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### Course: Class 11 Physics (India) > Unit 8

Lesson 12: Centripetal acceleration- Race cars with constant speed around curve
- Visual understanding of centripetal acceleration formula
- Deriving formula for centripetal acceleration from angular velocity
- Change in centripetal acceleration from change in linear velocity and radius: Worked examples
- Predicting changes in centripetal acceleration
- Centripetal acceleration review
- Relating angular and regular motion variables
- Calculus proof of centripetal acceleration formula

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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.

## Want to join the conversation?

- 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.(18 votes)
- Why the formulas of Centripetal Acceleration for the the Question 1 and 2 different?(6 votes)
- They both lead to the same answer, but the formula you use depend on what information you're missing.(7 votes)

- At2:39, 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)
- If you decrease something by a factor of 1/9 you are actually multiplying by 9. 10 devided by 1/9 equals 90. Similar to how 2-(-2) equals 4.(5 votes)

- 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)
- starting with the vector's signification then no explaining the other topics is a little bit exhausting for us..(2 votes)
- 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)
- 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.