arm nears +45 dgrees it is increasingly difficult to turn it via the rope and the Mass slows down to zero speed transfering all it's energy.
    Remember this is only a simple example which I used for the purposes of a simulation ( I did not try to build one ) and this was a couple of years back.
    You can stop the arm to determine angle -- but if you do energy will be lossed to heat and the above simple equation will not work.
    To find the range you must get the vert and horiz speeds from V and the angle. vert the projectile will rise but decellerate with gravity to zero and then fall . I assume the Vhoriz remains fixed in flight.
    For the beam mass it is preffered if the beam is balanced on the fulcrum
    Which avoids questions of potential energy for a beam ratio of 4 : 1
    the effective mass at the long end is about M/3.75 where M is the beam mass. But this requires a tapered beam to get the balance at the 4:1 point
    or at least close to it.
    I think people end up with fancy materials for the beam trying to get strength and lightness.
    Ray.
     

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    Last edited: Apr 22, 2005
  9. Apr 28, 2005 #8
    Can you explain how you got this equation, i see the kinetic energy part of it.

    M.d = 1/2 . ( m1 + m2 ) . V^2
     
  10. Apr 28, 2005 #9
    I just want to make sure it will work. So i need to know how you derived it.
     
  11. Feb 20, 2009 #10
    this might be an old comment.....but really when it comes to how do i build a catapult to launch a hey stack....really man come on...you don't need a caculated formula to figure out how much of the hey stack is really needed to be throne vs the length of the arm to launch the damn thing....because you can never find the right weight of a hey stack you need to if you build the damn thing right anyways....so i suggest you build something thing that can launch a "so called hey stack" over the distance you need and scale back from there...and if there are any more questions about how you did it from there....then just take all the weights and thrust from there and make them happy...cuz really you can't find a hey stack to find your needs to fit you equasions.

    so happy shootins and hope you hit your goal
     
  12. Feb 20, 2009 #11

    rcgldr

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    Look up teeterboard. The launch angle is mostly due to the positioning of center of mass at some height above and distance away from the end of the teeterboard. One limitation of an inanimate object is that it would not be easy to create a dynamic situation, such as a human leaning (falling) back a bit while being launched to control the angular and linear launch from the teeterboard. A real teeterboard is also flexible, much like a diving board, and is alternately called a springboard for that reason.

    For the inanimate object, a solution would be to restrict the travel of the teeterboard, which is normally 45 degrees. You might consider changing the angle and limiting it so that the teterboard stops short of horizontal to allow for a sideways component of linear velocity at launch.
     
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