# Block on loop-the-loop track

bulldog23

## 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. 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?

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:

Homework Helper
Gold Member
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:
bulldog23
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:
Homework Helper
Gold Member
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?

bulldog23
They are both constant? right?

Homework Helper
Gold Member
Yeah, the mechanical energy is constant and doesn't change. So, what is the mechanical energy at each point?

bulldog23
is that where you use KE_1+PE_grav?

Homework Helper
Gold Member
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).

bulldog23
How do I figure out a velocity to use in the equation?

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

Homework Helper
Gold Member
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.

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?

bulldog23
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?

Homework Helper
Gold Member
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 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?

bulldog23
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?

Homework Helper
Gold Member
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!

bulldog23
For part C I don't even know where to start. mv^2/r must be greater than or equal to mg, right?

Homework Helper
Gold Member
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?

bulldog23
I am not sure how to do that. Could you please explain it more to me?

Homework Helper
Gold Member
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?

bulldog23
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?

Homework Helper
Gold Member
Yes. Now what can you do from here?

bulldog23
I am not sure what to do from here. That's where I am stuck at

Homework Helper
Gold Member
Then does conservation of mechanical energy tell you anything?

Quoting myself from before...^

bulldog23
I understand that it is a constant, so does that mean KE+PE equals the answer from part A?

Homework Helper
Gold Member
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?

bulldog23
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.

Homework Helper
Gold Member
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.

bulldog23
So then can the lowest the height at Point P only be equal to the height of Point A which is 2r?

bulldog23
Can you show me how to set up the equations that I need?

Homework Helper
Gold Member
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:

$$ME_P=ME_A$$

Again, you should be able to find ME_P and ME_A, since you were able to do this in a) and b).

Last edited:
bulldog23
So I do 1/2mv_1^2+mgy_1=1/2mv_2^2+mgy_2? And then I am solving for y_1 right?

Last edited:
bulldog23
or do I just set mgh=204 and solve for h?

Homework Helper
Gold Member
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.