# How do i find the spring constant for a rope torsion spring

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1. Apr 22, 2017

### Buggsy GC

I have built a torsion wheel catapult (Mangonel) for an assignment,and I need to do some theoretical calculations about the displacement, velocity and time of the projectile, the problem is I don't know how to calculate the spring constant with out the displacement of the spring (x) and i dont Know how to measure x for a torsion wheel, I found one formula on wikipedia T= -k theta^2, https://en.wikipedia.org/wiki/Torsion_spring#Torsion_coefficient. Would this formula work for a catapult.

2. Apr 22, 2017

### Nidum

Torque = Torsional spring constant x angle of twist of one end of spring relative to other end .

So in principle to get a value for the torsional spring constant for your rope spring you need to measure the relative angle of twist which results when you input a range of known torques .

Can you see a practical way of doing this ?

3. Apr 22, 2017

### Nidum

I see in your other thread that you want to calculate the torsional spring constant rather than measure it .

Not impossible but very difficult . Sensibly I suggest that you just get your answers experimentally .

4. Apr 22, 2017

### Buggsy GC

I'm willing to do either, the theoretical calculations are for the projectile, I think I figured out the spring constant for different tightness of the rope, by using fishing weight, to weigh the arm down to firing position, and then using Newtons second law to calculate the force of the weights on the arm. I imagined that the force of the weights is equal to the force of the spring but in the opposing direction, it worked out that the less taught the rope the less weight were need as i expected. But I have no idea how to measure the displacement, or if the normal hooks law formula even applies for a torsion spring

5. Apr 22, 2017

### Buggsy GC

I thought maybe you could use the arc angle of the catapult, but I can't find any info to confirm this idea

6. Apr 22, 2017

### Staff: Mentor

My iinterpretation is that you only need calculations for time, velocity, angle, and distance for the projectile starting at the insurance of release. You don't need the spring constant or anything for the time prior to release.

If you write the solution for the trajectory given initial velocity, height and, and angle (neglect air drag), that gives you distance to hit the ground. Then, use the result backward, solve for the velocity and angle for distance 1m and 2m.

Then, ignore your calculations and calibrate your catapult experimentally. Say for example that you measure it's throw to be 2.2m. Then move the catapult back 0.2 meters and the next throw should land on the 2m mark. You can attach a 0.2m stick to the front to make your catapult come up exactly to the starting line. Then move the catapult back 1m, make the stick in front 1.2m long and the next throw should hit the 1m target. No need to readjust the catapult throw, just adjust the start position.

Your enemy to success might be variability. If ten similar throws give 10 different distances thrown, then you have a problem, and calculations become even more useless. Calibrating your catapult should include measuring the variability of distance with repeated throws,

In that case,the idea I suggested with a water balloon projectile may help, because the water will spread a half meter or so after bursting. That gives you a wide range of landing points that still hit the target.

7. Apr 22, 2017

### jack action

When you know the projectile initial velocity $v$, you know how much energy is required to get to that velocity, i.e. $\frac{1}{2}mv^2$, where $m$ is the projectile mass. So your spring must release at least this amount of energy for this to happen. The equation you presented represents the amount of energy stored in a spring. So the energy stored when the spring is loaded, $\frac{1}{2}K\theta_f^2$, minus the energy stored when the spring is fully released $\frac{1}{2}K\theta_i^2$ (You may preload the spring by an initial angle $\theta_i$) must at least be equal to the desired kinetic energy of your projectile or:
$$\frac{1}{2}mv^2 = \frac{1}{2}K\left(\theta_f^2 - \theta_i^2\right)$$
$$K = \frac{mv^2}{\theta_f^2 - \theta_i^2}$$
You will probably need more than that as some energy will be moving and deforming the catapult, but that is your starting point.

Normally, $\theta_f$ should be equal to the maximum torsion you can apply to your spring without breaking it (depends on the material and design of the spring) and $\theta_i$ will be equal to $\theta_f$ minus the angular displacement of your catapult. If the spring constant $K$ corresponds or is smaller to the one of the spring you analyzed, then it is a match. If it is smaller, then you can fine tune the preload angle $\theta_i$ to reduce the amount of energy stored and get the actual velocity you want.

8. Apr 22, 2017

### Dr.D

Do we have any real reason to believe that this system is linear? I would certainly question this.

I suspect that the "constant" is not really constant at all, but rather depends upon the tension in the rope spring.

9. Apr 22, 2017

### JBA

Would it be the rate that is affected by tension; or, simply the force at a given deflection?

10. Apr 22, 2017

### Dr.D

That's a tough question, one that I don't think we can even guess at without knowing a lot more details. I simply doubt that the whole system is linear and constant in all respects.

11. Apr 22, 2017

### CWatters

It might not be too bad. Usually the rope has a lot of twists even at the rest position and the difference between rest and firing position is only a small angle (<90 degrees) by comparison?

12. Apr 22, 2017

### CWatters

I think you need to measure the change in torque between the rest position and the firing position (in Newton Meters)

The displacement is the change in angle between the rest position and the firing position (in Radians).

The spring constant is one divided by the other.

13. Apr 22, 2017

### Buggsy GC

Going back to anorlundas statement: how do I find the initial velocity if the catapults arm is already moving before the projectile is released by the cross bar, do I need to do a previouse kinematic equation starting from the projectile at firing position. 2nd I thought I had to use the spring constant based off this source http://people.cs.ksu.edu/~nhb7817/ScratchCurriculum/Catapult/Physics Calculations for a Mangonel.pdf
Which states that PE=KErotational of the arm the main difference was these were the calculations for a rubber band mangonel instead of a torsion role model like mine.

14. Apr 22, 2017

### Buggsy GC

How do I measure the change in torque, I thought torque was the equivalent to Force except in the rotational axis e.g. F=ma, and τ=ια

15. Apr 22, 2017

### Buggsy GC

I've tested the catapult with marbles and it hits the target nearly every time or it's with in the accepted radius

16. Apr 22, 2017

### CWatters

If you are hanging weights on the arm the radius is the distance from the pivot to where the weights are attached.

17. Apr 22, 2017

### Buggsy GC

awesome so originally I just placed the weights on the firing cup, so does that mean I can just use the distance from the cup to the pivot point. and then once I have the torque of the spring I can rearrange the Equation τ=-kθ^2 to -τ/θ^2=k, and this will give me the spring constant for the torsion rope spring of my catapult

Last edited: Apr 22, 2017
18. Apr 23, 2017

### CWatters

Yes although strictly speaking you need to ensure the force (gravity acting on the weights) is at 90 degrees to the arm otherwise you need to calculate the lever arm length as ..

19. Apr 23, 2017

### CWatters

Eg measure the torque like this...

20. Apr 23, 2017

### Buggsy GC

thank CWatters and every one else this was a huge help. also even though we are dealing with rotational motion the angle can still be measured in degrees instead of radians correct, also would the PE of the spring 1/2kθ^2 still

Last edited: Apr 23, 2017