# Homework Help: Conservation of Mechanical Energy

1. Nov 3, 2007

### Luca169

[SOLVED] Conservation of Mechanical Energy

1. The problem statement, all variables and given/known data

A ride at a generic amusement park starts off by swinging like a simple pendulum until its amplitude becomes so great that it swings completely around. If the diameter of the circle is 30.0 m, what speed must the ship have at very bottom to just make it to the highest point and sit there with no residual speed.

So far, all I know is that the diameter of the circle is 30 m.

2. Relevant equations

I don't know how to use the fancy Latex thing to produce my equations, but I'll try my best without it.

Mi = Mf (Mechanical Energy, initial and final)
Mi = Ug + Ki (Mf would be equal to Ug + Kf)
Ug = m * g * h (Potential gravitational energy)
Ki = m*Vi^2 / 2 (Kf would be equal to m*Vf^2 / 2)

These are what I think, or rather have been using to solve the other problems associated with this problem in this particular section (Conservation of Mechanical Energy)

3. The attempt at a solution

I really don't know where to start, the lack of givens is confusing me. Perhaps, if I had the mass of the ship, then I could figure out the potential gravitational energy, and then find then kinetic energy equivalent, and then find the speed from there by rearranging the last equation I gave in part two. Any help with this would be greatly appreciated. This is the last problem in the section, and seems to be the most challenging, as I had no problems with the preceding problems.

2. Nov 3, 2007

### G01

You may not need the mass. Set up the conservation of energy equation for this problem, leaving mass just as a variable. You should not need the mass. Do you see why?

3. Nov 3, 2007

### Luca169

Not exactly, so I would set say x as the mass, and continue on from there?

Apparently, there is a definite answer to this problem, the answer being 24.2 m/s. :|

After plugging x into the equation, I did indeed get 24.2 m/s as my answer. Is this procedure correct?

I replicated my steps here..

Ug = mgh
Ug = x kg * 9.8 m/s * 30 m
Ug = x * 294

Mf = x * 294
Mf = m*Vf^2/2
X * 294 = x * Vf^2 / 2
2x * 588 = x * Vf^2
x * 588/x = Vf^2
Vf = √588
Vf = 24.2 m/s

Last edited: Nov 3, 2007
4. Nov 3, 2007

### G01

Yes, let mass just be a variable. Personally, I'd choose m, for mass, as the variable, but x works too.

Set up the equation for conservation of energy between the starting point and the end point, and you should see that you shouldn't need the mass. Once you do this, can you tell me why you don't need the mass?

Last edited: Nov 3, 2007
5. Nov 3, 2007

### Luca169

I think you don't need the mass because when you are determining the final speed, you will divide by the mass, which you multiplied by when finding the gravitational potential energy, so they will cancel out?

6. Nov 3, 2007

### G01

Exactly!

So, did you get the correct, answer, or is there something else confusing you?

7. Nov 3, 2007

### Luca169

No

No, I got the right answer, thanks again!