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Uniform Circular Motion and Gravitation

36 Centripetal Acceleration

Learning Objectives

  • Derive the expression for centripetal acceleration.

  • Explain how a centrifuge uses centripetal acceleration for practical applications.

In earlier chapters on kinematics, we learned that acceleration results from a change in velocity—either in magnitude, direction, or both. In uniform circular motion, an object moves at constant speed but continually changes direction. This change in direction means the object is always accelerating, even if its speed remains unchanged.

You may have experienced this type of acceleration while riding in a car that takes a sharp turn. Even if the car maintains a constant speed through the turn, you feel pushed sideways—this is a sensation caused by a change in the direction of motion, not speed. The tighter the curve or the faster the car, the more strongly this acceleration is felt.

 

Direction of Centripetal Acceleration

As shown in Figure 36.1, an object moving in a circle experiences a change in its velocity vector that always points toward the center of the circular path. This center-seeking acceleration is known as centripetal acceleration, denoted [latex]a_c[/latex] (from the Latin centrum petere, “to seek the center”).

The given figure shows a circle, with a triangle having vertices A B C made from the center to the boundry. A is at the center and B and C points are at the circle path. Lines A B and A C act as radii and B C is a chord. Delta theta is shown inside the triangle, and the arc length delta s and the chord length delta r are also given. At point B, velocity of object is shown as v one and at point C, velocity of object is shown as v two. Along the circle an equation is shown as delta v equals v sub 2 minus v sub 1.
Figure 36.1 An object moving in a circle has instantaneous velocity vectors pointing tangentially at every point. The change in velocity [latex]\Delta \vec{v}[/latex] between two points points inward toward the center of curvature. Therefore, the acceleration vector [latex]\vec{a}_c = \Delta \vec{v} / \Delta t[/latex] also points toward the center. (Vector triangle inset adapted from velocity vectors.)

Magnitude of Centripetal Acceleration

To derive the magnitude of [latex]a_c[/latex], we use geometric reasoning. Consider two positions of an object on a circular path and construct a triangle using the velocity vectors at those positions. Since both velocity vectors have the same magnitude [latex]v[/latex], this forms an isosceles triangle. A similar triangle can be drawn using the arc length [latex]\Delta s[/latex] and the radius [latex]r[/latex].

Using the similarity of the triangles, we find:

[latex]\frac{\Delta v}{v} = \frac{\Delta s}{r}[/latex]

Solving for [latex]\Delta v[/latex] and dividing by [latex]\Delta t[/latex]:

[latex]\Delta v = \frac{v}{r} \Delta s \quad \Rightarrow \quad \frac{\Delta v}{\Delta t} = \frac{v}{r} \cdot \frac{\Delta s}{\Delta t}[/latex]

Since [latex]\frac{\Delta s}{\Delta t} = v[/latex], we substitute to get:

[latex]a_c = \frac{\Delta v}{\Delta t} = \frac{v^2}{r}[/latex]

Thus, the magnitude of the centripetal acceleration is:

[latex]\boxed{a_c = \frac{v^2}{r}}[/latex]

This equation tells us that:

  • Acceleration increases quadratically with speed: doubling the speed quadruples [latex]a_c[/latex],

  • Acceleration increases inversely with the radius: tighter turns (smaller [latex]r[/latex]) result in larger [latex]a_c[/latex].

Centripetal Acceleration in Terms of Angular Velocity

Since the linear speed [latex]v[/latex] is related to the angular velocity [latex]\omega[/latex] by:

[latex]v = r\omega[/latex]

we substitute this into the expression for [latex]a_c[/latex]:

[latex]a_c = \frac{(r\omega)^2}{r} = r\omega^2[/latex]

So, we can also write centripetal acceleration as:

[latex]\boxed{a_c = r\omega^2}[/latex]

Use whichever form is more convenient—[latex]a_c = \frac{v^2}{r}[/latex] or [latex]a_c = r\omega^2[/latex]—depending on whether you are given linear or angular velocity.

Applications: The Centrifuge

A centrifuge (see Figure 36.2(b)) is a device that spins rapidly to generate very large centripetal accelerations. This acceleration allows materials to separate by density, mimicking the effect of gravity—but at much greater magnitudes (often hundreds of thousands of [latex]g[/latex]).

Centrifuges are widely used in:

  • Biology – to separate blood cells, bacteria, or viruses from liquid media.

  • Molecular biology – to isolate DNA, RNA, or proteins from solutions.

  • Aerospace and medicine – in human centrifuges, used to test how astronauts or pilots respond to high [latex]g[/latex]-forces.

The ability to produce large centripetal accelerations allows centrifuges to achieve rapid and efficient separation with very small sample volumes, making them essential tools across many scientific and medical disciplines.

How Does the Centripetal Acceleration of a Car Around a Curve Compare with That Due to Gravity?

What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure 36.2(a).

Strategy

Because [latex]v[/latex] and [latex]r[/latex] are given, the first expression in [latex]{a}_{\text{c}}=\frac{{v}^{2}}{r}\mathrm{; }{a}_{\text{c}}={\mathrm{r\omega }}^{2}[/latex] is the most convenient to use.

Solution

Entering the given values of [latex]v=\text{25}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m/s}[/latex] and [latex]r=\text{500 m}[/latex] into the first expression for [latex]{a}_{\text{c}}[/latex] gives

[latex]{a}_{\text{c}}=\frac{{v}^{2}}{r}=\frac{(25.0m/s)^2}{500 m}=1.25m/s^2[/latex]

Discussion

To compare this with the acceleration due to gravity [latex]\left(g=9\text{.}80\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)[/latex], we take the ratio of [latex]{a}_{\text{c}}/g=\left(1\text{.}\text{25}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)/\left(9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)=0\text{.}\text{128}[/latex]. Thus, [latex]{a}_{\text{c}}=0\text{.}\text{128 g}[/latex] and is noticeable especially if you were not wearing a seat belt.

In figure a, a car shown from top is running on a circular road around a circular path. The center of the park is termed as the center of this circle and the distance from this point to the car is taken as radius r. The linear velocity is shown in perpendicular direction toward the front of the car, shown as v the centripetal acceleration is shown with an arrow pointed towards the center of rotation. In figure b, a centrifuge is shown an object of mass m is rotating in it at a constant speed. The object is at the distance equal to the radius, r, of the centrifuge. The centripetal acceleration is shown towards the center of rotation, and the velocity, v is shown perpendicular to the object in the clockwise direction.
Figure 36.2 (a) The car following a circular path at constant speed is accelerated perpendicular to its velocity, as shown. The magnitude of this centripetal acceleration is found in the previous example. (b) A laboratory centrifuge rotates rapidly, causing small fluid samples to separate based on density due to extremely high centripetal acceleration.

How Big Is the Centripetal Acceleration in an Ultracentrifuge?

Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge spinning at

[latex]{\text{7.5 × 10}}^{\text{4}}\phantom{\rule{0.25em}{0ex}}\text{rev/min.}[/latex]

Determine the ratio of this acceleration to that due to gravity. See Figure 36.2(b)

Strategy

The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocity [latex]\omega[/latex]. Because [latex]r[/latex] is given, we can use the second expression in the equation

[latex]{a}_{\text{c}}=\frac{{v}^{2}}{r};\phantom{\rule{0.25em}{0ex}}{a}_{\text{c}}={\mathit{r\omega }}^{2}[/latex]

to calculate the centripetal acceleration.

Solution

To convert [latex]7\text{.}\text{50}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{rev}/\text{min}[/latex] to radians per second, we use the facts that one revolution is [latex]2\pi \text{rad}[/latex] and one minute is 60.0 s. Thus,

[latex]\omega =\text{7.50}×{\text{10}}^{\text{4}}\phantom{\rule{0.25em}{0ex}}\frac{\text{rev}}{\text{min}}×\frac{2\pi \phantom{\rule{0.25em}{0ex}}\text{rad}}{\text{1 rev}}×\frac{1\phantom{\rule{0.25em}{0ex}}\text{min}}{\text{60}\text{.}\text{0 s}}=\text{7854}\text{ rad/s.}[/latex]

Now the centripetal acceleration is given by the second expression in [latex]{a}_{\text{c}}=\frac{{v}^{2}}{r}\mathrm{; }{a}_{\text{c}}={\mathrm{r\omega }}^{2}[/latex] as

[latex]{a}_{\text{c}}={\mathrm{r\omega }}^{2}\text{.}[/latex]

Converting 7.50 cm to meters and substituting known values gives

[latex]{a}_{\text{c}}=(0\text{.}\text{0750 m})(\text{7854 rad/s}{)}^{2}=4\text{.}\text{63}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}.[/latex]

Note that the unitless radians are discarded in order to get the correct units for centripetal acceleration. Taking the ratio of [latex]{a}_{\text{c}}[/latex] to [latex]g[/latex] yields

[latex]\frac{{a}_{\text{c}}}{g}=\frac{4\text{.}\text{63}×{\text{10}}^{6}}{9\text{.}\text{80}}=4\text{.}\text{72}×{\text{10}}^{5}.[/latex]

Discussion

This last result means that the centripetal acceleration is 472,000 times as strong as [latex]g[/latex]. It is no wonder that such high [latex]\omega[/latex] centrifuges are called ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to cause the sedimentation of blood cells or other materials.

Centripetal Acceleration Requires a Net Force

As stated by Newton’s Second Law, a net external force is required to cause any acceleration. This principle applies to all forms of motion, including uniform circular motion, where the object’s speed is constant, but its direction is continuously changing, leading to centripetal acceleration.

Therefore, a net external force must be present to maintain circular motion. This force is always directed toward the center of the circle, causing the object to continuously deviate from a straight-line path.

In the next section—Centripetal Force—we will analyze the specific forces responsible for maintaining this motion in different physical contexts. Whether the force comes from tension, gravity, friction, or normal force, it must have a component pointing radially inward to sustain circular motion.

Without this net inward force, the object would move in a straight line, tangential to the circular path, in accordance with Newton’s First Law.

PhET Explorations: Ladybug Motion 2D

Explore how position, velocity, and acceleration vectors behave in different types of motion. Move the ladybug by setting its position, velocity, or acceleration, and observe how the corresponding vectors change in real time.

Choose between linear, circular, or elliptical motion paths. Use the record and playback features to analyze the motion frame by frame and better understand the relationships among these key vectors.

This interactive simulation helps reinforce core concepts in kinematics by allowing you to visualize how vectors evolve during various types of motion.

Section Summary

  • Centripetal acceleration [latex]a_{\text{c}}[/latex] is the acceleration an object experiences while undergoing uniform circular motion. This acceleration:

    • Always points toward the center of the circular path (it is “center-seeking”),

    • Is perpendicular to the object’s instantaneous linear velocity [latex]v[/latex],

    • Has magnitude given by:

    [latex]a_{\text{c}} = \frac{v^2}{r}; \quad a_{\text{c}} = r\omega^2,[/latex]

    where:

    • [latex]v[/latex] is the linear (tangential) speed,

    • [latex]r[/latex] is the radius of curvature,

    • [latex]\omega[/latex] is the angular velocity.

    The SI unit of centripetal acceleration is [latex]\text{m/s}^2[/latex].

Conceptual Questions

  1. Can centripetal acceleration change the speed of circular motion? Explain.

Problem Exercises

  1. A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity?
  2. A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If he completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration as he runs the curved portion of the track?
  3. Taking the age of Earth to be about [latex]4×{\text{10}}^{9}[/latex] years and assuming its orbital radius of [latex]1.5 ×{\text{10}}^{11}[/latex] has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).
  4. The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if it spins at 1200 rev/min? (b) What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac? (c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of [latex]g[/latex].
  5. An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min. (a) Calculate the magnitude of the centripetal acceleration at its edge in meters per second squared and convert it to multiples of [latex]g[/latex]. (b) What is the linear speed of a point on its edge?
  6. Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip. (a) Calculate the magnitude of the centripetal acceleration at the tip of a 4.00 m long helicopter blade that rotates at 300 rev/min. (b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s).
  7. Olympic ice skaters are able to spin at about 5 rev/s. (a) What is their angular velocity in radians per second? (b) What is the centripetal acceleration of the skater’s nose if it is 0.120 m from the axis of rotation? (c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since—at about 9 rev/s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius? (d) Comment on the magnitudes of the accelerations found. It is reputed that Button ruptured small blood vessels during his spins. d)The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That’s quite a lot of acceleration in itself. The centripetal acceleration felt by Button’s nose was 39.2 times larger than the acceleration due to gravity. It is no wonder that he ruptured small blood vessels in his spins.
  8. What percentage of the acceleration at Earth’s surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth?
  9. Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating: (a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min. (b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).
  10. A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of [latex]9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}[/latex] at the rim?
  11. At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m. (a) At how many rev/min are the tires rotating? (b) What is the centripetal acceleration at the edge of the tire? (c) With what force must a determined [latex]1\text{.}\text{00}×{\text{10}}^{-\text{15}}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex] bacterium cling to the rim? (d) Take the ratio of this force to the bacterium’s weight.
  12. Integrated Concepts Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity. (a) Assuming negligible friction, find the speed of the riders at the bottom of its arc, given the system’s center of mass travels in an arc having a radius of 14.0 m and the riders are near the center of mass. (b) What is the centripetal acceleration at the bottom of the arc? (c) Draw a free body diagram of the forces acting on a rider at the bottom of the arc. (d) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight. (e) Discuss whether the answer seems reasonable.
  13. Unreasonable Results A mother pushes her child on a swing so that his speed is 9.00 m/s at the lowest point of his path. The swing is suspended 2.00 m above the child’s center of mass. (a) What is the magnitude of the centripetal acceleration of the child at the low point? (b) What is the magnitude of the force the child exerts on the seat if his mass is 18.0 kg? (c) What is unreasonable about these results? (d) Which premises are unreasonable or inconsistent?

Glossary

centripetal acceleration
the acceleration of an object moving in a circle, directed toward the center
ultracentrifuge
a centrifuge optimized for spinning a rotor at very high speeds
definition

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College Physics 1 Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.