Physics of the catapult i made?

Question

We made a catapult from wood and springs. we attached the arm toa doorhinge and then attached springs further up. we pulled it back to release it and we need to knwo the physics of it. we are launching something just under a pound heavy and it has to go a distance of fifty meters. so we have delta x = 50m, m = 1 pound (needs to be in kilograms). we need the velocity and the angle we released it at was 45 degrees. is there any more information needed? please let me know if there is. if there isnt, what calculations to i need to make?

Answer 1

Well, the catapult launches the object at a certain velocity (v), directed 45 degrees upward. For projectiles, you have to break the velocity into an upward and a horizontal velociy. The upward velocity is v sin(45) and the horizontal velocity is v cos(45), so they are both .707v. The reason you do this is because the projectile travels to its highest point with its upward velocity in the same amount of time it travels half of its range with its horizontal velocity. You can find this time (t) using the formula v=vi+at: 0=.707v - 9.8(t), so t =.072v.
In this time, the object travels half of its range, so in twice this time the object travels its full range (50 meters). d = v*t, so 50 = .707v(2)(.072v), 50 = .1018 v^2, so v = 22.2 meters/second.

Now you have the velocity that the catapult must launch the object with. Now the other calculation you have to make is how much you have to compress the springs. This is a conservation of energy problem: when the cataput is released, the potential energy of the compressed springs becomes rotational kinetic energy that sets the catapult in motion, and launches the object.

The equations you use for this are:
Energy of compressed spring: E= 1/2k x^2
Rotational Kinetic energy: KE = 1/2 I w^2 ( I is the "moment of inertia", w is the angular velocity)
and
v=r*w (straight line velocity = angular velocity * radius of rotation)

So, you know that v = 22.2, and r is the length of the arm of your catapult (imagine that the catapult rotates in a circle with that radius), so you can solve for w.

now that you have w, you can find the rotational kinetic energy (1/2 I w^2). I in this case = 1/3 * r * m. r is the length of the arm, and m is the mass ( 1 pound = .45 kg).

Now that you have the rotational kinetic energy, you know what the spring energy has to be.

rotational ke = spring energy = 1/2 k x^2
(if you have more than one spring, your formula becomes rotational ke = n*1/2*k*x^2, where n is the number of springs)

If you don't know k, the spring constant of your springs, then what you do is tie an object (with mass m, in kg) to the end of your spring, hold it in the air, and measure how far the spring stretches because of it (in meters). Now use the formula F=k*x, where F is mg, and x is how much it is stretched), and you can find k.

Now you can solve for x, the distance you need to compress the springs for the catapult to get the right velocity.

Answer 2

The spring has to be "springy" enough to give the 0.45 kg weight enough accleration to traverse the 50 meters. The spring's action is related to the so-called Hookes Constant, Ks. The force developed by the spring is Ks * d, where d is the distance the spring is depressed prior to launch.
The force imparts an accleration to the weight in the angle of launch. If that is Al, then
50 m = 0.7 Al t^2, where t is the time the weight takes to traverse 50 m in the horizontal direction. Now you have to figure whether the weight will still be airborne at t, or if it will fall to earth first.

Answer 3

Non-metallic elastic bands have a rather poor efficiency; you shouldn't expect more than about 60% of the energy input to come out as output. Also, the rate at which the band contracts affects the efficiency. m/s² is not a unit for velocity You mentioned the range being 14 m. If you also knew the launch angle, you can calculate Vo very accurately using the range eq: R = Vo²*sin(2Θ)/g.........Solve it for Vo

Answer 4

the equation for the range (R=50) is
R=v^2sin(2theta)/g
=v^2/g since theta =45deg
so v=sqrt(gR)=sqrt(9.8*50)=22m/s neglecting air resistance