# Homework Help: Riding a Loop the Loop (velocity at the bottom of the loop)

1. Feb 25, 2008

### clope023

[SOLVED] Riding a Loop the Loop (velocity at the bottom of the loop)

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

A car in an amusement park ride rolls without friction around the track shown in the figure. It starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle.

part A - What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)?

found this to be 5/2R, since:

sigma Fy = Fc - N - mg = 0

N = 0 and Fc = mv^2/R

so Fy = Fc = mg and mv^2/R = mg and v^2 = GR

h = (1/2mgR + 2mgR)/mg = 1/2R + 2R = 5/2R

now I'm stuck on the velocity at the point C.

part B - If the car starts at height h = 4.50 R and the radius is R = 22.0m , compute the speed of the passengers when the car is at point C, which is at the end of a horizontal diameter.

2. Relevant equations

mgR = 1/2mv^2

v^2 = $$\sqrt{2mgR}$$

3. The attempt at a solution

got v = 29 which was wrong, I might be using the wrong formulas or the wrong radius, or the formulas in the wrong way; this is an energy conservation problem.

any help is appreciated.

Last edited: Feb 25, 2008
2. Feb 25, 2008

### clope023

another method I tried, was that:

h = 9/2R = 99m

so the conservation of energy equation would turn out to be:

mgh = 1/2mv^2

v^2 = 2mgh/m

so v = $$\sqrt{2gh}$$ = 44m/s, which also turned out to be wrong.

3. Feb 25, 2008

### PhanthomJay

Your h=2.5R looks good for part A. For part B, you are neglecting the fact that the coaster has speed at the top of the loop in your first attempt,; in your second attempt, you neglected it has PE at point C, if you used h=4.5R.

Last edited: Feb 25, 2008
4. Feb 26, 2008

### clope023

bump to this, I figured out how to solve the velocity, got 38.8m/s and found the radial acceleration to be 68.6m/s^2.

now I want to find the tangential acceleration, I don't know angular velocity formulas at this point, so not sure where to take the problem to find the tangential acceleration, the only thing I know would be that it would be pointing outside the loop, but not sure how to write that mathematically.

5. Feb 26, 2008

### clope023

forget it, lol, I solved it,

atan = v^2/R = gR/R = g

6. Oct 10, 2011

### Hollumber

Re: [SOLVED] Riding a Loop the Loop (velocity at the bottom of the loop)

erm...I just wanted to say that the reply above mine solving for the tangential acceleration is not proper.... v^2/r is for *centripetal* acceleration, or radial acceleration (pointing towards the center of the circle). The tangential acceleration is "g" because the vector is pointed straight down and the only force to accelerate the object of our interest is gravity. To find the tangential accel. in a situation like this, you must multiply the acceleration of gravity (g) by the cos(angle respect to vertical).

7. Mar 12, 2013

### zeralda21

I hope it is ok to bring up an old thread because I don't understand these calculations. I cannot understand what the force F_{c} represents and why it's direction is in the positive y-direction.

I do see that the loop exerts a normal force which is downward but why is this force equal to 0?

8. Mar 12, 2013

### zeralda21

Should not the force F_{c} be directed along the trajectory since it is just the mass(scalar)*acceleration(vector) and acceleration has two components; One to the center of the orbit and one along the trajectory.

9. Mar 12, 2013

### Staff: Mentor

That post is a bit confused. Fc stands for the centripetal force, which acts downward (at the top of the loop).

The analysis should be:
ƩFy = Fc = -N -mg
When finding the minimum speed to maintain contact, you want the normal force to just equal zero.

10. Mar 12, 2013

### zeralda21

I understand now why N has to go to zero. Good.But is Fc an actual force? Shouldn't it therefore be included in the analysis as ƩFy=-Fc-N-mg? Or is it just a representation of the forces acting?

It is something with the signs..If we proceed with ƩFy=Fc=-N-mg=-mg and use that Fc=mv^2/R we obtain that v^2=-gR which is wrong since (g,R)>0.

11. Mar 12, 2013

### Staff: Mentor

Fc is not a separate force and should not be included in ƩFy. Centripetal force is just the name we give to the net force that produces the centripetal acceleration. I advise not even using the term centripetal force and just stick to centripetal acceleration.

You just need to be consistent with signs. v^2/R is just the magnitude of the centripetal acceleration; The direction is downward so the sign will be negative:
ƩFy = ma
-N-mg = -mv^2/R

12. Mar 12, 2013

### zeralda21

Well put, I understand. One last concern: In the following step:

h=(1/2mgR + 2mgR)/mg is clearly equivalent to mgh=(1/2mgR + 2mgR) which means that the potential energy at beginning (mgh) is equal to mg2R(top of the circle)+mgR/2 (the difference from in energy from initial position and on top of the circle).

How do we know that difference is mgR/2?

13. Mar 12, 2013

### Staff: Mentor

What you know is that the initial PE must equal the PE (2mgR) + KE at the top.

And since you just determined that mv^2/R = mg at the top, you know that the KE = 1/2mv^2 = mgR/2.

14. Mar 12, 2013

### zeralda21

I see. Thank you DocAl. You have been of great help.