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Catapult/Seesaw Calculations, part 2?

  1. Apr 26, 2005 #1
    I have a seesaw style catapult.

    the arms are at a 4 to 1 ratio. I have all the masses and everything, the hacky is 50 grams, and The beam is 381 grams with a length of 50 cm (1 arm is 40 other is 10).

    My velocity to launch the hacky 2.5 meters away from the release point and have it land .76 m high is 6.4 m/s at a 64 degree angle off the floor.

    If i want to drop a 1 kg weight, how high would i need to drop it in order to achieve this velocity?
    Last edited: Apr 27, 2005
  2. jcsd
  3. Apr 26, 2005 #2
    any ideas? Anyone?
  4. Apr 27, 2005 #3
    IS this purely a matter of torques? Torque of gravity on the board, torque of hacky sack on the board, torque required to get the hacky sack to a certain velocity? Then countering all those torques?

    Would Torque for velocity be KE = 1/2 (m) v^2 so then W = torque (theta-sweep angle) ?
  5. Apr 27, 2005 #4
    How are you going to ask, in less than 3 hours, if anyone has solved your problem?
  6. Apr 28, 2005 #5
    I have been working on this for more than a week, and i still have no solution. Also no one seems to be able to help. I'm running out of time and i'm just trying to understand how the kinetic energy of the mass falling is converted into the kinetic energy of the hacky sack with a beam with the fulcrum at a 1 to 4 ratio on the beam.
  7. Apr 28, 2005 #6


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    There are too many unknowns in your problem. Is the kg mass going to bounce? Is the arm it lands on going to hit the floor? At what angle will the beam be postioned when the 1 kg hits it? Is friction in the fulcrum an issue? Will the beam bend like a springboard? Those come to mind real quick. I expect there are more issues, but those are enough to preclude a simple solution.
  8. Apr 28, 2005 #7
    I can answer all those questions i just need someone who will know how to solve it.
  9. Apr 28, 2005 #8


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    Qualitative answers will not do. You will have to quantify them. How much energy will be lost when the 1 kg hits the beam? How much more will be lost when the beam hits the floor? How high is the fulcrum? It sounds like you have determined only one possible path for you hacky. There may very well be no solution that will give that path. The physics may yield a result that cannot be solved analytically. It may require a computer simulation to get even an approximate solution
  10. Apr 29, 2005 #9


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    An oversimplified solution:

    If the beam starts out horizontal, the system will have angular momentum due to the dropping kg. Say it hits with speed v and sticks to the board (no bounce) exactly at the end, L/5 from the fulcrum and assume the board is rigid. Angular momentum will be conserved, and the initial angular momentum will be mvr. After that, the angular mometum will be in three parts: as long as the hacky is in contact with the board, its speed will be a multiple of the speed of the 1kg. The speed of the center of mass of the board will be a multiple of the speed of the 1kg. Figure out those speeds from the geometry and calculate the angular momentum of each of the three objects based on their speeds, masses, and distances from the fulcrum. From that you will find the speed of the hacky going straight up.

    Now you have to figure out the modifications to make this work the way you want. How far from the end of the board will the 1kg really hit? How can you make sure it sticks? What angle does the board have to be at to give the hacky the right launch angle? What is the angle of the board going to do to the angular momentum calculations? If you came within 30 to 40% of the intitial velocity needed, you have probably done well. Then you would have to build a prototype and tweak it. That's what engineers do.
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