Spring launching equation

  • #1
physicsdude2
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Homework Statement:
For this assignment, I will need to launch a spring from a launcher at an angle of Θ. Given the following equations, I need to combine them to find x, the distance that the spring needs to be pulled back to hit the target (fall inside a box). The provided inputs for the launch are the horizontal distance from the target to the launcher, the vertical distance from the target to the launcher, and the angle that I will be launching the spring at.
Relevant Equations:
The given equations are

Es=1/2(kx^2)
Eg=mgΔh
Ek=1/2(mv^2)
vh=vcosΘ
vv=vsinΘ
Δdh=vhΔt
Δdv=vvΔt+1/2(avΔt^2)

where vv is vertical velocity, vh is horizontal velocity, Δdv is vertical displacement, Δdh is horizontal displacement, t is time, x is the spring displacement (distance it needs to be pulled back, which is what i'm solving for), v is velocity, g is acceleration due to gravity (9.8m/s^2) and is also the same as av which is vertical acceleration, in this case 9.8 m/s^2 because this is dealing with projectile motion
I just need someone to check my calculations. There are no specific numbers, I just need a general equation. Look at my attached solution. I think I did everything right but the worrying detail is that gravitational energy is not included and I am not sure whether it should be.

Thanks in advance.
 

Attachments

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Answers and Replies

  • #2
pbuk
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  1. It is impossible to read, but from what I can make out it looks close.
  2. What is the sign of the denominator when the target and the launcher are at the same height (i.e. ## \Delta d_v = 0 ##)?
  3. Consider the simpler problem when we are launching a projectile at a known velocity and have to find the angle of elevation to hit a target: how do we take gravitational energy into account then?
 
  • #3
physicsdude2
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It is impossible to read, but from what I can make out it looks close.
Sorry, I didn't realize it doesn't get clearer as you zoom in. The original file I had would become more clear as you zoomed in on the words.
What is the sign of the denominator when the target and the launcher are at the same height (i.e. ## \Delta d_v = 0 ##)?
Please correct me if I'm mistaken. I think the denominator would be negative because the first term would be gone and you would be left with (-Δdhksin2Θ).
Consider the simpler problem when we are launching a projectile at a known velocity and have to find the angle of elevation to hit a target: how do we take gravitational energy into account then?
This is the part I am most unsure of. I was thinking that you wouldn't calculate gravitational energy but rather account for that by using -9.8 m/s^2 to calculate the acceleration downwards due to gravity as the projectile is in the air. Does that mean (in the original problem's context) that at the beginning there is only spring energy, which all gets converted into kinetic energy once the spring is launched? If that is true then the conservation of energy equation would be: Es=Ek
1/2(kx^2)=1/2(mv^2)
kx^2=mv^2
(kx^2)/m=v^2

This is what I initially had down and then after I rearranged the horizontal distance equation (Δt=Δdh/vcosΘ) and plugged it into the vertical distance equation [Δdv=vsinΘ(Δdh/vcosΘ)+1/2(av)(Δdh/vcosΘ)^2]. I then isolated for v^2 and subbed the energy equation into this and isolated the equation for x to get my final equation of x=√(avΔdh^2m)/(2kΔdvcos^2Θ-ΔdhksinΘcosΘ)
 
  • #4
haruspex
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[Δdv=vsinΘ(Δdh/vcosΘ)+1/2(av)(Δdh/vcosΘ)^2]. I then isolated for v^2 and subbed the energy equation into this and isolated the equation for x to get my final equation of x=√(avΔdh^2m)/(2kΔdvcos^2Θ-ΔdhksinΘcosΘ)
Isn’t Δdh the same as x? You have it on both sides of that last equation.
I suggest simplifying the notation, just x for Δdh and y for Δdv. I often just write s for sinΘ and c for the cos when there's no ambiguity.

Your image is nearly legible if opened in a new window, but the subscripts are still blurred. Per forum rules, images are for diagrams and textbook etc. extracts. Please take the trouble to type in equations, preferably in LaTeX.
 
  • #5
pbuk
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Please correct me if I'm mistaken. I think the denominator would be negative because the first term would be gone and you would be left with (-Δdhksin2Θ).
Yes: so how are you going to take the square root of something negative? Check through your workings making sure you are consistent; if you are taking ## g ## as positive then you need to measure vertical distances as positive in the same direction as ## g ##.

Does that mean (in the original problem's context) that at the beginning there is only spring energy, which all gets converted into kinetic energy once the spring is launched?
Strictly speaking that would only be true if the spring were horizontal: with a non-zero value for ## \theta ## then some of that energy is converted to gravitational potential. How much is this? Do you think the error it leads to is more or less significant than ignoring air resistance?

x=√(avΔdh^2m)/(2kΔdvcos^2Θ-ΔdhksinΘcosΘ)
I have converted this into ## \LaTeX##:
$$ x = \sqrt{m g \Delta d_h^2 )/(2k \Delta d_v \cos^2 \theta - \Delta d_h k \sin\theta\cos\theta}$$
you should learn how to do this, it is the digital language of maths.

$$ x = \sqrt{\frac{a_v \Delta d_h^2 m}{2k \Delta d_v \cos^2 \theta - \Delta d_h k \sin\theta\cos\theta}} $$

If ## a_v ## is positive (it is easier to group ## m a_v ## and replace with ## g ##) then the sign is wrong, also you should check the coefficient of ## \sin\theta \cos\theta ## (in your handwritten workings you had used the equivalence with ## \frac 1 2 \sin 2 \theta ## instead and I think that looked more correct).
 
  • #6
physicsdude2
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Isn’t Δdh the same as x? You have it on both sides of that last equation.
I suggest simplifying the notation, just x for Δdh and y for Δdv. I often just write s for sinΘ and c for the cos when there's no ambiguity.
Sorry if I didn't clarify, but Δdh is actually the horizontal distance from the target to the launcher, while x is the distance the spring needs to be pulled back to hit the target. Δdv is the vertical distance from the target to the launcher.
Your image is nearly legible if opened in a new window, but the subscripts are still blurred. Per forum rules, images are for diagrams and textbook etc. extracts. Please take the trouble to type in equations, preferably in LaTeX.
Sorry about that, I'm new to this forum (and the website overall) so I'm not overly familiar with the rules.
Yes: so how are you going to take the square root of something negative? Check through your workings making sure you are consistent; if you are taking ## g ## as positive then you need to measure vertical distances as positive in the same direction as ## g ##.
In my equation, I used av (acceleration vertical) instead of g, and is equal to -9.8 m/s^2.
Strictly speaking that would only be true if the spring were horizontal: with a non-zero value for ## \theta ## then some of that energy is converted to gravitational potential. How much is this? Do you think the error it leads to is more or less significant than ignoring air resistance?
So I should have included mgΔh for my gravitational energy. I think that if gravitational energy is supposed to be included, then the error it leads to will be more significant than ignoring air resistance.
I have converted this into ## \LaTeX##:
$$ x = \sqrt{m g \Delta d_h^2 )/(2k \Delta d_v \cos^2 \theta - \Delta d_h k \sin\theta\cos\theta}$$
you should learn how to do this, it is the digital language of maths.

$$ x = \sqrt{\frac{a_v \Delta d_h^2 m}{2k \Delta d_v \cos^2 \theta - \Delta d_h k \sin\theta\cos\theta}} $$

If ## a_v ## is positive (it is easier to group ## m a_v ## and replace with ## g ##) then the sign is wrong, also you should check the coefficient of ## \sin\theta \cos\theta ## (in your handwritten workings you had used the equivalence with ## \frac 1 2 \sin 2 \theta ## instead and I think that looked more correct).
Lastly, sin2Θ is equal to 2sinΘcosΘ, which was on the bottom and I simplified it to sin2Θ.

Once again, thanks for your help.
 
  • #7
pbuk
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So I should have included mgΔh for my gravitational energy.
No, you have already fully accounted for the effect of gravity during the flight of the projectile. The error I was talking about was the effect of gravity during the initial acceleration of the projectile by the spring. ## m g x \sin \theta ## of the spring's energy is converted to gravitational potential energy.

I think that if gravitational energy is supposed to be included, then the error it leads to will be more significant than ignoring air resistance.
Given the equation for the error I have given above would you like to reconsider that answer?

Lastly, sin2Θ is equal to 2sinΘcosΘ, which was on the bottom and I simplified it to sin2Θ.
Yes you did in your handwritten workings but in the typed solution you lost the 2 when you left it as ## \sin \theta\cos\theta ##. It (like the sign) is still wrong in my conversion to ## \LaTeX ##.
 

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