# Vertical Loop the Loop Problem

• Lilsilver
In summary, the problem is about finding the maximum radius of a vertical loop that a car with an initial speed of 4 m/s can still remain on the track. The key to solving this problem is understanding the conditions at the start point and at the top of the loop. By analyzing the forces and energies involved, it is determined that the maximum radius is 0.3265 m.

#### Lilsilver

Loop the Loop problem help.

The Problem says that there is a little car going at an initial speed of 4 m/s while entering a vertical loop the loop. Then all it wants is the largest possible radius of the loop that the car can still remain on the track at all times.
I don't understand, the speed is constantly changing and all you have is the initial. You can't find the centripital force at the top of the loop because the speed has changed from the 4m/s to get to that point. Anyone please help.!

Loop the Loop Problem

## Homework Statement

If a little car is traveling at an initial speed of 4m/s entering a vertical loop the loop, what is the largest possible radius the loop can have which the car remains in contact with the track at all times?

## Homework Equations

The only thing i can think of is (Centripital Force)=(mv^2)/r

## The Attempt at a Solution

Well all i have are 4 points of the loop the starting(Cf=(mv^2)/r=NormalForce-mg), 1/4 the loop(Cf=(mv^2)/r=Fn), At the top of the loop(Cf=(mv^2)/r= Normal Force+mg), And 3/4 the loop(Cf=(mv^2)/r=Fn).

## Homework Statement

If a little car is traveling at an initial speed of 4m/s entering a vertical loop the loop, what is the largest possible radius the loop can have which the car remains in contact with the track at all times?

## Homework Equations

The only thing i can think of is (Centripital Force)=(mv^2)/r

## The Attempt at a Solution

Well all i have are 4 points of the loop the starting(Cf=(mv^2)/r=NormalForce-mg), 1/4 the loop(Cf=(mv^2)/r=Fn), At the top of the loop(Cf=(mv^2)/r= Normal Force+mg), And 3/4 the loop(Cf=(mv^2)/r=Fn).

If the car can make it to the top of the loop without leaving the track, it can make it the rest of the way. You can prove that if you must be comparing the parabolic path of a projectile that reaches the top of the loop with the circular path of the car when the projectile and the car both have the same speed at the top. What is the smallest possible normal force you can have at the top if the car does in fact stay in contact with the track?

Well the only normal force that it could have at the top is the weight. But if it was only the weight at that point, there is no actual normal force so for that slight moment it is not in contact with the track.

I am extremely confused by this problem...

Lilsilver said:
Well the only normal force that it could have at the top is the weight. But if it was only the weight at that point, there is no actual normal force so for that slight moment it is not in contact with the track.

The weight is not the normal force. At the top of the track, the normal force is any force the track exerts on the car in addition to its weight. The least possible normal force is zero. That does not mean the car loses contact with the track, even for an instant. It just means that at that one moment the only force required to keep the car moving in the circular path is the weight of the car. In the next instant, more force will be required because the car will start speeding up and because gravity is not in the right direction to provide all of the centripetal force.

Oh ok. But since there is no mass and no given Velocity at that point, how can you find anything?

Can't do forces? How 'bout energy?

A common way of solving these problems is:
first, just assume the object has a mass m!
second, analyze what happens when the car falls off from the track.

If a car loses contact with the track, what would the normal force be? (yes, you guess it, the normal force would be less than zero!)

Third, now you got the condition, so what other variables are related to this condition? what is the normal force in terms of other informations (energy, momentum, height, net force...etc)?

Admin note: I have merged all the threads into one. Please do not post the same question multiple times.

- Warren

I think the key to this problem is understanding the conditions at the start point and at the very top of the loop. You start with kinetic energy. At the top of the loop you still have kinetic energy (the car doesn't just drop straight down once it reaches the top), maximum potential energy, and your gravity acceleration cannot exceed your centripetal acceleration (or the car will break contact).

start = final
(0.5)(m)(v1)^2 = (m)(g)(2)(r) + (0.5)(m)(v2)^2
m factors out, v1 is start velocity, g is gravity, 2r is your height (in terms of r because r is the final answer), v2 is velocity at the top (which we can calculate)
sub a few things into simplify it
(0.5)(4)^2 = (9.8)(2)(r) + (0.5)(v2)^2
finding (v2)^2 is done by comparing centripetal acceleration to gravity
(v2)^2/(r) = 9.8
(v2)^2 = 9.8r
Put that back in the original equation
(0.5)(4)^2 = (9.8)(2)(r) + (0.5)(9.8r)
Now just tidy it up
8 = 19.6r + 4.9r
8 = 24.5r

r = 0.3265m

I don't think you can just say it's when centripetal force equals the gravity force without taking into account the energies involved. Doing that does not take into account the way gravity slows the car on the way up the loop. Just take a quick run through the numbers.
(v^2)/r = 9.8
(v^2)/9.8 = r
(4^2)/9.8 = r
r = 1.6325m
This means the height would be 3.26m. Is it physically possible for the car to even go that high, let alone keep contact?
(0.5)(m)(v^2) = mgh
(0.5)(4^2) = (9.8)(3.26)
8 = 31.948 :yuck:
Nope, not possible

no, they would be in contact. think of it this way; if at the top of the track, the force of gravity is equal to the force of centrip. acceleration then there is no net force. this means there would be no movement in the y direction, which means the car would have to remain in it's original position; in contact with the track.

ShawnD said:
...r = 0.3265m...

hmmm, i got a different answer. it's close, but different.

6Stang7 said:
hmmm, i got a different answer. it's close, but different.

nope, 0.306m.
i have a feeling i made a mistake...

yup, it was me...little calculator error.

Thanks Lots guys! You were very helpful!

## What is a Vertical Loop the Loop Problem?

A Vertical Loop the Loop Problem is a physics problem that involves calculating the minimum speed required for a roller coaster car to complete a vertical loop without falling off the track. It is often used to demonstrate concepts such as centripetal force and conservation of energy.

## What are the key factors that affect the solution to a Vertical Loop the Loop Problem?

The key factors that affect the solution to a Vertical Loop the Loop Problem are the radius of the loop, the acceleration due to gravity, and the mass of the roller coaster car. These factors determine the amount of force needed to keep the car on the track.

## How is the minimum speed for a Vertical Loop the Loop Problem calculated?

The minimum speed for a Vertical Loop the Loop Problem can be calculated using the equation v = √(rg), where v is the minimum speed, r is the radius of the loop, and g is the acceleration due to gravity. This equation takes into account the forces acting on the roller coaster car as it goes through the loop.

## What is the role of centripetal force in a Vertical Loop the Loop Problem?

Centripetal force, which is the force that keeps an object moving in a circular path, is crucial in a Vertical Loop the Loop Problem. It is what allows the roller coaster car to stay on the track as it goes through the loop, and its magnitude must be equal to or greater than the gravitational force pulling the car down.

## How does the shape of the loop affect the solution to a Vertical Loop the Loop Problem?

The shape of the loop, specifically the radius, affects the solution to a Vertical Loop the Loop Problem because it determines the amount of centripetal force needed to keep the car on the track. A smaller radius requires a greater centripetal force, meaning the car needs to travel at a higher speed to successfully complete the loop.