# Complicated Catapult/Conservation of Energy Test Question

• ccmuggs13
In summary, the conversation is discussing a physics test problem involving a catapult and the calculation of elastic potential energy and velocity using conservation of energy. There is a discrepancy between the two methods used to calculate the elastic potential energy and the conversation also touches on the use of angular motion and the importance of considering the equilibrium position for the spring. The conversation concludes with the confirmation that the calculated results are correct and a plot showing the kinetic energy as a function of angle.
ccmuggs13
I just took a physics test today and there was one problem that I cannot figure out if I did correctly. The problem involves a catapult pictured here: http://i39.tinypic.com/a2t76h.jpg
The bar is 2m long and has a mass of 3kg. The rock at the end of it has a mass of 5kg. The spring is .3m from the pivot. As shown, when the spring is unstretched/uncompressed the bar makes an angle of 30° above the horizontal. It is then pushed down so it now makes an angle of 30° below the horizontal. My first question is what is the x for the spring when determining the elastic potential energy? Because I just set up triangles and figured it out to be .3m but my friend used the relationship between linear and angular distance, d=θr, and got x to be larger. But I believe that gives the arc length traveled by the end of the spring as it is compressed rather than how much it is actually compressed but I could be wrong. Here is a picture of what I think it is like: http://i44.tinypic.com/5zm6om.jpg

And so after you figure out what x is, what the problem asks for is the velocity of the rock as it passes back through the original position? (Using conservation of energy) So how would this be done and what is the answer? Because I ended up with a negative value while trying to find the answer meaning the spring did not have enough energy to get the catapult back to the starting position but this is probably wrong because that would be a dumb test question. Sorry this post is so long but please help! I've been going nuts trying to figure out what I did wrong!

The displacement would depend on where the other end of the spring is attached. If it's right bellow the point where it's attached to arm at equilibrium, your value of .3m is correct. I would go with that given no other information.

Do you happen to remember what the spring constant is? Also, is the equilibrium position equilibrium for the spring, equilibrium for unloaded catapult, or equilibrium for loaded catapult? Each of these would give you different results. In the final case, the speed will be maximum through equilibrium. But in the first two cases the kinetic energy might work out to be negative if the spring coefficient was too low.

The gravitational potential energy as function of angle looks like this.

$$U = sin(\theta)\frac{1}{2}mgL + sin(\theta)MgL$$

Kinetic energy as function of velocity.

$$T = \frac{1}{6}mv^2 + \frac{1}{2}Mv^2$$

Where L is the length of the arm, m is mass of the arm, M is mass of the boulder, and v is velocity of the boulder. The angle is measured from horizontal.

Assuming equilibrium is equilibrium for the spring, the rest is easy. Add spring potential to gravitational potential at lower point, subtract of gravitational potential at the equilibrium point, and that's your kinetic energy. Solve for v, and you're done. If some other equilibrium condition is meant, you'll have to correct for it.

Oh yea I completely forgot about the spring constant, it was 2000N/m. And the equilibrium position is equilibrium for the spring.

Also can you use the linear kinetic energy formula for this? Because a large portion of our test involved angular velocity and the problem gave us a hint reminding us of the formula for the moment of inertia of a bar about its end so I assumed we needed to use angular velocity and kenetic energy and such.

Oh I think I see that you did take angular motion into account as opposed to just linear which is why you have the kinetic energy of the bar start with 1/6 as opposed to 1/2

But unless I am doing my calculations wrong I am still getting a negative kinetic energy... Please check the calculations. This is a very dumb question if the spring really does not have enough energy to return to equilibrium

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I think x=2(.3)tan30° since Cos30°=-Cos30°

If you remembered all the numbers correctly, and the equilibrium is equilibrium for the spring, you are absolutely correct. Here is a plot of kinetic energy as function of angle. The points where the curve crosses zero are turning points. The angles are in radians. So the arm will get just a little past 0°, and then start coming down again, never reaching the equilibrium point. To get the arm to swing past equilibrium, you must pull the arm down past -60° before firing.

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## 1. What is a complicated catapult?

A complicated catapult is a type of machine that uses stored energy to launch or throw an object. It typically consists of a base or frame, a throwing arm, and a mechanism to release the arm. The design of a complicated catapult can vary, but it generally involves multiple components and moving parts, making it more complex than a simple catapult.

## 2. How does a complicated catapult work?

A complicated catapult works by using potential energy stored in the throwing arm to launch an object. The throwing arm is pulled back or loaded with tension, and when released, the potential energy is converted into kinetic energy, causing the arm to swing forward and launch the object. This is known as the principle of conservation of energy, where energy cannot be created or destroyed, only transferred from one form to another.

## 3. What is the conservation of energy test question?

The conservation of energy test question is a common physics problem that tests a student's understanding of the principle of conservation of energy. It typically involves a scenario where an object with a known mass and initial velocity is launched from a complicated catapult, and students are asked to calculate the final velocity or the height of the object at a specific point in its trajectory.

## 4. How is conservation of energy related to complicated catapults?

The conservation of energy is directly related to complicated catapults as they operate based on this principle. When a complicated catapult is loaded with potential energy, it has the potential to launch an object with a specific amount of kinetic energy. The total energy in the system remains the same, but the form of energy changes from potential to kinetic as the object is launched. This is why complicated catapults are great examples of the conservation of energy in action.

## 5. What are some real-world applications of complicated catapults?

Complicated catapults have been used throughout history for various purposes, from warfare to modern-day science experiments. Today, they are commonly used in science and engineering labs to teach students about potential and kinetic energy, as well as the principles of motion and mechanics. They are also used in some industries, such as aerospace, for testing and launching small objects or prototypes.

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