# Homework Help: Roller Coaster Height Problem

1. Nov 3, 2005

### Effitol840

I’m having trouble with my physics lab this week. Here is the problem which we are trying to solve:
What is the minimum height at which you can release a ball in order for the ball to just make it around a loop?
The ideal situation for this would be to drop it from a height at which the normal force at the top of the loop would equal 0. I’ve found an equation in which to figure out the height that would make the normal force 0. This is how I got that equation…
Since PE=KE I used the formula $$mgh=1/2v^2$$ where m is the mass, g is gravity, h is the height at which to drop it from and v is the velocity at the top of the loop. I figure that v is not important so what I did was I took the centripetal acceleration formula which states that $$a=v^2/r$$ and somehow decided that $$a=v^2/r$$ is the same thing as $$g=v^2/r$$ solved that for $$v^2$$ and came up with $$v^2=gr$$. This made my equation $$mgh=1/2gr$$ and then I solved that for my h to get the height which would be $$h=(1/2r)/m$$.
Now I don’t know if this is at all right or not that is why I am posting this. I’m also having another problem. Since that is when n=0 then it’s the absolute extreme case for which the ball to travel around the loop neglecting friction. Now I need to devise a formula or a way to come up with a more reasonable height in which to drop it from. The trick is that I have to state it physically how I came up with that height. If anyone could help me with this too that would be great. Thanks.

2. Nov 3, 2005

### kp

dont you also have potential energy at the top of the loop?

3. Nov 3, 2005

### Effitol840

The potential energy at the top of the loop is mgh where h in this is the height of the loop. It doesnt matter what the potential energy is at the top of the loop i just need to figure out how to find an equation to show where to pick a point along the ramp so that the makes it fully through the loop but goes the shortest distance possible.

4. Nov 3, 2005

### Staff: Mentor

You've almost got it, but you made an error in your first step. That should be $mgh=1/2mv^2$; if you would have checked the dimensions (units) of your answer, you would have caught this.

Applying Newton's 2nd law at the top of the loop:
$$mg = m v^2/r$$

Thus the KE at the top of the loop must be:
$$1/2 m v^2 = mgr/2$$

Set that equal to PE to find the minimum height above the top of the loop that the ball must be released.

5. Nov 3, 2005

### Effitol840

Ok I get that. That gives me the minimum distance that i have to drop the marble in order for it to fully travel around the loop. The next thing i need to figure out is what point i can drop it factoring in friction and other things. I dont want to use another formula with friction in it i just need to figure out a reason as to why i picked a point a little more than the minimum and back it up using physics. This is where i get stuck. i dont actually even know where to start.

6. Nov 3, 2005

### Staff: Mentor

Use an energy argument. Dropping the ball from the given height gives it exactly enough energy to stay in contact with the loop assuming no energy is lost to friction. If energy is lost to friction, the speed will be insufficient to keep the ball on the track. Thus, to compensate for that energy loss, start it at a higher point.

7. Nov 3, 2005

### Effitol840

that much i understand, the thing is my lab ta is an ass and he grades hard as hell so i need some sort of equations dealing with variables no numbers. its hard to explain what i need to figure out. maybe saying that adding a constant to the height at which you drop or something. i dont know if im asking too much or what but if you could help that would be great.

8. Nov 3, 2005

### kp

hmm...I thought you where trying to define "h" (start height) in terms of height of the loop.

9. Nov 4, 2005

### Effitol840

I am but I know how to get it. The only problem is that it is the minimum height at which to drop the marble in relation to the height of the loop in order for it to travel fully through the loop. That is neglecting friction. If I take the minimum height and try and drop it its not going to make it through, thats why im asking about friction and a way to physically explain choosing a higher point so that it just barely makes it through the loop.

10. Nov 4, 2005

### kp

Couldn't you define the height of the loop as "2r"
and its potential energy as mg2r + its K.E. which you have already defined as 1/2mgr.
all without friction.

I think i'm just taking a different path to the same place.