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