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

A worked example finding the change in centripetal acceleration from the change in linear speed, and an example finding the change in centripetal acceleration from the change in radius.

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  • blobby green style avatar for user mkhan66
    I understand how to tackle these problems, it's fairly straightforward. However, I started this course (College Phyiscs 1) from the beginning and, no, we did not derive the formulas for centripetal acceleration (whether using linear speed or angular) in previous videos. But I hope I'll see those videos, wherever they are. I really think this particular course needs to be organized better so that we are asked questions that we've actually been prepared for.
    (17 votes)
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  • leaf green style avatar for user Sivalingam Manasan
    Why the formulas of Centripetal Acceleration for the the Question 1 and 2 different?
    (6 votes)
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  • marcimus pink style avatar for user IZH1
    At , why would you answer the question saying the second centripetal force's magnitude decreases by a factor of nine, instead of saying the second centripetal force's magnitude decreases by a factor of 1/9?
    (5 votes)
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  • blobby green style avatar for user 400904
    Couldn't you just set acceleration proportional to v^2 because 1/r is technically a constant therefore it can be taken out cuz this is a ratio problem anyways
    (3 votes)
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  • aqualine seed style avatar for user maiden
    starting with the vector's signification then no explaining the other topics is a little bit exhausting for us..
    (2 votes)
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  • blobby green style avatar for user orelimereshka
    according to "a=v^2/r," acceleration is inversely proportional to the radius. but why in the second example when the radius is 2r, acceleration also increases by a factor of 2, rather than decreases?
    (1 vote)
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    • blobby blue style avatar for user joshua
      You forgot velocity changes as well.

      It's easy to show it.
      v₁ = rω
      When we have 2r and ω stays constant,
      v₂ = 2rω
      v₂ = 2v₁

      Initially, we have
      a₁ = rω²

      When we have 2r and ω stays constant,
      a꜀ = v² / r
      a꜀ = (v₂)² / (2r)
      a꜀ = (2rω)² / (2r)
      a꜀ = 4(rω)² / (2r)
      a꜀ = 2rω²
      a꜀ = 2a₁
      (1 vote)

Video transcript

- [Instructor] 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 1/3 v. And 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, and 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. Alright, now let's work through this together. So let's first think about the first curve. So the first curve the magnitude of our centripetal acceleration for curve one, I have another subscript one here, this is around the first curve. They tell us that our linear speed is v. So we have v squared over and the radius of that curve is r. This is going to be a straight up v squared over r for that first curve. The magnitude of our centripetal acceleration. Now what about the second curve? So the magnitude of our centripetal acceleration around second curve, that's what that two is, is going to be equal to, they tell us we now have a linear speed of 1/3 v. So in our numerator we're gonna square that, 1/3 v squared all of that over the curve of the same radius. So our radius is still r. So let's just do a little algebraic simplification. 1/3 v times 1/3 v is just going to be 1/9 v squared. So it's going to be 1/9 v squared over r. All I did is square this numerator here. Or I could write this as 1/9 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 1/9 times, instead of writing v squared over r, I could say hey that's r the magnitude of our centripetal acceleration around the first curve. 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 1/9 the magnitude of centripetal acceleration. So we could say the magnitude, or I could just say, they already asked us how does the magnitude change so we could say decreases, decreases by a factor factor of nine. And I wrote it in this language. You could say it got multiplied by a factor of 1/9 or you could say decreases by a factor of nine because on the Khan Academy exercises that deal with this, they use language like that. Let's do another example. Here we are told a father spins 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 two r 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 at a previous video, is to realize that the magnitude of centripetal acceleration is equal to r times our angular speed squared. And so initially so the magnitude of our centripetal acceleration initially, I'll do that with a sub i. That 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. And then when we think about the father, he extends his arms. So then you have the magnitude of your centripetal acceleration, I could say final or extended or I'll just say final sub f. What does that going to be equal to? Well now our radius, the radius of our circle is two r. So it's going to be two r and they say the same angular speed. So our angular speed is still omega. Two r omega squared. Well this part right over here r omega squared, that was just the magnitude of our initial centripetal acceleration. That was the magnitude of our initial centripetal acceleration. And so you see that the magnitude of our centripetal acceleration, has increased by a factor of two. Increased, increased by a factor of two. And we are done.