Uniform Circular Motion

6.2 Uniform Circular Motion

Section Learning Objectives

By the end of this section, you will be able to do the following:

  • Describe centripetal acceleration and relate it to linear acceleration
  • Describe centripetal force and relate it to linear force
  • Solve problems involving centripetal acceleration and centripetal force

Teacher Support

Teacher Support

The learning objectives in this section will help your students master the following standards:

  • (4) Science concepts. The student knows and applies the laws governing motion in a variety of situations. The student is expected to:
    • (C) analyze and describe accelerated motion in two dimensions using equations, including projectile and circular examples.
    • (D) calculate the effect of forces on objects, including the law of inertia, the relationship between force and acceleration, and the nature of force pairs between objects.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Circular and Rotational Motion, as well as the following standards:

  • (4) Science concepts. The student knows and applies the laws governing motion in a variety of situations. The student is expected to:
    • (C) analyze and describe accelerated motion in two dimensions using equations, including projectile and circular examples.

Section Key Terms

centrifugal force centripetal acceleration centripetal force uniform circular motion

Centripetal Acceleration

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Teacher Support

[BL][OL] Review uniform circular motion. Ask students to give examples of circular motion. Review linear acceleration.

In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.

You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes.

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Teacher Support

[BL][OL][AL] Demonstrate circular motion by tying a weight to a string and twirling it around. Ask students what would happen if you suddenly cut the string? In which direction would the object travel? Why? What does this say about the direction of acceleration? Ask students to give examples of when they have come across centripetal acceleration.

Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine Δs Δs becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration ac because centripetal means center seeking.

The diagram shows a formula: change in v is equal to v2 minus v1. There are three rays positioned on a triangle with points P, Q, and R. The ray between P and Q is labeled v1. The ray between P and R is labeled v2 and the ray opposite P between Q and R is labeled change in v. Below the formula and triangle is a circle with center A and two radii labeled r with the angle between them labeled change in theta. The points on the circumference for the two radii are labeled B and C with a line segment of change in r indicated for the distance between the two points. The arc going from B to C along the circumference is labeled change of s and the velocity arrow at point B is labeled v1 and at C is labeled v2.
Figure 6.7 The directions of the velocity of an object at two different points are shown, and the change in velocity Δv Δv is seen to point approximately toward the center of curvature (see small inset). For an extremely small value of Δs Δs, Δv Δv points exactly toward the center of the circle (but this is hard to draw). Because a c =Δv/Δt a c =Δv/Δt, the acceleration is also toward the center, so ac is called centripetal acceleration.

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Teacher Support

Consider Figure 6.7. The figure shows an object moving in a circular path at constant speed and the direction of the instantaneous velocity of two points along the path. Acceleration is in the direction of the change in velocity and points toward the center of rotation. This is strictly true only as Δs Δs tends to zero.

Now that we know that the direction of centripetal acceleration is toward the center of rotation, let’s discuss the magnitude of centripetal acceleration. For an object traveling at speed v in a circular path with radius r, the magnitude of centripetal acceleration is

a c = v 2 r . a c = v 2 r .

Centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you may have noticed when driving a car, because the car actually pushes you toward the center of the turn. But it is a bit surprising that ac is proportional to the speed squared. This means, for example, that the acceleration is four times greater when you take a curve at 100 km/h than at 50 km/h.

We can also express ac in terms of the magnitude of angular velocity. Substituting v=rω v=rω into the equation above, we get a c = (rω) 2 r =r ω 2 a c = (rω) 2 r =r ω 2 . Therefore, the magnitude of centripetal acceleration in terms of the magnitude of angular velocity is

a c =r ω 2 . a c =r ω 2 .
6.9

Tips For Success

The equation expressed in the form ac = rω2 is useful for solving problems where you know the angular velocity rather than the tangential velocity.

Virtual Physics

Ladybug Motion in 2D

In this simulation, you experiment with the position, velocity, and acceleration of a ladybug in circular and elliptical motion. Switch the type of motion from linear to circular and observe the velocity and acceleration vectors. Next, try elliptical motion and notice how the velocity and acceleration vectors differ from those in circular motion.

Grasp Check

In uniform circular motion, what is the angle between the acceleration and the velocity? What type of acceleration does a body experience in the uniform circular motion?

  1. The angle between acceleration and velocity is 0°, and the body experiences linear acceleration.
  2. The angle between acceleration and velocity is 0°, and the body experiences centripetal acceleration.
  3. The angle between acceleration and velocity is 90°, and the body experiences linear acceleration.
  4. The angle between acceleration and velocity is 90°, and the body experiences centripetal acceleration.

Centripetal Force

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Teacher Support

[BL][OL][AL] Using the same demonstration as before, ask students to predict the relationships between the quantities of angular velocity, centripetal acceleration, mass, centripetal force. Invite students to experiment by using various lengths of string and different weights.

Because an object in uniform circular motion undergoes acceleration (by changing the direction of motion but not the speed), we know from Newton’s second law of motion that there must be a net external force acting on the object. Since the magnitude of the acceleration is constant, so is the magnitude of the net force, and since the acceleration points toward the center of the rotation, so does the net force.

Any force or combination of forces can cause a centripetal acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, the friction between a road and the tires of a car as it goes around a curve, or the normal force of a roller coaster track on the cart during a loop-the-loop.

The component of any net force that causes circular motion is called a centripetal force. When the net force is equal to the centripetal force, and its magnitude is constant, uniform circular motion results. The direction of a centripetal force is toward the center of rotation, the same as for centripetal acceleration. According to Newton’s second law of motion, a net force causes the acceleration of mass according to Fnet = ma. For uniform circular motion, the acceleration is centripetal acceleration: a = ac. Therefore, the magnitude of centripetal force, Fc, is F c =m a c F c =m a c .

By using the two different forms of the equation for the magnitude of centripetal acceleration, a c = v 2 /r a c = v 2 /r and a c =r ω 2 a c =r ω 2 , we get two expressions involving the magnitude of the centripetal force FcFc. The first expression is in terms of tangential speed, the second is in terms of angular speed: F c =m v 2 r F c =m v 2 r and F c =mr ω 2 F c =mr ω 2 .

Both forms of the equation depend on mass, velocity, and the radius of the circular path. You may use whichever expression for centripetal force is more convenient. Newton’s second law also states that the object will accelerate in the same direction as the net force. By definition, the centripetal force is directed towards the center of rotation, so the object will also accelerate towards the center. A straight line drawn from the circular path to the center of the circle will always be perpendicular to the tangential velocity. Note that, if you solve the first expression for r, you get

r= m v 2 F c . r= m v 2 F c .

From this expression, we see that, for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

The figure shows a semicircle with an arrow indicating a counterclockwise Path with radius labeled r, and arrow pointing from the outside of the circle towards the radius arrow. The arrow opposite the radius is labeled large r and has an f indicating force. The arrow tangential to the circle between the radius and large r arrows is labeled v (velocity). Below the first semicircle is a formula: f equals Fc is parallel to ac since Fc equals m times a alpha. Another semicircle with the same directional path is drawn below formula with a radius r prime, velocity v, large Fc pointing toward the r prime radius arrow and a label below the large Fc showing smaller r prime and same v.
Figure 6.8 In this figure, the frictional force f serves as the centripetal force Fc. Centripetal force is perpendicular to tangential velocity and causes uniform circular motion. The larger the centripetal force Fc, the smaller is the radius of curvature r and the sharper is the curve. The lower curve has the same velocity v, but a larger centripetal force Fc produces a smaller radius r ′ r ′ .

Watch Physics

Centripetal Force and Acceleration Intuition

This video explains why centripetal force, when it is equal to the net force and has constant magnitude, creates centripetal acceleration and uniform circular motion.

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Teacher Support

Misconception Alert

Some students might be confused between centripetal force and centrifugal force. Centrifugal force is not a real force but the result of an accelerating reference frame, such as a turning car or the spinning Earth. Centrifugal force refers to a fictional center fleeing force.