# 5th Gear Loop the Loop

Maybe this is a little old (in internet age), but it is a great example. Here is the Loop-the-loop stunt from the show Fifth Gear.

I like this. First, it is a bold stunt. But also, there is some good physics here. Though, most importantly, the Fifth Gear producers were kind enough to include a shot that was very compatible with video analysis.

I went to the official site of this stunt – http://looptheloop.dunlop.eu. From here I found some useful info:

• Loop is 40 feet tall
• The car is a Toyota Aygo
• Some physics-y guy calculated that the car must go 36 mph to do the loop (I think that is calculated as the speed at the bottom).
• The wheel base of the car is 2.34 meters – (needed for scaling of the video)

Let me get something off my mind that was bothering me. If you watch the videos on looptheloop.dunlop.eu there is this physics guy explaining how it will work (well calculating the speed needed). A couple of times he said “oh, there is a formula for that” – as in there is a formula for a car going around a track or something. Maybe it is not a big deal, but he is promoting the idea that physics is a whole bunch of formulas. Really, there are only a few that can be applied in lots of cool ways. Ok, I feel better now.

Now for some graphs. What is better in an analysis than graphs? A free body diagram is cool, but not as good as a graph. First graph is the trajectory of the car. Just because.

Where am I trying to go? Well, I think the important questions are:

• What is the acceleration at the top of the circle?
• How fast is the car going?
• Does the car slow down, or does it maintain a constant-ish speed?

To look at the acceleration, I will plot the x- and y-components of velocity as a function of time. To determine the y-velocity as a function of time, think of a series of y positions. Let me call them y1, y2, y3 etc. Each of these y’s have the same time difference between them. In general, to calculate the y-velocity, I can say:

This would work. But, it would say the velocity at time 2 would only depend on what happens between time 1 and 2. That isn’t really fair, is it? So, Tracker Video uses the following formula:

And here is a plot of the y-velocity as a function of time:

I fit a linear function to the highlighted region as a means of obtaining the y-acceleration. Since this data looked linear (and that interval covers the point where the car is at the highest point), a function like this is a good way to get the acceleration. The other method would be similar to the way the velocity was found, but it would be messy – like this:

So, the slope of a y-velocity plot will be the y-acceleration. For this interval, that is -18.7 m/s2. What about the x-velocity and acceleration? I will come back to the y-acceleration at the top. Here is a plot of the x-velocity:

Again, I fit a linear function to a set of the data. This interval covers the time that the car was at the top of the circle (around 1.2 seconds). The acceleration during this time is about 0.9 m/s2. If you look at the video frame by frame, you can tell the car is more difficult to see (because part of the track is in the way). This is probably why that data is not as ‘smooth’.

Here is a plot of the speed of the car as a function of time. By speed, I mean the magnitude of the velocity.

So, it appears the car does slow down as it goes around the loop.

Now for the physics. Really there are two important physics ideas here. The work-energy principle and acceleration due to circular motion. First, work-energy says that:

Here is a much more detailed look at work-energy. For this case, I will take the car plus Earth as the system. This means that the energy is a combination of kinetic energy and gravitational potential energy. The work on the car will be from the road pushing in the same direction as the car. The normal force from the track will not do any work on the car since it (the force) is perpendicular to the displacement. So, let me assume that the car is not “driving” so that the work done on the car is zero. If this is the case, then the total energy at the bottom and the top of the track is the same. I will call the energy at the bottom E1 and the energy at the top E2. Let me also say that there is zero gravitational potential energy at the bottom of the track.

Now, solving for the velocity at the top of the track:

Now, what about the motion at the top of the track? Let me start with a free body diagram for the car at the top.

Now, I can use Newton’s second law along with the acceleration of an object moving in a circle. Newton’s second law says that:

And if the car is moving in a circle, then its acceleration is (just due to circular motion)

Here, the acceleration is towards the center of the circle. In this case that would be in the negative y-direction. Let me put stuff together now. The radius of the circle is h/2 and the velocity at the top is v2. This means that the acceleration at the top (in terms of the starting speed at the bottom) would be:

Now to calculate the force the track exerts on the car. At that instant, in the y-direction, Newton’s second law says:

Hopefully, it is clear that I am calling FN the force the track exerts on the car. Let me solve for that:

There is only one important point from this equation. What if v12 is less than 5gh? That would make the force the track exerts on the car in the opposite direction that I assumed. Thus the track would have to pull up on the car. This particular kind of car and track can’t do that. That means the car would fall if the initial speed was smaller than the square root of 5gh. For this case, I would even go faster than that.

Update: A big thanks to reader Carlos (see comments below) for spotting my error. I had replaced r with 2h when in fact,r = h/2. I changed the equations that had the incorrect value for r in them. Maybe I could say I made the mistake on purpose to see if you were paying attention.

1. #1 Cleon Teunissen
September 2, 2009

I notice the track in the video is a perfect circle. I suppose that was the best solution with a limited budget.

In rollercoaster design the track of a loop-the-loop section is rarely circular (if ever). Instead the curvature of the track varies smoothly, with the strongest curvature at the top of the loop. Obviously that makes for a smoother ride.

Hm, that sounds paradoxical, aren’t rollercoaster rides supposed to be scary and rough? As I understand it, rollercoaster designers do want the ride to have places with high G’s and some places with negative G. But I suppose that sudden changes in G-load (jerks) are the wrong kind of rough.

Cleon Teunissen

2. #2 Rhett
September 2, 2009

@Cleon,

That would make a great homework question: what shape would the curve need to be to make a constant force from the track?

3. #3 Joseph Smidt
September 3, 2009

Another great post.

4. #4 Carlos
September 6, 2009

Very nice post. I wish I had met this blog before; there is a lot of interesting stuff here.

I have a question about Tracker, if you don’t mind. Is it possible to analyze a video directly from YouTube, without first downloading it? I have never used Tracker, but will definitely give it a try if it has this feature.

A comment on the post: you seem to have taken r = 2h into the calculation, instead of r = h/2. This changes the minimum speed necessary to complete the loop.

5. #5 Rhett
September 9, 2009

@Carlos,

It looks like you are correct about the mistake on the 2 – I will fix that.

I don’t think there is a way to analyze straight from youtube. There are some other video analysis programs, and maybe one of them could do it. I played around with some that were listed in the last issue of The Physics Teacher journal, but they all pretty much were inferior to tracker and not really that useful.

Sorry for the long delay in the reply.

6. #6 Emma
November 17, 2009

on the free body diagram, would there be another force vector pointing to the left, which would be the centripetal force?

7. #7 Rhett
November 18, 2009

@Emma,

No. The centripetal force is the force that makes something move in a circle. In this case, the centripetal force is the force from the track and the weight together. Both of these pull the car towards the center of the circle and thus both together are the centripetal force.

8. #8 Jack
December 7, 2009

Hi, just wondering what adjustments would have to be made if work is done in the transverse direction? ie if we take the track to be rough and there is a driving force.

9. #9 raghav
May 30, 2010