Block on loop-the-loop track

In summary, the block will reach point A on the loop without leaving the track if it has a speed greater than or equal to g.
  • #1
bulldog23
120
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.
 
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  • #2
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?
 
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  • #3
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?
 
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  • #4
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?
 
  • #5
They are both constant? right?
 
  • #6
Yeah, the mechanical energy is constant and doesn't change. So, what is the mechanical energy at each point?
 
  • #7
is that where you use KE_1+PE_grav?
 
  • #8
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).
 
  • #9
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?
 
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  • #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?
 
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  • #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 centripital 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?
 
  • #31
Can you show me how to set up the equations that I need?
 
  • #32
bulldog23 said:
Can you show me how to set up the equations that I need?

You should be able to do this if you made it this far. You already set them up in parts a and b. What is the mechanical energy at P? What is the mechanical energy at A? Then, set these two quantities equal to each other, since you know they are equal, and you should be able to solve for h. As I said, you already did this in part a and b, but there you were just solving for something else. In simplest form then, the equation you want is:

[tex]ME_P=ME_A[/tex]

Again, you should be able to find ME_P and ME_A, since you were able to do this in a) and b).
 
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  • #33
So I do 1/2mv_1^2+mgy_1=1/2mv_2^2+mgy_2? And then I am solving for y_1 right?
 
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  • #34
or do I just set mgh=204 and solve for h?
 
  • #35
bulldog23 said:
So I do 1/2mv_1^2+mgy_1=1/2mv_2^2+mgy_2? And then I am solving for y_1 right?

This is correct.
 

1. What is a block on a loop-the-loop track?

A block on a loop-the-loop track is a small object, usually made of wood or plastic, that is placed on a track that forms a loop. It is used to demonstrate the principles of potential and kinetic energy.

2. How does a block on a loop-the-loop track work?

The block on a loop-the-loop track works by converting potential energy into kinetic energy as it moves down the track. When the block reaches the bottom of the loop, it has enough kinetic energy to complete the loop and return to its starting point.

3. What factors affect the speed of a block on a loop-the-loop track?

The speed of a block on a loop-the-loop track is affected by the height of the loop, the mass of the block, and the force of gravity. The higher the loop, the faster the block will go, and the heavier the block, the slower it will go. Gravity also plays a crucial role in the speed of the block.

4. How does friction affect a block on a loop-the-loop track?

Friction can affect the speed of a block on a loop-the-loop track by slowing it down. Friction is the resistance between two surfaces, and it can cause the block to lose some of its kinetic energy as it moves along the track. This can result in the block not having enough energy to complete the loop.

5. What safety precautions should be taken when using a block on a loop-the-loop track?

When using a block on a loop-the-loop track, it is essential to wear safety goggles and to keep the track clear of any obstacles. It is also crucial to make sure the track is stable and secure to prevent it from tipping over. Additionally, it is recommended to have a designated area for the track to prevent any accidents from occurring.

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