1. The problem statement, all variables and given/known data
A mass m starts at rest on a frictionless track a distance H above the floor. It slides down to the floor-level where it encounters a loop of radius R.

A) What is the minimum height H for which the object makes around the loop without losing contact with the track?

B) What is the height H for which the normal force on the block by the track at the top of the loop is equal to the weight of the block?

2. Relevant equations
[tex]F_cp =\frac{mv^2}{R}[/tex]
Conservation of mechanical energy

3. The attempt at a solution

A) I found another thread on here discussing this question (plus, most of this I could've worked out by myself), (w_ww.physicsforums.com/showthread.php?t=218700&highlight=Roller+coaster[/url]), however I need an explanation of this section:

B) This is the problem. I've been discussing it with my group and we're all at a loss. Help?

Oh, and I'm sorry if I don't know the all the correct signs for everything. I go to a fairly small school in Sweden, so I don't know what's used outside our school, really.

You get that relationship by drawing a force diagram for the object when it's upside down at the highest point of the loop. For most speeds, you have gravity and the normal force acting on the object, and you know the acceleration if [itex]v^2/r[/itex] since the object is moving in a circle.

But when the problems says the object is just about to lose contact with the track, what do we know about the forces? Once you have the diagram, what equation do you get by applying Newton's law to the diagram?

I'm starting to think that the issue is that I don't entirely understand centripetal force & normal force in this case.

Just to clarify one thing for me: (this is formulated in a really bad way, but I'm a bit too tired to be bothered with my English) In this case, is centripetal force the normal force? That's the bit of information I'm a bit unclear of. Because in that case I could probably solve it, since it'd just be weight=centripetal force, which just gets me to the height 2R.

(subscript t is at the top of the loop and subscript i is the initial height, meaning on the ramp.)

[tex]mg2R + mv^2_t = mgH_i[/tex]
Which goes to:
[tex]2Rv^2_t + mv^2_t = H_i v^2_t[/tex]

Because mass cancel out and;

[tex]mg = mv^2_t [/tex]

mass cancel out;

[tex]g = v^2_t [/tex]

If I cancel the velocity out from before, I get that the minimum height is 2R.

Oh dear, I sorry if I ask stupid questions.. our teachers thought it'd be funny to give us a project involving a force we've never studied. I don't quite understand it just yet. And I hate feeling that I don't understand things..

I think you are missing a factor of 1/2 for the kinetic energy here.

No. If a particle is going in uniform circular motion, the centripetal force is equal to the net force. (If it's speed is changing in circular motion, the centripetal force is the compoent of the net force in the direction toward the center of the circle of motion.)

So in general, when the object is at the top of the ramp, we have (using down as the postive direction):

[tex]
N + m g = m a_c
[/tex]

and we know [itex]a_c= v^2/r[/itex]. This formula would be true for any speed as long as the object is still on the track. However, for the special case of the object going slow enough that it is just losing contact with the surface, what is the value of the normal force?

First of all, yes, I am missing 1/2 for the kinetic energy. My bad. Thanks for pointing it out.

And you just explained so much! When the object just loses contact with the track normal force is obviously 0. (it's not actually supposed to lose contact with the track, but I'm guessing that it doesn't, sort of. I mean, it's not just hanging in the air.)

Anyway, that centripetal force is the net force was just what I needed to be told. It's been said in a few info pages I've read, but not as clearly. Thank you lots. B suddenly became clear. Since normal force is supposed to be equal to the weight, it's just 2mg=centripetal force. :) Thank you loads, really.

About the object losing contact: No, it's not just hanging in the air. But the question is it following a circular path?

There is a certain specific speed at the top of the loop for which the object will just lose contact with the track. (That's the speed that it has in part A).

If it moves faster than that (like it is doing in part B) the track will need to put a force on it at the top to keep it going in a circle. If it's going slower, the path may look like a kind of circle flattened somewhat on the top; it may have enough speed to keep going to get back on the other side of the track at some point, but it does not have enough speed to get there by following the circular track.

Just to clarify (my teachers are going to be asking very nasty questions, so I really need to have it clear in my mind):

For part A; It does follow the circular motion and therefore the track, but just at the top the track doesn't need to exert a force on it to keep it in the path - because it just loses contact at that really short time, which gives you the minimum height..

I do hope that it can lose contact just at that tiny point, because otherwise I'm in huge trouble. I'll have to ask them tomorrow, but I really think that this is the solution they want to the problem.

It's difficult to talk about what happens when something goes right at that speed, because that speed if more of a dividing line between two cases. It's better to talk about what happens if something is going faster or slower than that speed.

However, if it is going at that speed, and if it is losing contact, then like you say it would only lose contact at that very highest point.

The answer, in terms of R, is actually 5/2R. We just worked this similar problem during lecture.

Arad acceleration is = V^2/R. Let R be solved and swap Arad for g gravity and you have minimal velocity at the highest point of the loop [v=(R*g)^(1/2)]. This can also be written as V^2 = R*g.

Using the Work-Energy Theorem for a frictionless surface, K_1 + U_1 = K_2 + U_2, we have no initial velocity. Solving for K_2 we have U_1-U_2=K_2.

U_1 will equal mgy_1. Let y_1 = h.

U_2 will equal mgy_2. Let y_2 = 2R. We could use the diameter D for the height of the loop, but we wish to solve in terms of R.

K_2 will equal 1/2mv^2. As we mentioned above, V^2 = R*g.

Putting this together we have mg(h-2R) = 1/2m(R*g)

This problem is independent of mass, so the m's cancel as does gravity, leaving;

h = 2R +1/2R = 5/2R

Hope this helps to whomever might be reading. Gig' Em!