Build a Coin-Throwing Catapult in 2 Weeks

  • Thread starter Thread starter mcoy
  • Start date Start date
  • Tags Tags
    Catapult
Click For Summary
SUMMARY

The discussion focuses on constructing a catapult capable of launching a 5-cent Australian coin exactly 2 meters within a two-week timeframe. Participants emphasize the importance of calculating the angle of launch, energy changes in the catapult arm, and the use of springs or elastic bands for energy storage. A specific formula for estimating range is provided: Range = 2 * h * (M_{cw}/m_{p}), where h is the height the counterweight falls. Additionally, users share insights on calibrating the initial velocity of the arm and measuring energy stored in elastic bands.

PREREQUISITES
  • Understanding of basic physics concepts such as energy conservation and projectile motion.
  • Familiarity with catapult mechanics, specifically counterweight systems and elastic potential energy.
  • Ability to perform calculations involving angles, heights, and energy transformations.
  • Knowledge of measuring spring constants and load distances for elastic materials.
NEXT STEPS
  • Research the principles of projectile motion to optimize launch angles for desired distances.
  • Learn about energy conservation in mechanical systems, focusing on potential and kinetic energy transformations.
  • Study the mechanics of trebuchets and their design to understand counterweight dynamics.
  • Explore methods for measuring the spring constant and calculating energy stored in elastic bands.
USEFUL FOR

This discussion is beneficial for students, hobbyists, and educators involved in physics projects, particularly those focused on mechanical design and energy calculations for catapults and similar devices.

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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 21 ·
Replies
21
Views
10K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
35K
Replies
7
Views
13K
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 73 ·
3
Replies
73
Views
10K