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### Course: Physics library > Unit 4

Lesson 1: Circular motion and centripetal acceleration- Race cars with constant speed around curve
- Centripetal force and acceleration intuition
- Visual understanding of centripetal acceleration formula
- What is centripetal acceleration?
- Optimal turns at Indianapolis Motor Speedway with JR Hildebrand
- Calculus proof of centripetal acceleration formula
- Loop de loop question
- Loop de loop answer part 1
- Loop de loop answer part 2

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# Loop de loop answer part 1

Figuring out the minimum speed at the top of the loop de loop to stay on the track. Created by Sal Khan.

## Want to join the conversation?

- At1:50, why is that we take acceleration at the topmost point to be equal to gravity? Why is the normal reaction taken as zero?(68 votes)
- To keep the car in a circular path, the centripetal acceleration (v^2/r) has to be greater than or equal to the acceleration due to gravity (g = 9.81 m/s^2). We set a = 9.81 because this gives us the minimum speed the car must have to stay in a circular path. As soon as the car goes slower than this, g will be greater than the centripetal acceleration, so the car will fall off the track.

At the top of the loop:

Fnet = ma

Fgravity + Fnormal = ma, and because a = centripetal acceleration = v^2/r, then

Fgravity + Fnormal = mv^2/r

So at the minimum speed for the car to stay in a circular path, Fnormal = 0. As the car's speed increases, however, so does Fnormal (because Fgravity stays constant).(61 votes)

- According to the video on fifth-gear website a Professor at Cambridge University calculated that he had to be going 36 miles per hour. Was he just building in a safety factor?(37 votes)
- I think the 36 mph is how fast you'd have to be going when you hit the ramp, assuming you weren't able to keep the engine running with enough traction on the wheels to keep the speed constant as you climbed up. So if you hit the ramp at 36 mph, the car will slow down but still be going faster than the minimum required (16 mph) when it gets to the top.(54 votes)

- It might sound stupid to ask this question, but I've got this question in mind that if the direction of the centripedal force is toward the center of the loop de loop, when the car was at the top, the centripedal force and the gravity force have the same direction, so the car will be pulled down with the two forces! What is the force that exert the car out of the center?(43 votes)
- For the car to not fall, it needs to stay in a circular path. Every object that moves in a circular path, with a constant speed, experiences a centripetal acceleration. So if something is moving along a circle it's experiencing centripetal acceleration, or the other way around, if something is experiencing centripetal acceleration, then it has to move in a circular path. So if we want to make sure that the car does not fall, we have to maintain it's circular motion, which means that all the forces on the car have to sum up to a centripetal force. For a minimal speed, the only force that the car is experiencing (at the top) is the force of gravity, so by simply saying that the force of gravity is a centripetal force, the car has to stay in a circular motion.

The centripetal force always points towards the center, but that does not mean that the car will go towards the center. Remember that the car also has forward velocity, so the centripetal force affects the direction of the forward velocity in such a way that it makes the car go in a circle. That's literally the definition of centripetal acceleration. If the car did not go in a circular path, then it would not experience centripetal acceleration.(36 votes)

- I'm having trouble with an initial concept Sal makes around3:05- Why does AC = 9.81m/s^2? The 9.81 is from the force to due gravity, not centripetal acceleration. The 9.81 is in effect on the car at all points around the circle, not just at the top.(8 votes)
- I've had to revisit this topic because I realised that I didn't really understand it all!

I understand that at the top of the loop de loop, centripetal acceleration is attributed by gravity and normal force, but what EXACTLY IS PREVENTING the car from falling off? Gravity and normal force are pointed towards the centre of the loop de loop, and if centripetal acceleration needs to be equal to or greater than these combined forces, shouldn't the car just fall off the track because there is an overall net force towards the centre?

The only "logical deductions" that I've come up with are:

1) The vehicle experiences inertia in that it would otherwise travel on the tangential if it wasn't for the inward force attributed by centripetal acceleration

2) As per Newton's third law, the vehicle pushes on the track with a force that is equal in magnitude, but opposite in direction to the which the track pushes on the vehicle

I still, however, cannot grasp how the vehicle stays on the track in spite of the above mentioned factors. Any help would be greatly appreciated! :)(7 votes) - So does the weight of the car not matter? I mean even if the car would weight 5 tons it would still make and loop de loop?(4 votes)
- Yes because it would have the kinetic energy associated with a 5 ton weight having been lifted to the launch height(4 votes)

- My text book derives the formula for critical speed as root(5gr) ..... If i calculate it by that formula i get a different answer.. And yes the derivation is very convincing :p(2 votes)
- sqrt(5gr) is the speed required at the BASE of the loop de loop.

So the kinetic energy at the base is 1/2 mv^2 = 5/2 mgr

Using the conservation of mechanical energy,

we get 5/2 mgr -2mgr = 1/2 mgr at the TOP of the loop. (this is kinetic energy)

and again using Ek = 1/2 mv^2,

The speed at the TOP is sqrt(2gr), as Sal has mentioned.(6 votes)

- How much harder do these kinds of problems become if the track has the more general shape of an ellipse?

In other words, how would you calculate this exact same problem if the width and height were different, and not perfectly equal as they are for a circle?

(Of course, a circle is a type of ellipse)(3 votes)- I found your question very interesting!

Let's work with an ellipse of semi major axis a, semi minor axis b and b is parallel to the gravity.

First, think about what type of movement is naturally elliptical. The best example, I believe, is the planetary motion. So, from that we can withdraw the following results:

At a given point, the planet feels a central force directed towards the focus of the type

(GM).m/r² , where G is the gravitational constant and M is the star's mass;

The planet's speed v = sqrt[GM.(2/r - 1/a)] ------ This is a planet's speed equation as a function of the distance from the star;

So, writing out the constant GM, we can obtain

F = mv²/a

That's what happens for a planet in elliptical motion

We also need the following fact:

For any body running on a specified path, the perpendicular force acting upon it equals mv²/R , where R is the curvature radius of the path at that point.

So, what is the ellipse's curvature radius at the top? Here we use the planetary motion's result. The perpendicular force is F.(b/a) , since (b/a) is the cosine of the angle formed between the force directed towards the focus and the line perpendicular to the ellipse's surface at the top (which is vertical). Thus, mv²b/a² = mv²/R , so R = a²/b at the top of the ellipse.

Knowing the curvature radius R, we can easily find P + N, because of the fact mentioned above: "For any body running on a specified path, the perpendicular force acting upon it equals mv²/R , where R is the curvature radius of the path at that point" . Since the perpendicular force equals P + N, then P + N = mv²/(a²/b) , for minimum required speed N = 0, so v = a.sqrt(g/b)

Notice that this result coincides for what Sal found if a = b (a.k.a. circle)

EDIT : I changed my last comment because I believe the explanation with analogy is unsufficient, one must also include the curvature radius, making the result much more general (the car can have any value for its acceleration parallel to the surface) and mathematically accurate.(2 votes)

- At5:26, Sal says something about the traction of the road. What does he mean by traction?(2 votes)
- Traction - maximum frictional force that can be produced by two surfaces without slipping(2 votes)

- Why doesn't the car fall down on the highest point of the loop? Irrespective of its linear velocity at that point, isn't the net force on the car only it's weight acting downwards? How does the speed affect the car being able to stay in the loop?(1 vote)
- The car IS falling, but it is also moving sideways, and the track is curving downward faster than the car is falling.(3 votes)

## Video transcript

What I want to do
now is figure out, what's the minimum
speed that the car has to be at the top of this
loop de loop in order to stay on the track? In order to stay in
a circular motion. In order to not
fall down like this. And I think we
can all appreciate that is the most difficult part
of the loop de loop, at least in the bottom half
right over here. The track itself
is actually what's providing the centripetal force
to keep it going in a circle. But when you get
to the top, you now have gravity that is
pulling down on the car, almost completely. And the car will have to
maintain some minimum speed in order to stay in
this circular path. So let's figure out what
that minimum speed is. And to help figure
that out, we have to figure out what the radius of
this loop de loop actually is. And it actually does not
look like a perfect circle, based on this little screen
shot that I got here. It looks a little
bit elliptical. But it looks like the radius
of curvature right over here is actually smaller
than the radius of the curvature of the
entire loop de loop. That if you made
this into a circle, it would actually be maybe
even a slightly smaller circle. But let's just assume, for
the sake of our arguments right over here, that this
thing is a perfect circle. And it was a perfect
circle, let's think about what that
minimum velocity would have to be up here at the
top of the loop de loop. So we know that the magnitude
of your centripetal acceleration is going to be equal
to your speed squared divided by the
radius of the circle that you are going around. Now at this point
right over here, at the top, which is
going to be the hardest point, the magnitude
of our acceleration, this is going to be 9.81
meters per second squared. And the radius,
we can estimate-- I copied and pasted
the car, and it looks like I can get it to stack
on itself four times to get the radius of this
circle right over here. And I looked it up on the
web, and a car about this size is going to be about
1.5 meters high from the bottom of the
tires to the top of the car. And so it looks like--
just eyeballing it based on these copying
and pasting of the cars, that the radius of this
loop de loop right over here is 6 meters. So this right over
here is 6 meters. So you multiply both
sides by 6 meters. Or actually, we could keep
it just in the variables. So let me just rewrite
it-- just to manipulate it so we can solve for v. We have
v squared over r is equal to a. And then you multiply
both sides by r. You get v squared is
equal to a times r. And then you take the principal
square root of both sides. You get v is equal to
the principal square root of a times r. And then if we plug
in these numbers, this velocity that we
have to have in order to stay in the circle is going
to be the square root of 9.81 meters per second
squared, times 6 meters. And you can verify that
these units work out. Meters times meters is meter
squared, per second squared. You take the square
root of that, you're going to get
meters per second. But let's get our calculator
out to actually calculate this. So we are going to get
the principal square root of 9.81 times 6 meters. It gives us-- now here's
our drum roll-- 7.67. I'll just round to three
significant digits, 7.67 meters per second squared. And significant digits
is a whole conversation, because this is just a very,
very rough approximation. I'm not able to measure
this that accurately at all. But I get roughly 7
point-- I'll just round, 7.7 meters per second. So this is approximately
7.7 meters per second. And just to give a sense
of how that translates into units that we're used
to when we're driving cars, we can convert 7.7
meters per second. If we want to say how many
meters we go to an hour, well, there's 3,600
seconds in an hour. And then if you want to
convert that into kilometers-- this will be in meters--
you divide by 1,000. One kilometer is
equal to 1,000 meters. And you see here,
the units cancel out. You have meters, meters,
seconds, seconds. You're left with
kilometers per hour. So let's actually
calculate this. And so we get our
previous answer. We want to multiply
it times 3,600 to figure out how many
meters in an hour. And then you divide
by 1,000 to convert it to kilometers per hour. So you divide by 1,000. And we get 27.6
kilometers per hour. So this is equal
to 27.6 kilometers per hour, which is
surprisingly slow. I would have thought
it would have to be much, much, much faster. But it turns out, it does not
have to be much, much faster. Only 27.6 kilometers per hour. Now the important
thing to keep in mind is this is just fast
enough, at this point, to maintain the circular motion. But if this were a perfect
circle right over here, and you were going at exactly
27.6 kilometers per hour, you would not have much
traction with the road. And if you don't have much
traction with the road, the car might slip
and might not be able to actually
maintain its speed. So you definitely
want your speed to be a good bit larger
than this in order to keep a nice
margin of safety-- in order to especially
have traction with the actual loop
de loop, and to be able to maintain your speed. Now what I want to
do in the next video is actually time the car to
figure out how long does it take it to do this loop de loop. And we're going to assume
that it's a circle. And we're going
to figure it out. And we're going to
figure out how fast it's actual average velocity
was over the course of this loop de loop.