Build a Coin-Throwing Catapult in 2 Weeks

  • Thread starter Thread starter mcoy
  • Start date Start date
  • Tags Tags
    Catapult
AI Thread Summary
The discussion centers on creating a catapult that can accurately launch a 5-cent coin a distance of 2 meters within a two-week timeframe. Participants emphasize the importance of calculations related to energy, angle of launch, and the mechanics of the catapult, suggesting that a counterweight design may be overly complex for this task. Suggestions include using springs or elastic bands, with advice on measuring spring constants and energy stored. There is also mention of using a trebuchet design, although some argue it's unnecessary for launching a coin. Overall, the focus is on finding a simple, effective design and the necessary calculations to achieve the desired launch distance.
mcoy
Messages
5
Reaction score
0
I need to make a catapult in less than 2 weeks... it has to be able to throw a 5 cent coin (aust coin) on exactly 2 metres... its not much but i can't figure out what design I'm going to make, since i need to be able to do calculations beforehand and make the catapult using my calcs...
so now i need a good and easy-to-make design... i.e. making a catapult that doesn't need any gears... i was thinking of a counter-weight type but that would be too much for a 5 cent coin... i think...

somebody help me please??
 
Physics news on Phys.org
Doesn't that rather depend on the size of the counterweight, the weight of the arm, and the moment arms involved?
 
mcoy said:
I need to make a catapult in less than 2 weeks... it has to be able to throw a 5 cent coin (aust coin) on exactly 2 metres... its not much but i can't figure out what design I'm going to make, since i need to be able to do calculations beforehand and make the catapult using my calcs...
so now i need a good and easy-to-make design... i.e. making a catapult that doesn't need any gears... i was thinking of a counter-weight type but that would be too much for a 5 cent coin... i think...
somebody help me please??
Calculations will only give you an approximate result. I would suggest that you use an energy approach.

You have to determine the angle at which the coin leaves the catapult. Then determine how high the coin has to go. Calculate the change in energy of the catapult arm and counterweight between initial and final positions. Assume the energy loss of the arm is gained by the coin.

If you use a spring or elastic band, you have to measure the spring constant and measure the load distance to find the energy stored in the spring. You also have to take into account the kinetic energy of the arm after the coin is released.

AM
 
Once I was really into building catapults with counter-weights. they are called trebuchets. you can see a movie of one of the big ones (6-feet+) fireing at my site. I also made small ones out of lego and cardboard that shot for small distances (though if you plan it right I know from experience that you can make a 1-foot trebuchet that can shatter a marble on a wall from over 10-feet) a rough approximation for the distace that a trebuchet can shoot is:
Range = 2 * h * (M_{cw}/m_{p}) were h is the distance that the CW falls. You can start with that and then make small changes until it works just right. You can learn a lot about building them on the web.
 
daniel_i_l said:
Once I was really into building catapults with counter-weights. they are called trebuchets. you can see a movie of one of the big ones (6-feet+) fireing at my site. I also made small ones out of lego and cardboard that shot for small distances (though if you plan it right I know from experience that you can make a 1-foot trebuchet that can shatter a marble on a wall from over 10-feet) a rough approximation for the distace that a trebuchet can shoot is:
Range = 2 * h * (M_{cw}/m_{p}) were h is the distance that the CW falls. You can start with that and then make small changes until it works just right. You can learn a lot about building them on the web.

yeah but firing a coin doesn't need a trebuchet setup... I've done some calculations, but i just need the design, and to callibrate it... i don't have any idea how create and calibrate the initial velocity of the arm (the throwing bit) of the catapult...
 
Pls help me Calculation on distance

I'm doing a Assignment is to conststruce a captapult that will launch a marshmallow(Large size) at target on the floor at the other end of the classroom.
I designed catapult. the rules saing the energy;; to launch the marshmallow cannot come directly from the motion of a human being.so i used elastic band .
but, they saing to calculate the distance.
so, i use an energy approach. so I determine the angle at which the mashmallow leaves the catapult.then I assume then energy loss of the arm is gained by the marshmallow.but I ued elastic band, but, i don't know how to measure the load distance to find the energy stored in the elaastic band.how to i take into this measure ments...coz, I don't know the formula of the elasticband energy stored.
just i used this formula to find the velocity.
Mgh = 1/2*M*V^2

but I don't consider the elastice band energy :frown:
please anyone could me help to solve this probelm
I need the full foumula how to fine the calculation...
pls sent me to desitamil@gmail.com
Thanks for look through my post.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top