# Need to find the spring constant to achieve Max Velocity

• MikeyDoubleDEE
In summary, the student has calculated the kinetic energy of a softball at 22 meters per second, using the law of conservation of energy and the spring constant for a given distance and launch angle.
MikeyDoubleDEE
Hello All.
I am mentoring a high school student in my area with his class project for school. He has chosen he wants to launch an object (in our case, a softball) into a 5' diameter area. The idea is to build basically an oversized slingshot using an extension spring as the source of energy.

We arbitrarily picked a distance of 50yds (45.7 m) with a launch angle of 30°. Using projectile motion fundamentals we came up with the following:

d=(Vi²/g)*sin2Θ

Plugging in:
d = 45.7m
Vi = ?
Θ = 30°
g = 9.81m/s²Solving for Vi, I got 22m/s (Which will serve as the final Velocity in my forthcoming spring calcs)

So now we migrate into spring fundamentals. I initially thought to use an engineer's best friend F=ma but quickly realized that the acceleration is not constant in the context of springs and that calculus would need to be leveraged for a diminishing a value. So scratch that.

Settling on the law of conservation of energy, I sought to find the Kinetic Energy of the mass (Softball = .195kg) at my final Velocity of 22m/s.

K=1/2(mv²)

Plugging in:
m = .195kg
v = 22m/s

Solving for K, I got 47.19 Joules

This is where I start to doubt myself:
I found this equation in my old College Physics textbook
Vmax=√(k/m) * A
(A standing for displacement, and k standing for spring constant)

Plugging in:
Vmax = 22m/s
m = .195kg
A = .61m (24" is what we arbitrarily selected as a starting point for our spring selection)

Solving for the spring constant k, I got 2507 N/m (14.315 lbs/in : My mind thinks in pounds and inches, not Newtons and meters)

Is this the correct approach? I don't want to start buying hardware to build our prototype unless I can get some feedback from the physics community that our approach to solving this design problem is valid. Thank you so much for your time!

Disclaimer: I realize that this approach does not take into account gravity during the acceleration of the mass "uphill" at 30 deg. Also, this assumes a frictionless slingshot. And we decided to ignore wind resistance in the projectile motion portion of this problem.

Last edited:
Welcome to the PF.
MikeyDoubleDEE said:
We arbitrarily picked a distance of 50yds (45.7 m) with a launch angle of 30 yards.
I assume that's a typo, and you mean 30 degrees. Is there a reason you didn't pick the more traditional launch angle of 45 degrees? A low launch angle like that will make it harder to hit the target.
MikeyDoubleDEE said:
And we decided to ignore wind resistance in the projectile motion portion of this problem.
That's probably not a good assumption. It makes the calculations easier, but you may want to consider making a first order correction to that once you get your initial numbers nailed down.

Also, if you don't impart some spin on the softball, it will act like a knuckle ball in the air, which will impair your accuracy some, IMO. If you can somehow give it some backward spin during the launch, that could improve your accuracy a fair bit.

Yes that was a typo. I've edited the mistake.

We may recalculate to include wind resistance. That could be a good exercise.

I have thought of the knuckle ball effect and completely agree with you. As we get into prototyping, we will start to troubleshoot some of our findings. It will make for some more meat in his paper that accompanies the project

berkeman
As a quick reality check on your calculations, you could take a softball out to a local football field and the student could take a video of you throwing the softball the 50 yard distance to hit the center of the field from the goal line. There are several ways that you can analyze the video to see how close your calculation of 22m/s is for Vo. Fun project.

MikeyDoubleDEE
I like that idea!

Of course, if the student has a better arm, maybe you should shoot the video...

anorlunda
Can anyone enlighten on my approach to this problem? In particular did I use the correct equation to determine a spring rate with a given Vmax?

## 1. What is the spring constant and why is it important?

The spring constant, also known as the force constant, is a measure of the stiffness of a spring. It determines how much force is needed to stretch or compress a spring by a certain distance. In the context of achieving maximum velocity, the spring constant is important because it affects the speed at which an object can move when attached to a spring.

## 2. How is the spring constant calculated?

The spring constant is calculated by dividing the force applied to a spring by the distance the spring is stretched or compressed. This can be represented by the equation k = F/x, where k is the spring constant, F is the applied force, and x is the distance the spring is stretched or compressed.

## 3. What units is the spring constant measured in?

The spring constant is typically measured in units of newtons per meter (N/m) in the metric system or pounds per inch (lb/in) in the imperial system. These units represent the amount of force needed to stretch or compress the spring by one meter or one inch, respectively.

## 4. How does the spring constant affect maximum velocity?

The spring constant directly affects the maximum velocity that can be achieved by an object attached to a spring. A higher spring constant means that more force is needed to stretch or compress the spring, resulting in a greater acceleration and therefore a higher maximum velocity. Conversely, a lower spring constant will result in a lower maximum velocity.

## 5. How can the spring constant be adjusted to achieve maximum velocity?

The spring constant can be adjusted by changing the stiffness of the spring or by changing the amount of force applied to the spring. To achieve maximum velocity, the spring constant should be set to a value that allows the object to accelerate quickly without causing the spring to break or become too difficult to stretch or compress. This may require some trial and error or mathematical calculations.

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