Visual understanding of centripetal acceleration formula | Physics | Khan Academy
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Visual understanding of centripetal acceleration formula | Physics | Khan Academy


Let’s say I have
some object that’s traveling in a circular
path just like this. And what I’ve drawn here
is its velocity vector at different points
along that path. And so this right over
here is going to be v1, velocity vector 1. This is going to be
velocity vector 2. And this right over here is
going to be velocity vector 3. And what we’re going to
assume, in this video, is that the magnitude of these
velocity vectors is constant. Or another way to think about it
is that the speed is constant. So I’ll just say lowercase
v without the arrow on top– so this is going to be a scalar
quantity– I’ll call this the speed. Or you could call this the
magnitude of these vectors. And this is going
to be constant. So this is going to be equal
to the magnitude of vector 1, which is equal to the
magnitude of vector 2. The direction is
clearly changing, but the magnitude is going
to be the same, which is equal to the
magnitude of vector 3. And we’re going to assume
that it’s traveling in a path, in a circle with radius r. And what I’m going
to do is, I’m going to draw a position
vector at each point. So let’s call r1– actually I’ll
just do it in pink– let’s call r1 that right over there. That’s position vector r1. That is position vector r2. So the position is
clearly changing. That’s position vector r2. And that is position vector r3. But the magnitude of
our position vectors are clearly the same. And I’m going to call the
magnitude of our position vectors r. And that’s just the
radius of the circle. It’s this distance
right over here. So r is equal to
the magnitude of r1, which is equal to the
magnitude of r2, which is equal to the magnitude of r3. Now what I want to
do, in this video, is prove to you visually,
that given this radius and given this speed,
that the magnitude of the centripetal
acceleration– and I’ll just write that as a sub c,
I don’t have an arrow on top, so this is a scalar quantity. So the magnitude of the
centripetal acceleration is going to be equal
to our speed squared, our constant speed
squared, divided by the radius of the circle. I want you to feel good
that this is indeed the case by the
end of this video. And to understand
that, what I want to do is I want to re-plot
these velocity vectors on another
circle and just think about how the vectors
themselves are changing. So let’s copy and paste this. So let me copy and paste v1. So copy and paste. So that is v– actually I
want to do it from the center –so that is v1. Then let me do the
same thing for v2. So let me copy and paste it. That is v2. And then let me
do it also for v3. I’ll just get the vector part;
I don’t have to get the label. So copy and paste it. And that right over
there is vector v3. And let me clean
this up a little bit. So that’s clearly v2. I don’t think we have
to label anymore. We know that v2 is in orange. And what is the
radius of this circle going to be right over here? Well, the radius
of this circle is going to be the magnitude
of the velocity vectors. And we already know the
magnitude of the velocity vectors is this quantity
v, this scalar quantity. So the radius of this circle is
v. The radius of this circle, we already know, is equal to r. And just as the
velocity vector is what’s giving us the
change in position over time, the change in
position vector over time, what’s the vector
that’s going to give us the change in our
velocity vector over time? Well, that’s going to be
our acceleration vectors. So you will have
some acceleration. We’ll call this a1. We’ll call this a2. And I’ll call this a3. And I want to make
sure that you get the analogy that’s
going on here. As we go around in this
circle, the position vectors first they point out
to the left, then the upper, kind of in a maybe
11 o’clock position, or I guess the top left,
and then to the top. So it’s pointing in these
different directions like a hand in a clock. And what’s moving it
along there is the change in position vector
over time, which are these velocity vectors. Over here, the velocity
vectors are moving around like the hands of a clock. And what is doing
the moving around are these acceleration vectors. And over here, the
velocity vectors are tangential to the
path, which is a circle. They’re perpendicular
to a radius. And you learned
that in geometry– that a line that is
tangent to a circle is perpendicular to a radius. And it’s also going to be the
same thing right over here. And just going back
to what we learned when we learned
about the intuition of centripetal acceleration, if
you look at a1 right over here, and you translate this vector,
it’ll be going just like that. It is going towards the center. a2, once again, is going
towards the center. a3, once again, if
you translate that, that is going
towards the center. So all of these are actually
center-seeking vectors. And you see that
right over here. These are actually centripetal
acceleration vectors right over here. Here we’re talking about
just the magnitude of it. And we’re going to
assume that all of these have the same magnitude. So we’re going to
assume that they all have a magnitude of
what we’ll call a sub c. So that’s the magnitude. And it’s equal to
the magnitude of a1. That vector, it’s equal
to the magnitude of a2. And it’s equal to
the magnitude of a3. Now what I want to think
about is how long is it going to take for
this thing to get from this point on the circle
to that point on that circle right over there? So the way to think about it
is, what’s the length of the arc that it traveled? The length of this arc that
it traveled right over there. That’s 1/4 around the circle. It’s going to be 1/4
of the circumference. The circumference is 2 pi r. It Is going to be 1/4 of that. So that is the
length of the arc. And then how long will
it take it to go that? Well, you would
divide the length of your path divided by the
actual speed, the actual thing that’s nudging it
along that path. So you want to divide that by
the magnitude of your velocity, or your speed. This is the magnitude of
velocity, not velocity. This is not a vector right
over here, this is a scalar. So this is going to be the
time to travel along that path. Now the time to
travel along this path is going to be the exact
same amount of time it takes to travel along this
path for the velocity vector. So this is for the position
vector to travel like that. This is for the velocity
vector to travel like that. So it’s going to be
the exact same T. And what is the
length of this path? And now think of it in the
purely geometrical sense. We’re looking at a circle here. The radius of the circle
is v. So the length of this path right over
here is going to be 1/4. It is going to be– I’ll
do it in that same color so you see the analogy–
it’s equal to 1/4 times the circumference of the circle. The circumference of
this circle is 2 pi times the radius of the
circle, which is v. Now what is nudging
it along this circle? What is nudging it
along this path? What is the analogy for
speed right over here? Speed is what’s nudging it
along the path over here. It is the magnitude of
the velocity vector. So what’s nudging it along
this arc right over here is the magnitude of the
acceleration vector. So it is going to be a sub c. And these times are going
to be the exact same thing. The amount of time it
takes for this vector to go like that, for
the position vector, is the same amount of
time it takes the velocity vector to go like that. So we can set these 2
things equal each other. So we get, on this side,
we get 1/4 2 pi r over v is equal to 1/4 2 pi v over the
magnitude of our acceleration vector. And now we can simplify
it a little bit. We can divide both sides by 1/4. Get rid of that. We can divide both sides
by 2 pi, get rid of that. Let me rewrite it. So then we get r/v
is equal to v over the centripetal acceleration. And now you can cross multiply. And so you get v
times v. So I’m just cross multiplying right
over here. v times v, you get v squared, is
equal to a c times r. And cross multiplying, remember,
is really just the same thing as multiplying both sides
by both denominators, by multiplying both
sides times v and ac. So it’s not some magical thing. If you multiply both sides times
v and ac, these v’s cancel out. These ac’s cancel out. You get v times v is v squared,
is equal to a sub c times r. And now to solve
for the magnitude of our centripetal
acceleration, you just divide both sides by r. And you are left
with– and I guess we’ve earned a drum
roll now– the magnitude of our centripetal
acceleration is equal to the magnitude,
our constant magnitude of our velocity. So this right here
is our speed, divided by the radius of the circle. And we’re done!

About James Carlton

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80 thoughts on “Visual understanding of centripetal acceleration formula | Physics | Khan Academy

  1. @riimzo There is only one velocity vector on the circle for every position vector. If you know it's position, you know it's velocity. So if the position changes at a rate, the velocity vectors much change at the same rate.

  2. @PmQable1 It's useful to describe the change in direction of bodies. Especially when they have paths like circles and ellipses. No?

  3. This is awesome. Now I understand where this equation comes from. For those of you guys who will take Dynamics in the future, you better learn this stuff clearly cuz your gonna bump into it again in Curvilinear motion in Dynamics. I didn't understand anything about that kind of motion at first cuz my professor skipped the Centripetal Acceleration section when we were in Phys 1… I think i got it now 😀

  4. i like sal he not only explains thing good but being able to have a clear and concise illustration of what your talking about as you mention it is important in my opinion

  5. I get it now thanks!. Also, during class, I always have to stop listening to my professor to think about what she is saying. Ex. "The direction of velocity is moving in clockwise rotation." But when this guy says "like hands on a clock," less thinking needs to be done on my part. Many thanks.

  6. AM I CORRECT?
    if we assume that angle is small so that arc lenght is flat hence equal to Δv so that velocity x θ = Δv and then dividing both sides with t gives Δv / t = velocity x θ/ t as θ / t = w hence Δv/t = a = velocity x w

  7. So if there is a change in direction there is an acceleration??
    How come there is an acceleration, if the magnitude is constant? aren't you gonna get zero or there is an exception in circular motion?

  8. Hi Sal, can you explain me why does the velocity vector and radius vector takes equal time period to travel 1/4circle ??

  9. Sorry, this didn't help at all, too many questions during the video, it might be my lack of knowledge on the subject

  10. What does the second circle represent? In what case do velocities come from the origin? Or does the second circle just represent the first circle with the forces redrawn?

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  12. Wow thank you so much, I was staring at my notebook trying to visualize why would that make sense, and I couldn't get anything at all, this is perfectly what I tried to look up

  13. physics is crazy bro. I understand this video but it also got me thinking why does nature work this way? like think about it… you have an object moving at a constant speed in a circle and that object has a force pushing it towards the center of the circle. The only reason why it doesn't come crashing towards the center is because the force tangent to the circle is greater. Like bruh if that isn't crazy then idk what is.

  14. All the centripetal acceleration is is the vertical component of a parabolic motion. To understand, let me use satellites as an example. The curvature of earth drops about 16 feet every 5 miles. Every single object which falls on earth, (excluding air resistance), falls almost exactly 16 feet after the first second of free fall. If an object is launched at 5 mi/s just above the surface of earth, it will cover 5 miles horizontally after one second and 16 feet vertically after one second. Any other circular motion is just the same. The earth is so big and the horizontal speed is so small relative to its size that the vertical componet/ centripetal acceleration is noticeable. Zooming in on any circular motion object, and slowing down time, the vertical component is noticeable.

  15. Why is the time (T) in the left circle is the same as the one in the right? Like, how can we compare the change in the velocity vector direction and the acceleration vector direction?

  16. r, v and a are considered by only magnitude (they are constant) and the direction change that respect to time is neglected. In this sense, the change of r , v are just position changes, in the track of motion, which is arc length. It is very good interpretation for the uniformly centripetal acceleration formula. It is indeed a miss for physics lectures that are supposed to cover. Thank you for teaching me something new that is an important formula derive in physics.

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