# Roller Coaster Problem

1. May 27, 2004

### e(ho0n3

Hi everyone,

Suppose I have a track that starts at a height h, goes down (not a vertical fall, but gradually in a slope), enters a circular loop of radius r (the loop is 'laying' on the ground so the top of the loop is at a height 2r from the ground) and then continues at ground level.
• Calculate the minimum height h so that the mass remains on the track at all times (assume no friction).
• Let 2h be the release height. What is the normal force on m at the bottom of the loop (before entering it), at the top of the loop, and when the block exits the loop onto the flat section.
• Show that the difference in the apparent weight of the mass at the top of the loop and the bottom of the loop is 6mg, and that this answer does not depend on the size of the loop or the speed of the mass as long as the speed is above the minimum required to go through the loop.

Quite a quesiton huh? Anyways, I need confirmation for my answers to a and b, and I need some help with c.
• As I figure it, as long as the mass can make it through 3/4 of the loop, it should make it through the entire loop (i.e. if the velocity of the mass is zero when it goes through 3/4 of the loop, then gravity should help push it through the entire loop at that point). Using energy methods, I can equate the total mechanical energy at the top of the track (at a height h) with the total mechanical energy when it's gone through 3/4 of the loop. So,
$$mgh = mgr \Rightarrow h = r$$
• For the normal force N1 on m before entering the loop:
$$mv^2/r = N_1 - mg \Rightarrow N_1 = mv^2/r + mg$$​
$$2mgh = mv^2/2 \Rightarrow mv^2 = 4mgh$$​
$$N_1 = mv^2/r + mg = 4mgh/r + mg$$​
For the normal force N2 on m at the top of the loop:
$$mv^2/r = N_2 + mg \Rightarrow N_2 = mv^2/r - mg$$​
$$2mgh = mv^2/2 + 2mgr \Rightarrow mv^2 = 4mg(h - r)$$​
$$N_2 = mv^2/r + mg = 4mg(h - r)/r + mg$$​
For the normal force N3 when the mass exists the loop:
$$0 = N_3 - mg \Rightarrow N_3 = mg$$​
• Suppose I have a spring scale used for measuring the weight of the mass when it's at the top of the loop. Let w be the force that the spring measures (i.e. the force in which the spring pulls up on the mass). Applying Newton's 2nd law yields:
$$mv^2/r = N_2 + mg - w \Rightarrow w = N_2 + mg - mv^2/r$$​
$$mv^2 = 4mg(h - r) \Rightarrow w = N_2 + mg - 4mg(h - r)/r$$​
This is as far as I can go. Maybe this approach is wrong. Any tips?

e(ho0n3

2. May 27, 2004

### Staff: Mentor

question a

This makes no sense. You are saying that if it stops at 3/4 of the loop, then it will keep going through the entire loop? If it stops, well, that's it.
I have no idea what you are calculating here, but do a reality check on your answer. Setting h = r (half the height of the loop) cannot possibly get the mass to 3/4 of the loop, never mind all the way.

The way to think of this problem is to find what minimum speed is required at the top of the loop to barely maintain contact. That minimum speed will just let the normal force go to zero. Then calculate how high above the top of the loop (height = 2r) h must be to give the mass that speed.

Last edited: May 27, 2004
3. May 28, 2004

### e(ho0n3

I was thinking that maybe the velocity would be enough to carry the mass 3/4 through the loop, reach 0 and then pick up again as it finishes the loop. So much for my physical intuition. Sigh...

Is the velocity of the mass as it goes through the loop constant? Doesn't it slow down as it goes through parts of the loop. What you're saying gives me the impression that the velocity of the mass slows down as it goes through the first half of the loop, the picks up again after reaching the top. Hence I would need to find the minimum speed to get the mass to the top of the loop to find the minimum h.

e(ho0n3

4. May 28, 2004

### Staff: Mentor

Since on the top half of the loop, all forces act down... How could it pick up speed?
Of course not!
Absolutely. Mechanical energy is conserved.
I don't know why you think the mass is somehow able to pick up speed as it rises. As the mass rises, it gains PE and loses KE.
There is a minimum speed necessary to keep the mass in contact with the loop.

5. May 28, 2004

### e(ho0n3

Maybe you misunderstood me. But anyways, I understand now. I'm still stumped with problem c though.

e(ho0n3

6. May 29, 2004

### Staff: Mentor

Perhaps I misunderstood you, but for problem a your answer (h = r) and reasoning, as far as I could see, was incorrect.
You need to redo it. (And then use that answer to redo problem b.)

For problem c realize that the normal force is the apparent weight.

7. May 29, 2004

### e(ho0n3

For problem a I get h = 5/2r. Hmm...I thought problems a and b are unrelated. How is the answer to part a going to help me. Do you mean I need to set 2h = 5r?
I see. I don't know how one would realize this. I guess you've helped enough with these kinds of problems that you know what to look for. Maybe the concept of an 'apparent weight' is escaping me.

Thanks,
e(ho0n3

8. May 29, 2004

### Staff: Mentor

Yes, that's how I read the problem.
"Apparent" weight is distinguished from "real" weight. Real weight is the pull of the earth, w = mg; apparent weight is what a scale would read if you were standing on one. The scale reads the normal force exerted between you and the scale. In everyday (unaccelerated) life, your apparent weight happens to equal mg. But jump out the window and your apparent weight goes to zero (ignoring air resistance): that's what being "weightless" means.

Of course, it would be nice if the professor explained what the terms meant before assigning the problems...

9. May 29, 2004

### Chen

I get the feeling you don't quite understand why this is not true, so I'll explain it... if you do understand, ignore this and forgive me.

After three quarters of the loop the mass would be R high above the ground, and you're saying that if it reaches that point with no kinetic energy (i.e it momentarily stops), gravity will be able to keep it going through the rest of the loop. That's correct, if you take the mass by hand and lay it on the three quarters position and leave it, gravity will kick in and the mass will complete the rest of the loop.

However, you forgot two crucial points. The first is that before reaching the 3/4 point, the mass must also pay a visit to the 1/4 point. Since the height of those two points above the ground is the same (R), the kinetic energy will be the same at both locations. And if you assume that the kinetic energy at the 3/4 point is zero, then it should also be zero at the 1/4 point. But if that is the case, the mass stopped at the 1/4 point without going any further. Not good!

The second point is that the mass will also be passing through higher points in its journey, such as the 1/2 point which is 2R high above the ground or R above the 3/4 point. But as we all know the potential energy decreases whenever you go lower, and as a result the kinetic energy increases. This implies that the kinetic energy at the 3/4 point must be bigger than the kinetic energy at the 1/2 point, because of the height difference. But if you want the kinetic energy to be zero at the 3/4 point, the kinetic energy at the 1/2 point must be negative. Not good!

10. May 29, 2004

### Staff: Mentor

1/4? 3/4?... D'oh!

So that's what you mean by 3/4 of the loop? No wonder I thought e(ho0n3 had slipped a gear there: I assumed that 3/4 of the loop meant that you were 3/4 of the way to the top of the loop. D'oh! Now I see what he was saying... (I still disagree, though).

Thanks for jumping in, Chen!

11. Jun 13, 2004

### e(ho0n3

A somewhat new problem

Consider my original problem, but let the mass m be a marble with radius R that rolls down (without slipping) instead of sliding. What is the minimum value of h if the marble is to reach the highest point of the loop without leaving the track?

Using energy methods,
$$mg(h+R)$$​
is the initial energy of the system and
$$mg(r-R) + mv^2/2 + I\omega^2/2$$​
is the energy at the highest point on the loop.
I can solve for v by applying Newton's second law when the marble is at the highest point in the loop. I is given and $\omega$ can be derived from v. Solving for h gave me
$$h = \frac{17}{10}r - 2R$$​
The book where I got this problem insists that
$$h = \frac{27}{10}r - \frac{17}{10}R$$​
What gives!?

12. Jun 13, 2004

### Staff: Mentor

The book assumes that h is the initial height of the marble's center (apparently)
The height of the marble's center at that point is "2r - R", not "r - R".
You calculated the centripetal force wrong: it should be mg = mv^2/(r-R), not mg = mv^2/r. (The circle traced by the marble's center has a radius of r - R.)
Assuming the initial height of the marble's center is h (not h + R), the book is correct.

13. Jun 14, 2004

### e(ho0n3

It seems that way doesn't it.
AH! Your right! What was I thinking!?
Apparently I wasn't all there when I did this problem.

Thanks for clearing it up.