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What is centripetal acceleration?

Learn what centripetal acceleration means and how to calculate it.

What is centripetal acceleration?

Can an object accelerate if it's moving with constant speed? Yup! Many people find this counter-intuitive at first because they forget that changes in the direction of motion of an object—even if the object is maintaining a constant speed—still count as acceleration.
Acceleration is a change in velocity, either in its magnitude—i.e., speed—or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the speed might be constant. You experience this acceleration yourself when you turn a corner in your car—if you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion. What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we'll examine the direction and magnitude of that acceleration.
The figure below shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation—the center of the circular path. This direction is shown with the vector diagram in the figure. We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration a, start subscript, c, end subscript; centripetal means “toward the center” or “center seeking”.
The directions of the velocity of an object at two different points, start text, B, end text and start text, C, end text, are shown, and the change in velocity, delta, v, is seen to point roughly toward the center of curvature. To see what happens instantaneously, the points start text, B, end text and start text, C, end text must be very close and delta, theta very small. We'll then find that delta, v points directly toward the center of curvature.
Because a, start subscript, c, end subscript, equals, start fraction, delta, v, divided by, delta, t, end fraction, the acceleration is also toward the center. Because delta, theta is very small, the arc length delta, s is equal to the chord length delta, r for small time differences. Image credit: Openstax College Physics
The direction of centripetal acceleration is toward the center of the circle, but what is its magnitude? Note that the triangle formed by the velocity vectors and the triangle formed by the radii r and delta, s are similar. Both the triangles A, B, C and P, Q, R are isosceles triangles with two equal sides. The two equal sides of the velocity vector triangle are the speeds v, start subscript, 1, end subscript, equals, v, start subscript, 2, end subscript, equals, v. Using the properties of two similar triangles, we obtain start fraction, delta, v, divided by, v, end fraction, equals, start fraction, delta, s, divided by, r, end fraction.
Acceleration is start fraction, delta, v, divided by, delta, t, end fraction, so we first solve the above expression for delta, v:
delta, v, equals, start fraction, v, divided by, r, end fraction, delta, s
If we divide both sides by delta, t we get the following:
start fraction, delta, v, divided by, delta, t, end fraction, equals, start fraction, v, divided by, r, end fraction, times, start fraction, delta, s, divided by, delta, t, end fraction
Finally, noting that start fraction, delta, v, divided by, delta, t, end fraction, equals, a, start subscript, c, end subscript and that start fraction, delta, s, divided by, delta, t, end fraction, equals, v, the linear or tangential speed, we see that the magnitude of the centripetal acceleration is a, start subscript, c, end subscript, equals, start fraction, v, squared, divided by, r, end fraction.
This is the acceleration of an object in a circle of radius r at a speed v. So, centripetal acceleration is greater at high speeds and in sharp curves—smaller radii—as you have noticed when driving a car. But it is a bit surprising that a, start subscript, c, end subscript is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/hr than at 50 km/hr. A sharp corner has a small radius, so a, start subscript, c, end subscript is greater for tighter turns, as you have probably noticed.

What is a centrifuge?

A centrifuge is a rotating device used to separate specimens of different densities. High centripetal acceleration significantly decreases the time it takes for separation to occur and makes separation possible with small samples. Centrifuges are used in a variety of applications in science and medicine, including the separation of single cell suspensions such as bacteria, viruses, and blood cells from a liquid medium and the separation of macromolecules—such as DNA and protein—from a solution.
A particle of mass m in a centrifuge is rotating at constant speed. It must be accelerated perpendicular to its velocity or it would continue in a straight line. Image credit: Openstax College Physics
Centrifuges are often rated in terms of their centripetal acceleration relative to acceleration due to gravity, g; maximum centripetal acceleration of several hundred thousand g is possible in a vacuum. Human centrifuges, extremely large centrifuges, have been used to test the tolerance of astronauts to the effects of accelerations larger than that of Earth’s gravity.

What do solved examples involving centripetal acceleration look like?

Example 1: Curving car

What is the magnitude of the centripetal acceleration of a car following a curve, see figure below, of radius 500 m at a speed of 25 m/s—about 90 km/hr? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed.
Image credit: Openstax College Physics

Example 2: Ultracentrifuge

Calculate the centripetal acceleration of a point 7.5 cm from the axis of an ultracentrifuge spinning at 7, point, 5, times, 10, start superscript, 4, end superscript revolutions per minute.

Want to join the conversation?

  • spunky sam green style avatar for user Taha Anouar
    how can deltaS equal deltaR?
    (10 votes)
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    • leaf green style avatar for user qrrqtx
      That's a good question. Notice how the article says: when Δθ is very small, Δs = Δr. That's calculus at work - by very small it means infinitesimally small. On a bigger scale, it's obvious that an arc and a chord are not equal, but the bigger scale is just an illustration of the ideas, an approximation. As θ gets smaller and smaller, Δs and Δr (the arc length and chord length) get closer and closer to being the same length. By using limits, you can actually prove that as Δθ approaches 0, Δs = Δr. It seems to defy "common sense", but it's true and it works. To get a little more intuitive picture of what's going on, start with a circle of radius 1 and a 90° arc, figure out the arc length and the chord length. Then figure out arc and chord for a 9° angle, then 0.9°, etc. See how close arc and chord get in just a few steps? Imagine you did that infinitely many times.
      (45 votes)
  • blobby green style avatar for user Archi130679
    what is the real forces that provide centripetal acceleration
    (7 votes)
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    • blobby green style avatar for user Rajeev Agarwal
      centripetal actually means - towards the center .So centripetal force is not a new type of force .Any force which is acting towards center can be called as centripetal force. To understand it better think of gravitational force , it acts in downwards direction so we call it downwards force because of its direction .There are only four real forces in nature i.e. Gravitational , Electromagnetic ,weak nuclear and strong nuclear .
      whenever any of these four forces acts towards center that force is called centripetal(towards center) . In case of earth and sun system ,real force acting is Gravitational but as it is always pointing towards center we call it centripetal . In case of an electron revolving around nucleus ,the real force between electron and positive nucleus is electromagnetic but because of its direction always pointing towards center ,it is centripetal force. Friction too is electromagnetic force.
      (20 votes)
  • leafers ultimate style avatar for user Esha
    why is the triangle ABC and triangle PQR similar?
    (14 votes)
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  • aqualine ultimate style avatar for user Riya Mahajan
    If an object has a centripetal acceleration towards the center, why does it not go towards the center?
    (7 votes)
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  • blobby green style avatar for user zqiu
    Why does centripetal force does not work?
    (4 votes)
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  • boggle green style avatar for user Mister  Owl
    Can centrifugal force be thought of as the "equal and opposite force" to centripetal force? When turning in a car, it seems as if one tends away from the turn (away from the center). Maybe centrifugal force is just a vernacular term for Newton's first law when moving in a circle. The object is "trying" to maintain its fixed velocity, and when centripetal force acts on the object, it tends to stay in motion at its fixed velocity.
    (3 votes)
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  • leaf red style avatar for user Surbhi Kavishwar
    what is meant by utlracentrifuge?
    (3 votes)
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  • blobby green style avatar for user siddharth kashyap
    why is centripetal acceleration equal to negative of v^2/r
    (2 votes)
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  • boggle blue style avatar for user Ishan Saha
    How would you consider an object with changing magnitude and direction for centripetal acceleration?

    a=V^2/r, but doesn't address change in speed
    (3 votes)
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    • blobby green style avatar for user caleyandrewj
      Ishan, the direction is already changing because the acceleration is towards the center but the velocity is tangential, so it travels in a circle constantly changing direction as mentioned. If the speed, or magnitude, weren't constant and changed, in order to plug in for V in the formula, you would take the average. This can be done by finding the initial speed and final speed and dividing by 2. Assuming the acceleration is constant.
      (1 vote)
  • piceratops seed style avatar for user theo.pierik2927
    In the example, how does it got from deltaV/V=DeltaS/r to DeltaV=r/v x delta s?
    (3 votes)
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