How Do You Calculate Kinetic Energy and Acceleration on a Loop-the-Loop Track?

  • Thread starter Thread starter bulldog23
  • Start date Start date
  • Tags Tags
    Block Track
Click For Summary

Homework Help Overview

The problem involves a block sliding along a loop-the-loop track, focusing on calculating kinetic energy and acceleration at specific points. The context includes concepts from mechanics, particularly energy conservation and circular motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss energy conservation principles, questioning how to relate potential and kinetic energy at different points on the track. There are attempts to derive relationships between energy and velocity, as well as considerations of centripetal acceleration.

Discussion Status

Participants are actively engaging with the problem, exploring various approaches to calculate kinetic energy and acceleration. Some have reached partial conclusions regarding energy conservation, while others are still clarifying their understanding of the relationships involved.

Contextual Notes

There is an emphasis on showing work to receive help, and participants are navigating through the implications of mechanical energy being constant throughout the motion. Some are uncertain about specific values, such as velocity at different points, which affects their calculations.

bulldog23
Messages
120
Reaction score
0

Homework Statement


A small block of mass m = 1.3 kg slides, without friction, along the loop-the-loop track shown. the block starts from the point P at rest a distance h = 52.0 m above the bottom of the loop of radius R = 18.0 m.
prob17a.gif

A)What is the kinetic energy of the mass at the point A on the loop?
B)What is the magnitude of the acceleration of the mass at the point A of the loop?
C)What is the minimum height h for which the block will reach point A on the loop without leaving the track?

Homework Equations


F=ma
KE=1/2mv^2
PE=mgh

The Attempt at a Solution


I am unsure how to go about this problem. Can someone help me out and walk me through it Please! I guess we can just start with part A and work our way from there.
 
Last edited:
Physics news on Phys.org
In order to get help on this forum, you have to show some work. What are your thoughts on this problem? You knew enough to notice that the energy relations will be important here. This is correct.

HINT:

Is anything conserved?
 
Last edited:
Well what I tried to do was set 1/2mv^2=2mgh, and solve for a velocity. Then I tried to plug that into KE=1/2mv^2, but I don't think that is right. Is the total mechanical energy conserved?
 
Last edited:
Yes, the total mechanical energy is conserved. What does this tell us about the values of the total mechanical energy at P and at A then?
 
They are both constant? right?
 
Yeah, the mechanical energy is constant and doesn't change. So, what is the mechanical energy at each point?
 
is that where you use KE_1+PE_grav?
 
Yeah, that quantity is the mechanical energy at a point. Can you find that quantity at P and A? If you can, you should be able to see how this leads you to your answer for part a).
 
How do I figure out a velocity to use in the equation?
 
  • #10
You shouldn't need it. The quantity 1/2mv^2 at A is the kinetic energy at A. You want to know what this whole quantity is. So, just solve for that quantity. Now what about the 1/2mv^2 and P? Do you know v at P?
 
Last edited:
  • #11
I am confused on point A. How would I solve for KE if I don't have a velocity? How do I solve for that quantity?
 
Last edited:
  • #12
I don't know v at P...to get that do I set KE=2PE
 
  • #13
bulldog23 said:
I am confused on point A. How would I solve for KE if I don't have a velocity? How do I solve for that quantity?

Write out the equation you have for mechanical energy at both points, but instead of writing 1/2mv^2 for the kinetic energy at A, write KE or something. This "KE" is what you want to find for part A, correct? So solve for it.

bulldog23 said:
I don't know v at P...to get that do I set KE=2PE

For the speed at P, the problem statement tells you that the block starts at rest. What does this tell you about its initial speed and thus the kinetic energy at P?
 
  • #14
Alright so I got part A. I found the PE at point P and subtracted the PE at point A. For part B, wouldn't the acceleration just be equal to g?
 
  • #15
bulldog23 said:
Alright so I got part A. I found the PE at point P and subtracted the PE at point A. For part B, wouldn't the acceleration just be equal to g?

Good:smile:

Now, the acceleration at A may be equal to g, but it may be greater depending on how fast you are moving around the loop. Now, what do you know about circular motion? The acceleration would be a centripetal acceleration, correct? What relationships can you use to find a centripetal acceleration?
 
  • #16
For part B, I just take the KE that I solved for and set it equal to 1/2mv^2 and solve for v. Then v^2/r=a, right?
 
  • #17
bulldog23 said:
For part B, I just take the KE that I solved for and set it equal to 1/2mv^2 and solve for v. Then v^2/r=a, right?

Correct. Good job!
 
  • #18
For part C I don't even know where to start. mv^2/r must be greater than or equal to mg, right?
 
  • #19
For part, C, the minimum speed at which the ball can make it over the loop is when the centripetal acceleration is equal to g. Now, using this fact, can you work backwards, using the methods from parts a) and b) to find the height at which it should be released?
 
  • #20
I am not sure how to do that. Could you please explain it more to me?
 
  • #21
bulldog23 said:
I am not sure how to do that. Could you please explain it more to me?

Well, once you know that the centripetal acceleration at A is g, can you find the speed at A? Once you find the speed at A, kinetic energy is no problem, right? Then does conservation of mechanical energy tell you anything?
 
  • #22
So when I take the velocity from the centripetal acceleration, and plug it into KE=1/2mv^2, I get 114.63 J. Is this right so far?
 
  • #23
Yes. Now what can you do from here?
 
  • #24
I am not sure what to do from here. That's where I am stuck at
 
  • #25
G01 said:
Then does conservation of mechanical energy tell you anything?

Quoting myself from before...^
 
  • #26
I understand that it is a constant, so does that mean KE+PE equals the answer from part A?
 
  • #27
We are changing the centripetal acceleration, so the mechanical energy isn't the same as it is in part A, but it is constant from one point to another. What does this tell you about the mechanical energy at point P in this part?
 
  • #28
I haven't done this before so I am really confused right now. I am not sure what it tells about the mechanical energy at point P.
 
  • #29
It is constant from one point to another, so the total mechanical energy at point A has to be equal to the total mechanical energy at point P.
 
  • #30
So then can the lowest the height at Point P only be equal to the height of Point A which is 2r?
 

Similar threads

Loop problem with skier on ramp going into a loop-the-loop
  • · Replies 13 ·
Replies
13
Views
2K
Starting height of marble rolling around a loop the loop
  • · Replies 12 ·
Replies
12
Views
13K
Minimum h for Car to Remain on Track Around Loop
  • · Replies 6 ·
Replies
6
Views
2K
Loop-the-loop with potentional and kinetic energy
  • · Replies 8 ·
Replies
8
Views
3K
Loop-the-Loop and velocity problem
  • · Replies 8 ·
Replies
8
Views
3K
Solving a Physics Problem: Find Velocity of Block at Bottom of Loop
  • · Replies 1 ·
Replies
1
Views
2K
How Do You Calculate Speed and Normal Forces in a Loop-the-Loop Problem?
  • · Replies 2 ·
Replies
2
Views
4K
How Fast Must a Block Slide to Circle a Loop After Inelastic Collision?
  • · Replies 1 ·
Replies
1
Views
9K
What is the normal force on a block at the top of a loop?
  • · Replies 3 ·
Replies
3
Views
3K
How Much Should the Spring Be Compressed to Keep the Mass on Track?
  • · Replies 13 ·
Replies
13
Views
2K