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Spring launcher equation 
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#1
Jan411, 05:09 PM

P: 28

1. The problem statement, all variables and given/known data
A spring is fired from a launcher angled Θ degrees from the floor. The spring has force constant k and mass M. The target is Δd metres away from the launcher and the target is elevated Δy metres. When the spring is launched, it travels at initial speed v0 towards the target, Θ degrees from the horizontal. The acceleration due to gravity is g = 9.8 m/s^{2}. What is the distance the spring needs to be pulled back, x? 2. Relevant equations t = time x = deformation of spring Ee = 0.5kx^{2} Ek = 0.5mv0^{2} Δy = (t)(v0)(sinΘ) + 0.5(g)(t^{2}) Δd = (v0)(cosΘ)(t) 3. The attempt at a solution I just need someone to check my calculations. I need to figure out how far to pull back the spring based on a given angle, and a given horizontal distance and vertical distance to the target. There are no specific numbers, I just need a general equation. The elastic potential energy in the spring is approximately equal to the kinetic energy the spring has after the launch. Air resistance and friction are negligible. Therefore, 0.5kx^{2} =0.5mv0^{2} kx^{2} = mv0^{2} v0^{2} = (kx^{2}) / m Rearranging Δd = (v0)(cosΘ)(t) t = (Δd) / (v0 cosΘ) Substitute t into Δy = (t)(v0)(sinΘ) + 0.5(g)(t^{2}) Δy = (Δdv0sinΘ/v0cosΘ)(v0)(sinΘ) + 0.5(g)(Δd/v0cosΘ)^{2} Δy = ΔdtanΘ + (gΔd^{2})/(2v0^{2}cos^{2}Θ) Δy  ΔdtanΘ = (gΔd^{2})/(2v0^{2}cos^{2}Θ) 2v0^{2}cos^{2}Θ = (gΔd^{2})/(Δy  ΔdtanΘ) v0^{2} = (gΔd^{2})/(2Δycos^{2}Θ2ΔdsinΘcosΘ) v0^{2} = (gΔd^{2})/(2Δycos^{2}ΘΔdsin2Θ) Substitute v0^{2} = (kx^{2}) / m (kx^{2}) / m = (gΔd^{2})/(2Δycos^{2}ΘΔdsin2Θ) x^{2} = (gΔd^{2}m)/(2kΔycos^{2}ΘΔdksin2Θ) x = √(gΔd^{2}m)/(2kΔycos^{2}ΘΔdksin2Θ) 


#2
Jan411, 06:55 PM

P: 28

bummmppppp



#3
Jan411, 07:04 PM

P: 33

what type of physics course is it?
just wondering 


#4
Jan411, 07:21 PM

P: 28

Spring launcher equation
senior high school physics



#5
Jan411, 07:30 PM

P: 33




#6
Jan411, 07:49 PM

P: 28

Nope. Just normal physics. But I'm selfstudying for the AP physics exam. We were assigned a project to build a spring launcher. The teacher places a target somewhere, specifies the height and the distance, and also specifies the angle of launch. The point of the project is to hit the target. Right now I'm working on the equation to calculate the distance I need to pull the spring back.



#7
Jan411, 08:20 PM

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Or is the distance Δd defined as the spring's center, after the spring is already compressed? If you wish to be precise, the geometry is important here. If the launcher and target do not move, the spring's total distance to the target changes as the spring is compressed (changes by some function of x). So it's important to precisely model how Δd, x, and the spring's total distance to the target (after compression) all fit together. A diagram would prove very useful. (Without backing this up with any math, I did a quick and dirty calculation that came out as only about 1/2 of the original potential energy gets translated into the linear kinetic energy of the center of mass of the spring.) [Edit: this assumes that the spring is not attached to any masses, and it is only the very spring itself that has a mass M.] 


#8
Jan411, 08:46 PM

P: 28

Just to clarify, the spring is the projectile.
"What part of the launcher? The top of the launcher? And if so how is the top of the launcher defined?" The horizontal distance Δd from the target to the launcher is defined as the distance along the floor from the target to the base of the launcher (the launcher is always on the floor). This is in the xdirection only. The vertical distance Δy is the height of the target above the floor. Sometimes the target is elevated, sometimes it isn't. In the diagram, the target isn't elevated. "What is the length of the spring at equilibrium?" The length of the spring at equilibrium is 7.5 cm, but this shouldn't matter. All that matter is x, which is the amount of deformation. The distance from the spring to the target doesn't change as the spring is pulled back. The front of the spring stays stationary. The distance from the centre of mass of the spring to the target changes as the spring is deformed, but this small distance is negligible. "I don't think that's a good assumption. Much of the initial potential energy remains in the massive spring, after the launch, in the form of internal oscillations." So what would be your version of the modified equation to figure out the distance to pull the spring back in terms of Δd, mass of the spring, Δy, k spring constant, and theta, the angle of launch. 


#9
Jan411, 10:13 PM

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But before I let it drop, I just want you to realize that you are introducing a source of error into the math by doing so. Even in perfect conditions (ideal springs, launched in a vacuum, etc), it will cause the spring to be off target by an amount comparable to x (it also depends on Θ too). So if you calculate that x should be 4 cm, don't be surprised if the spring misses its target by a couple cm or so, even in the best of conditions. But what you do need to do is estimate the spring's approximate center of mass in relation to the dot on the floor indicating where the base of the launcher is supposed to go. You need to find this displacement, in both the horizontal and vertical directions. Perhaps you can estimate this once, and use the same value for all launches. But you should at least measure it once. All of the kinematics equations need to relate to the spring's center of mass. So you should have at least a rough idea of where it is in relation to the target. What you are given is the displacement between the target and the base of the launcher. You will need to do is convert this to the displacement between the target and the spring's approximate center of mass. You can fit all this right into your spring stretching equation. If instead of a massive spring, you were to launch a mass from the end of an ideal massless spring, 100% of the spring's potential energy would end up as the mass's kinetic energy. So let's model the massive spring by approximating it as a mass and separate massless spring combination. Consider a massless spring of spring constant k. Break the spring into two equal pieces, and glue a point mass M, right in the middle. When you break a spring in half, the spring constant is approximately double what the original spring constant was. So let's call this new subspring constant j j = 2 kNow let's go back to two subspring system with the mass in the middle. Imagine compressing (or stretching) the entire thing by the amount x. How far does the mass in the middle move? Let's call this displacement of the mass in the middle z, z = ½ xOnce compressed (or stretched), and immediately after being released, we can ignore the subspring that's no longer pushing or pulling on the mass. Only one of the subsprings make a difference anymore. So the total potential energy, that will end up getting converted to the mass's kinetic energy, is, E = ½ j z^{2}How does this energy relate to k and x? (And if you're wondering where the rest of the energy went, it went into the other subspring that we can ignore for this. You can think of it as the internal oscillation part if you want to. But it doesn't affect the center of mass.) 


#10
Jan511, 03:23 PM

P: 28

Alright that helped a lot. So, the new equation would be
x = √ [(2gΔd^{2}m) / (2k(y_{2}0.3sinΘ)cos^{2}ΘΔdksin^{2}Θ)] By the way, the length of the launcher platform is 30 cm, so the original height of the spring above the floor would be 0.3sinΘ. And your suggestions about the elastic potential energy and kinetic energy altered the equation by a factor of root 2. This means that, if I plug in mass of spring, horizontal displacement, spring constant, vertical displacement, and angle theta, I will get the distance I should pull back? And if I pull the spring back by that amount, it would hit the target? 


#11
Jan511, 08:58 PM

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Hello tobywashere,
I have a few separate comments on your new equation, based on the equation that I came up with. I'll take them one at a time. But are you sure you want to use the entire length of the vertical rod for the 0.3 in the 0.3sinΘ? What you're interested in the distance from the pivot point to the distance to the approximate center of mass of the spring. I can't tell what this is without a better diagram or description of the launcher. I just want you to be careful on this. Drawing a detailed diagram of the launcher, and calculating the height from that might be better. (And it might turn out that the difference in height of the spring due to the tilt Θ, is small enough to ignore. I mean we're already making a few approximations anyway. But that really depends on the design of the launcher, and how much error is acceptable.) Right now you have something in the form (y  h_{0}),but it should be of the form (h_{0}  y).where h_{0} is the initial height of the spring above the ground. Consider the case where Θ = 0. Making the correction (and substitutions), the equation simplifies to: x = √[(2gd^{2}m)/(2k(h_{0}  y))]which makes sense. If Θ = 0, the target better be below the initial height of the spring, otherwise, no value of x will ever cause the spring to hit the target. But in your equation, the value of x will be an imaginary number, unless y_{2} is large and positive. That doesn't make much sense to me. What should happen is that x becomes imagnary if y is too large and positive. Or if instead of switching them around, you could put a negative sign on the whole term. But one way or another something doesn't look right with the negative signs. [Edit: corrected a couple typo mistakes.] 


#12
Jan511, 09:51 PM

P: 28

Thanks for all your help!
Also, y_{2} is the height of the target above the floor, which is the final height of the spring projectile. The expression is Δy = y_{2}  0.3sinΘ, which I used to replace Δy in the equation. In the case that the target is on the floor and below the spring, Δy would be negative. In the case that the target is elevated and above the spring, Δy would be positive. 


#13
Jan511, 11:03 PM

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But see my comments below about g The constant g, by convention, is always a positive 9.8... m s^{2}. If, for the specific problem at hand, you define your position, velocity and acceleration vectors pointing up, then the acceleration due to gravity is g = 9.8 m s^{2}. But the value of the constant g itself is still positive! In other words, depending on the problem at hand, feel free to put a negative sign in front of g, if the acceleration vector points in the opposite direction of gravity. But don't change the value of g itself! Using this convention will save you immeasurable confusion in the future, such as a future college course is physics. The convention is that g is always positive. The acceleration itself isn't necessarily positive (such as g); but the constant g itself is positive. http://en.wikipedia.org/wiki/Standard_gravity I probably should have mentioned that earlier, since you indicated that in your original post, but I didn't really think about it then. So your equation should work okay as it is if you (erroneously) define g as being negative, (except for the sin^{2}Θ vs the sin(2Θ) difference). But I suggest reworking the problem using a positive g from the beginning; particularly if you are going to show this equation to your instructor. Let me present you with an equation (that uses a positive g). What do you think of this? [tex] x = \sqrt{\frac{2g(\Delta d)^2 m}{2 k (0.3 \sin \theta  y_2) \cos^2 \theta + \Delta dk \sin (2 \theta)}} [/tex] x = √[(2gΔd^{2}m)/(2k(0.3sinΘ  y_{2})cos^{2}Θ + Δdksin(2Θ))] Do you think that will work? Does it match your version (with a positive g) or do you find any mistakes in it? 


#14
Jan611, 07:52 PM

P: 28

That's perfect. Thank you so much for your help. What do you do? Are you a physics teacher? Professor?



#15
Jan611, 08:03 PM

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Good luck on your project! 


#16
Jan611, 08:14 PM

P: 28

One last thing. About your statement that only half of the elastic potential energy is eventually converted into kinetic energy. I asked my physics teacher and he didn't believe it. He said that he has been doing the project for years and all of his students have calculated that all of the elastic potential energy gets converted to kinetic energy. Are you absolutely sure? I can follow your line of reasoning, but my teacher isn't usually wrong.



#17
Jan711, 02:56 AM

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The entire time, I was thinking that they spring to be launched was one that, starting from its equilibrium state, can be both stretched and compressed, while remaining in its linear region. I was also thinking that the spring had a very low internal damping coefficient, such that once it started to oscillate, it would keep going for awhile. Imagine a spring that can be squeezed together when a compressive force is applied, and can be pulled apart when a stretching force is applied. (Image courtesy of this site: http://www.shutterstock.com/pic2320...ackground.html) Consider a spring such as the one discussed above. Secure one end of the spring to a solid, immobile platform. Then either stretch it or compress it some distance x and let go. The spring will oscillate back and forth for some time. This is the energy which gets trapped in the oscillations that I was discussing earlier. So for a spring such as this, if launched from a launcher, I would guess that only roughly half the kinetic energy would end up as the kinetic energy in the center of mass. ============================ On the other hand, I hadn't even considered springs that are already at or near their maximum compression at equilibrium! Imagine a spring, starting from equilibrium, that you can stretch apart, but cannot compress (at least not very much, before reaching saturation). (Image courtesy of this site: http://www.shutterstock.com/pic4496...alspring.html That type of spring might change the centerofmass kinetic energy ratio significantly. Essentially, what happens is that each loop on the spring hit's the adjacent loop at the moment the entire spring fully collapses, and it ends up being one big inelastic collision. I did some fairly detailed calculations regarding the ratio of kinetic energy after vs the total kinetic energy before. (Normally, I would not just give these calculations away to the original poster of a thread, but you mentioned that your instructor is already recommending using a given value; and besides my calculations are a bit beyond most highschool physics anyway.) Here is my model. In this model we are only concerned with the inelastic collisions themselves, and kinetic energies before and after the collisions. Nothing so far deals with spring characteristics (more on the below). Springs are modeled by distributed masses. Suffice it to say, that all the spring's potential energy is converted to the mass's kinetic energies immediately before the collisions (kinetic energy is not conserved after the collision however).
//////// Two mass example. //////// Consider a "spring" simply modeled by a mass at the tail end, and an equal mass in front (at rest) before the collision. Conservation of momentum: [tex] mv_1 + 0mv_1 = (2m)v_2 [/tex] [tex] v_2 = \frac{1}{2}v_1 [/tex] Kinetic energy comparison: [tex] \mathrm{before} \ \Leftrightarrow \ \mathrm{after} [/tex] [tex] \frac{1}{2}mv_1^2 \ \Leftrightarrow \ \frac{1}{2}(2m) \left(\frac{1}{2}v_1 \right)^2 [/tex] simplifying, [tex] \frac{1}{2}mv_1^2 \ \Leftrightarrow \ \frac{1}{4}mv_1^2 [/tex] [tex] \frac{(\mathrm{K.E. \ after})}{(\mathrm{K.E. \ before})} = \frac{1}{2} [/tex] //////// Three mass example. //////// Conservation of momentum: [tex] mv_1 + m\left( \frac{1}{2} v_1 \right) = (3m)v_2 [/tex] [tex] v_2 = \frac{1}{2}v_1 [/tex] Kinetic energy comparison: [tex] \mathrm{before} \ \Leftrightarrow \ \mathrm{after} [/tex] [tex] \frac{1}{2}mv_1^2 + \frac{1}{2}m \left( \frac{1}{2}v_1 \right)^2 \Leftrightarrow \frac{1}{2}(3m) \left(\frac{1}{2}v_1 \right)^2 [/tex] simplifying, [tex] \frac{1}{2}mv_1^2 + \frac{1}{8}mv_1^2 \ \Leftrightarrow \ \frac{3}{8}mv_1^2 [/tex] [tex] \frac{5}{8}mv_1^2 \ \Leftrightarrow \ \frac{3}{8}mv_1^2 [/tex] [tex] \frac{(\mathrm{K.E. \ after})}{(\mathrm{K.E. \ before})} = \frac{3}{5} [/tex] //////// Four mass example. //////// Conservation of momentum: [tex] mv_1 + m\left( \frac{2}{3} v_1 \right) + m\left( \frac{1}{3} v_1 \right)= (4m)v_2 [/tex] [tex] v_2 = \frac{1}{2}v_1 [/tex] Kinetic energy comparison: [tex] \mathrm{before} \ \Leftrightarrow \ \mathrm{after} [/tex] [tex] \frac{1}{2}mv_1^2 + \frac{1}{2}m \left( \frac{2}{3}v_1 \right)^2 + \frac{1}{2}m \left( \frac{1}{3}v_1 \right)^2 \ \Leftrightarrow \ \frac{1}{2}(4m) \left(\frac{1}{2}v_1 \right)^2 [/tex] simplifying, [tex] \frac{1}{2}mv_1^2 + \frac{1}{2}m\frac{4}{9}v_1^2 + \frac{1}{2}m\frac{1}{9}v_1^2 \ \Leftrightarrow \ \frac{1}{2}mv_1^2 [/tex] [tex] \frac{7}{9}mv_1^2 \ \Leftrightarrow \ \frac{1}{2}mv_1^2 [/tex] [tex] \frac{(\mathrm{K.E. \ after})}{(\mathrm{K.E. \ before})} = \frac{9}{14} [/tex] //////// n mass example. //////// Conservation of momentum: [tex] \sum_{j=1}^{n1} \frac{mvj}{n1} = nmv_2 [/tex] simplifying, [tex] \frac{nmv_1}{2} = nmv_2 [/tex] [tex] v_2 = \frac{1}{2}v_1 [/tex] Which is now shown to be true for all n. Kinetic energy comparison: [tex] \mathrm{before} \ \Leftrightarrow \ \mathrm{after} [/tex] [tex] \frac{1}{2} \sum_{j=1}^{n1} \frac{k^2}{\left(n1 \right)^2}mv_1^2 \ \Leftrightarrow \ \frac{n}{8}mv_1^2 [/tex] simplifying, [tex] \frac{1}{2}mv_1^2 \frac{n(2n1)}{6(n1)} \ \Leftrightarrow \ \frac{n}{8}mv_1^2 [/tex] [tex] \frac{(\mathrm{K.E. \ after})}{(\mathrm{K.E. \ before})} = \frac{3(n1)}{2(2n1)} [/tex] [tex] \mathrm{maximum} \ \frac{(\mathrm{K.E. \ after})}{(\mathrm{K.E. \ before})} = \lim_{n \to \infty} \frac{3(n1)}{2(2n1)} = \frac{3}{4} [/tex] Here is the results: the spring's maximum final kinetic energy is at most 3/4 the original total potential energy (it is assumed that the springs original, total potential energy is completely converted to kinetic energy immediately before the collision [but not after]). The rest is doomed to be lost to heat (friction). Remember, with the other type of spring (which is the ideally elastic version), I mentioned that only 1/2 the original energy gets transferred to the centerofmass kinetic energy. With the new type of spring, you can bring that up to 3/4. But according to my math, you'll never get any more than that. ============================================ Special note regarding springs that are fully compressed at equilibrium: Springs that are already fully compressed with no tension need to be treated a little differently. Hooke's law still applies, but it needs to be modified: F = k(x + x_{0}); for x > 0.where x is measured at the point where the spring is fully compressed. The constant x_{0} cannot be measured directly, but it can be measured by hanging calibrated weights on the spring and measuring x; then determined either by graphing or solving a couple simultaneous equations. Your instructor should know how to do this. If you don't account for x_{0}, the potential energy stored in the spring might be underestimated. Underestimating the potential energy stored in the spring, combined with overestimating the kinetic energy of the spring when it flies off the launcher (100% vs. 3/4) might together account for why nobody (even your instructor) ever noticed anything in past classes (I'm only speculating on that point). ================================================ Here is what I recommend moving forward.



#18
Jan711, 04:30 PM

P: 28

My instructor isn't providing springs. The students are responsible for providing their own springs. The springs I'm using are identical to the springs in the second picture. The springs can't be compressed at equilibrium.
So I edited the equation assuming 3/4 of the elastic potential energy is converted into kinetic energy. I measured the k constant and the mass of the spring today. I tested the launcher out, and the equation proved to be surprisingly accurate. The launcher wasn't perfect, but there were many factors that could've caused inaccuracies. I don't understand the part about applying hook's law to my type of spring. Isn't F = k(x + x0) just the k constant times the change in the spring's length (deformation)? I've attached the new equation and a picture of my launcher below. 


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