How to Use Conservation of Energy to Solve a Loop-the-Loop Problem?

In summary, the conversation is discussing a problem involving the conservation of energy in a roller coaster loop. The participants are trying to find the amount of kinetic energy needed at the top of the loop to successfully complete the loop. The solution involves setting the potential energy at the top of the loop equal to the kinetic energy, and also taking into account the centripetal acceleration at the top of the loop. The correct answer is 2.5R.
  • #1
Goofball Randy
21
0

Homework Statement



vlQD6eH.png

Homework Equations



Conservation of Energy (Potential + Kinetic = Potential + Kinetic)

The Attempt at a Solution



At the start of the ramp, potential energy is mgh (gravitational potential) and kinetic is 0, since it's not moving.
At the bottom of the loop, potential energy is 0 (I consider the bottom to be the zero of gravitational potential to make things easy), and kinetic is 0.5 mv^2.

The ms cancel to leave me with 2gh = v^2, but that's as far as I got...
 
Physics news on Phys.org
  • #2
Well keep in mind that it will only have to have enough kinetic energy to get to the top of the roller coaster and also since we need to neglect friction, then at the top of the loop, we can have KE equal to zero, since it will only need to have very minimal Kinetic energy to get over the top of the loop (since the top of the loop is actually a very very small straight line, we know this if we examine the tangent line at the top of the circle but it isn't essential for the problem to be honest).

So take your equation for PE, and set it equal to KE, and make sure you have your h correct. (think about what R means in this case)
 
Last edited:
  • #3
RaulTheUCSCSlug said:
Well keep in mind that it will only have to have enough kinetic energy to get to the top of the roller coaster and also since we need to neglect friction, then at the top of the loop, we can have KE and PE equal to zero, since it will only need to have very minimal Kinetic energy.

So take your equation for PE, and set it equal to KE, and make sure you have your h correct. (think about what R means in this case)

Err, KE and PE are BOTH zero at the top of the loop? How does that make sense when considering the conservation of energy? (mgh + 0 = 0 + 0)

Also, I'm not sure I get what you're saying...set PE equal to KE, isn't that what I did above?
 
Last edited:
  • #4
Goofball Randy said:
Err, KE and PE are BOTH zero at the top of the loop? How does that make sense when considering the conservation of energy? (mgh + 0 = 0 + 0)
Woops sorry I meant that we only need just enough kinetic energy to get you through the top of the roller coaster so kinetic energy basically is zero at the top, therefore it is maximum potential energy. So PE is KE. Got ahead of myself ha
 
  • #5
RaulTheUCSCSlug said:
Woops sorry I meant that we only need just enough kinetic energy to get you through the top of the roller coaster so kinetic energy basically is zero at the top, therefore it is maximum potential energy. So PE is KE. Got ahead of myself ha

I'm still confused :(
How does finding the energy at the top help me to solve the problem? If KE is basically zero at the top of the loop, that would imply that it's all potential energy...are you saying that PE at the top of the loop is the same as the PE at the very start? Because that would imply that the answer is 2R...
Or am I missing another type of PE?
 
  • #6
Goofball Randy said:
I'm still confused :(
How does finding the energy at the top help me to solve the problem? If KE is basically zero at the top of the loop, that would imply that it's all potential energy...are you saying that PE at the top of the loop is the same as the PE at the very start? Because that would imply that the answer is 2R...
Or am I missing another type of PE?

You have the right answer! The answer is infact 2R! Do you get why though? The PE has to be at the very least the same PE from the beginning so that it may go through the loop.
 
  • #7
RaulTheUCSCSlug said:
You have the right answer! The answer is infact 2R! Do you get why though? The PE has to be at the very least the same PE from the beginning so that it may go through the loop.

But the answer key says it's "D", or 2.5R. I don't understand why :(
 
  • #8
Oh sorry then I don't know how to do it either : /
 
  • #9
RaulTheUCSCSlug said:
Oh sorry then I don't know how to do it either : /

Thank you for the help anyway :) I figured out the correct answer.
 
  • #10
Oh I forgot about the centripetal acceleration! The cars can not have 0 KE at the top or else it would just fall straight down! So you want KE energy to be equal to the centripetal acceleration, which is V^2/R, so you want to solve for V, and that will tell you how much KE you need. See where you can go from there.
 
  • #11
RaulTheUCSCSlug said:
you want KE energy to be equal to the centripetal acceleration, which is V^2/R,
An energy cannot equal an acceleration. Consider the force balance at the top of the loop (##\Sigma F_{y} = ma_{y}##).
 
  • Like
Likes RaulTheUCSCSlug

Related to How to Use Conservation of Energy to Solve a Loop-the-Loop Problem?

What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant.

Why is the conservation of energy important?

The conservation of energy is important because it helps us understand and predict the behavior of physical systems. It also allows us to make efficient use of energy and reduce waste, which is crucial for sustainable development.

How does the conservation of energy apply to everyday life?

The conservation of energy applies to everyday life in many ways. For example, when we turn on a light switch, electrical energy is converted into light and heat energy. When we ride a bike, our muscles convert chemical energy into kinetic energy. The law of conservation of energy governs all such energy transformations.

Can energy ever be completely conserved?

According to the first law of thermodynamics, energy cannot be created or destroyed. However, in real-world systems, some energy is always lost as heat or other forms of energy. This is known as energy dissipation, and it means that energy cannot be completely conserved.

What are some practical applications of the conservation of energy?

The conservation of energy has many practical applications in various fields, including engineering, physics, and environmental science. Some examples include the design of efficient engines, the development of renewable energy sources, and the calculation of the energy efficiency of buildings. The law of conservation of energy is also used in designing energy-saving devices and systems.

Similar threads

Conservation of energy problem: Ball rolling down inclined plane and then through a loop-the-loop
    • Introductory Physics Homework Help
Replies
10
Views
582
Is potential energy defined only for internal conservative forces?
    • Introductory Physics Homework Help
Replies
15
Views
439
Loop problem with skier on ramp going into a loop-the-loop
    • Introductory Physics Homework Help
Replies
13
Views
1K
Challenging problem about an impact with a smooth frictionless surface
    • Introductory Physics Homework Help
2
Replies
55
Views
2K
Problem with when to use Force and Energy. What compresses spring/
    • Introductory Physics Homework Help
Replies
3
Views
515
Conservation of Energy with Gravitational Potential Energy
    • Introductory Physics Homework Help
Replies
6
Views
1K
Confused about a conservation of energy problem
    • Introductory Physics Homework Help
Replies
4
Views
559
Energy Conservation of a Vertical Spring
    • Introductory Physics Homework Help
Replies
5
Views
1K
Using energy considerations to analyse particle motion
    • Introductory Physics Homework Help
Replies
3
Views
314
Loop the Loop Problem Solution | N > 0, Energy Conservation Method
    • Introductory Physics Homework Help
Replies
8
Views
2K

Hot Threads

Recent Insights

Back
Top