yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

Change in centripetal acceleration from change in linear velocity and radius: Worked examples


3m read
·Nov 11, 2024

We are told that a van drives around a circular curve of radius r with linear speed v. On a second curve of the same radius, the van has linear speed one third v. You could view linear speed as the magnitude of your linear velocity.

How does the magnitude of the van's centripetal acceleration change after the linear speed decreases? So pause this video and see if you can figure it out on your own. I'll give you a little bit of a hint: we know that the magnitude of centripetal acceleration, in general, is equal to linear speed squared divided by radius, the radius of the curve.

All right, now let's work through this together. So let's first think about the first curve. The magnitude of our centripetal acceleration for curve one (I have another subscript one here, of course, this is around the first curve) tells us that our linear speed is v. So we're gonna have v squared over the radius of that curve, which is r. So, it's just going to be straight up v squared over r for that first curve: the magnitude of our centripetal acceleration.

Now, what about the second curve? The magnitude of our centripetal acceleration around the second curve (that's what that 2 is) is going to be equal to... They tell us we now have a linear speed of one third v. So, in our numerator, we're going to square that one third v: (1/3 v)², all of that over the curve of the same radius.

Our radius is still r, and so let's just do a little algebraic simplification: one third v times one third v is just going to be one ninth v squared. So it's going to be one ninth v squared over r. All I did is square this numerator here, or I could write this as one ninth times v squared over r.

The reason why I wrote this in green is because this is the exact same thing as this, and so this is going to be equal to... This is equal to one ninth times, instead of writing v squared over r, I can say, "Hey, that's our centripetal acceleration, the magnitude of our centripetal acceleration around the first curve."

So, how does the magnitude of the van's centripetal acceleration change after the linear speed decreases? Well, around the second curve, we have one ninth the magnitude of centripetal acceleration. So we could say the magnitude (mag) or I could just say, well, they already asked us how does the magnitude change. So we could say decreases, decreases by a factor of 9.

And I wrote it in this language: you could say it got multiplied by a factor of one ninth, or you could say decreases by a factor of 9 because on the Khan Academy exercises that deal with this, they use language like that.

Let's do another example here. We're told a father spends his daughter in a circle of radius r at angular speed omega. Then the father extends his arms and spins her in a circle of radius 2r with the same angular speed.

How does the magnitude of the child's centripetal acceleration change when the father extends his arms? Once again, pause this video and see if you can figure it out. Well, the key realization here, and we derived this in a previous video, is to realize that the magnitude of centripetal acceleration is equal to r times our angular speed squared.

So initially, the magnitude of our centripetal acceleration (I'll do that with a sub i) is going to be equal to... Well, they're using the same notation; we have omega as our angular speed and our radius is r. So it's just going to be r omega squared.

Then, when we think about the father extending his arms, we can say the magnitude of our centripetal acceleration (I could say final or extended, well, I'll just say final, sub f) what is that going to be equal to? Well, now our radius, the radius of our circle, is 2r. So it's going to be 2r, and they say the same angular speed. So our angular speed is still omega: 2r omega squared.

Well, this part right over here, r omega squared, was just the magnitude of our initial centripetal acceleration. That was the magnitude of our initial centripetal acceleration.

And so you see that our the magnitude of our centripetal acceleration has increased by a factor of two: increased, increased by a factor of two. And we're done.

More Articles

View All
Jessica Livingston Introduces Startup School SV 2014
Good morning! Hello everyone, welcome! I’m Jessica Livingston with Y Combinator. We’re going to get started now, and today’s a really special [Applause] day. Today is the 10th Startup School we’ve ever done. It’s pretty amazing to think we’ve been doing t…
EPIC NOSE PICKING and why Football RULES -- IMG! #20
Master Chief loves football, and the most confused face ever. It’s a special football episode of IMG North American football. It gives you everything a guy could want: kicks to the face, kicks to the nuts, and heads up your butt. You get to pick; you can …
🎄 Grey Grades Canada's Flags! (And Merry Xmas!) 🎄
Merry Christmas, internet! I’ve unwrapped one of the Bonnie B bonus videos for the holidays. It’s the Canadian flag tier list, and even if you’ve seen it before, I’ve decorated it with, uh, decorations. Hope you like it, and I’ll be back after it’s over. …
Introduction to the possessive | The Apostrophe | Punctuation | Khan Academy
Hello Garans, hello Paige, hi David in the driver’s seat. So Paige, today, uh, it is my understanding that we are going to talk about the possessive. That’s right. Um, what even is the possessive in English? What does that mean? When we say that, like, w…
LearnStorm Growth Mindset: Animation Director on setting goals
My name is Lisa Labraccio. I’m 32 years old. I am an animation director at Ted Ed. I’ve always wanted to do animation, so it just, at whatever point in high school, when they tell you to start looking at colleges and what you might, where you might want t…
Introduction to experiment design | Study design | AP Statistics | Khan Academy
So let’s say that I am a drug company and I’ve come up with a medicine that I think will help folks with diabetes. In particular, I think it will help reduce their hemoglobin A1c levels. For those of you who aren’t familiar with what hemoglobin A1c is, I …