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Seesaw Physics

  1. Apr 20, 2005 #1
    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
    No one can help? My calculus teacher didn't really know either, he knew it was a ratio of radii but that's obvious. My physics teacher hasn't helped me since it's part of my final project, I've done some extensive research online into catapults and such, but hav been unable to find what I need. So I turned here hoping for a response. Any help would be appreciated!
     
  4. Apr 21, 2005 #3
    Should I move this to a more advanced board? I asked my physics teacher and he said I need to talk to my calc teacher, whom also wasn't sure, so I have one more teacher who might know...
     
  5. Apr 21, 2005 #4
    You need to be more specific,

    "but the hacky sack needs to land .76 m to the left, on a stool .76 m high."

    Left of where?

    "I figured my launch angle to be approx.... 80 degrees"
    relative to what? 80 degrees towards the other end of the seesaw, from the ground?

    What exactly are you trying to find? The force downwards? The lengths of the arms? The weight of the projectile?
     
  6. Apr 21, 2005 #5
    "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"

    Isnt necessarily true, the projectile can fly before the seesaw reaches equilibrium (i assume by equilibrium you mean parallel to the ground) depending on the friction between the two objects.
     
  7. Apr 21, 2005 #6
    Sorry maybe I should have rephrased that, the force on the left needs to over power the force(s) on the right before it can move.

    I'll make a picture real quick hold on.
     
    Last edited: Apr 21, 2005
  8. Apr 21, 2005 #7
    What initial position is the seesaw in? What are the relative masses of each arm? Uniformity?
     
  9. Apr 21, 2005 #8
    Last edited: Apr 21, 2005
  10. Apr 21, 2005 #9
  11. Apr 21, 2005 #10
    V = I_0 Theta(dot).

    The dot signifies the time derivative of theta, which is omega (angularvelocity).

    If your going to actually do this demonstration, itwill be much harder to do the calculations as youwillhave to take several more factors into account, such as the inertia of the beams, air resistance perhaps, and more specifically the places where the object falls and where the hackysack is.

    Why dont you go ahead and spell out what you've got, we'll work from there.
     
  12. Apr 21, 2005 #11
    Okay the weight will be landing on one extreme of the beam and the hacky will be on the other extreme of the beam. The Length of the beam on the side with the hacky will be maybe 4 times larger than the side with the weight falling. I'm going to ignore air resistance as it wont have enough of an impact to make a huge difference. If you can tell me what calculations I need to do in order to solve for the Angular Velocity then I should be okay. Lets not work with numbers right now, and instead work with variables thatI can plug in later.

    My biggest problem is that I don't understand how the object falling relates to what goes on once it impacts. Aside from the obvious of course. Yet the force of the object falling will always be the same.... however different heights affect the beam differently. So what involves velocity and mass, Momentum, however I don't see how I can use momentum to calculate anything I need either. I'm very lost with this. I'd really like to be able to figure out how I Can do this.
     
  13. Apr 21, 2005 #12
    Consider the energy of the falling object, [itex] PE = mgh [/itex]. As h increases, the potential energy increases, correct? This PE is then converted to KE as it falls, and with given h, the KE will impart a certain amount of energy into the beam. You'll need to know the mass of the entire beam system, aside from the shaft that holds it up, to find the clockwise torque due to the right side being 4x longer. Don't forget to include the projectile.

    Once you can find that, you'll find the torque required to put the beam in motion.
     
  14. Apr 21, 2005 #13
    So once the mass impacts the beam most of the energy has been converted into kinetic energy with a small amount of potential energy still left. So this kineteic energy is then transfered over to the hacky sack, but you're right, I would need to find the torque first, lets give some theoretical numbers first. Lets say the beam is equally distrubted in mass, with a total mass of 1 kg. The mass falling is 3kg, and the hacky sack is .05 kg. Lets say the height of the mass falling is .5m. Left arm 5 cm right arm 20 cm

    GPE would therefore be: 14.7

    Kinetic energy at impact would be (approx): 14.7 ( a little less)

    net Torque: 1.461Nm
     
  15. Apr 21, 2005 #14
    You want the launch angle to be 80 degrees. Find the displacement from the original position the seesaw will be in to its final position. You'll need to specify the seesaw height too.
     
    Last edited: Apr 21, 2005
  16. Apr 21, 2005 #15
    Lets change the launch angle to 60 degrees becausethe seesaw is so small, the fulcrum would only be .8 cm high with an 80 degree launch. With a 60 degree launch the height of the fulcrum would be 4.33 cm. Also I'm thinking of starting with the seesaw level to make calculations a little easier. So the arc length would be .1047 m of travel distance for the hacky.
     
  17. Apr 22, 2005 #16
    If you start the seesaw level, then your going to need a force to balance the seesaw initially, since the long side is significantly heavier than the short. Consider this.

    Otherwise, now that you found the height you should be able to find the different torques that the seesaw will experience when a certain amount of energy (from a falling object) is imparted on it. If you're having difficulty try energy analysis. Some assumptions and simplifications will greatly help you here.

    Find the new flight time with the new launch angle, from there you can use energy analysis again to find the amount of energy needed to bring the ball to the peak of its trajectory, and thats the energy you'll want to drop on the seesaw.
     
  18. Apr 22, 2005 #17
    How would the Kinetic Energy tranfer from one end of the beam to the other if you knew the n et torque and everything and the kinetic energy at impact. Are they both the same? Is the kinetic energy of the hacky the same ? Or is it affected depending on the torque?
     
  19. Apr 22, 2005 #18
    [tex] KE = W = \tau \theta [/tex]

    The work done on it will equal the torque applied to it times the angle it sweeps, or

    The work done to it will equal the kinetic energy of the ball as it impacts the system.
     
  20. Apr 22, 2005 #19
    The torque on the left would be the same as the torque of the right if they were the same length. Find the moment of inertia of each arm, if you know the torque of one end, you can find the torque of the other end.

    [tex] T = I\alpha = \frac{1}{12}mr^2\alpha [/tex]

    [tex] r_{right} = 4r_{left} [/itex]
     
  21. Apr 22, 2005 #20
    so would the kinetic energy be the same for both masses, just one would be in the opposite direction?
     
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