# Loop the Loop (Conservative Forces and Potential Energy)

## Homework Statement

A mass m = 71 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 16.1 m and finally a flat straight section at the same height as the center of the loop (16.1 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)

In the problem, I've already found:

What is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track? 12.561

What height above the ground must the mass begin to make it around the loop-the-loop? 40.25

I'm stuck on the next few parts which are

a. If the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop?

b. If the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (16.1 m off the ground)?

c. Now a spring with spring constant k = 15400 N/m is used on the final flat surface to stop the mass. How far does the spring compress?

d. It turns out the engineers designing the loop-the-loop didn’t really know physics – when they made the ride, the first drop was only as high as the top of the loop-the-loop. To account for the mistake, they decided to give the mass an initial velocity right at the beginning.

How fast do they need to push the mass at the beginning (now at a height equal to the top of the loop-the-loop) to get the mass around the loop-the-loop without falling off the track?

e. The work done by the normal force on the mass (during the initial fall) is: Positive, Negative, or Zero.

## Homework Equations

F=MA and F=mv^2/r

## The Attempt at a Solution

a. I think it is the same speed I found before, (12.561), but I'm not sure.

b. I thought I had to use .5mv^2=.5m1v^2+ghm, but that makes the velocity negative, so I don't think that can be right.

c. I need the velocity from part b to do this one.

d. I don't even know where to start on this one.

e. I think it's zero, because the normal force didn't really do anything, only gravity did, but I don't know if this is right.

## The Attempt at a Solution

Doc Al
Mentor
a. I think it is the same speed I found before, (12.561), but I'm not sure.
So the speed doesn't change as it falls a distance of 2R? What about energy conservation?

Okay, that makes sense, all I was thinking about was that there wasn't friction to slow it down. Thank you!

I actually spend some more time with this problem and I have all of the answers except for e.

I know it a really big conceptual thing to know, but I don't really understand it. The normal force would be perpendicular to the plane of motion as it goes down the incline, which would mean that it wouldn't do any work, right?

Doc Al
Mentor
I know it a really big conceptual thing to know, but I don't really understand it. The normal force would be perpendicular to the plane of motion as it goes down the incline, which would mean that it wouldn't do any work, right?
Right! Simple as that. (Why do you say you don't really understand it?)

Oh, okay! Thank you so much. I'm just really sketchy on my understanding of everything. Thanks again! :)