6.2 Uniform Circular Motion

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

    Teacher Support

    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.

    Teacher Support

    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 Δ s Δ s 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 a_c because centripetal means center seeking.

    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." width="450" height="539" />

    Figure 6.7 The directions of the velocity of an object at two different points are shown, and the change in velocity Δ v Δ v is seen to point approximately toward the center of curvature (see small inset). For an extremely small value of Δ s Δ s , Δ v Δ v points exactly toward the center of the circle (but this is hard to draw). Because a c = Δ v / Δ t a c = Δ v / Δ t , the acceleration is also toward the center, so a_c is called centripetal acceleration.

    Teacher Support

    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 Δ s Δ s 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 a_c 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 a_c in terms of the magnitude of angular velocity . Substituting v = r ω v = r ω into the equation above, we get a c = ( r ω ) 2 r = r ω 2 a c = ( r ω ) 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 .

    Tips For Success

    The equation expressed in the form a_c = rω 2 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

    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

    Teacher Support

    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 F_net = ma. For uniform circular motion, the acceleration is centripetal acceleration: a = a_c. Therefore, the magnitude of centripetal force, F_c, 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 F c F c . 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 = m r ω 2 F c = m r 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.

    

    Figure 6.8 In this figure, the frictional force f serves as the centripetal force F_c. Centripetal force is perpendicular to tangential velocity and causes uniform circular motion. The larger the centripetal force F_c, 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 F_c 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.

    Teacher Support

    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.

    Imagine that you are swinging a yoyo in a vertical clockwise circle in front of you, perpendicular to the direction you are facing. If the string breaks just as the yoyo reaches its bottommost position, nearest the floor. What will happen to the yoyo after the string breaks?

    The yoyo will fly inward in the direction of the centripetal force. The yoyo will fly outward in the direction of the centripetal force. The yoyo will fly to the left in the direction of the tangential velocity. The yoyo will fly to the right in the direction of the tangential velocity.

    Solving Centripetal Acceleration and Centripetal Force Problems

    To get a feel for the typical magnitudes of centripetal acceleration, we’ll do a lab estimating the centripetal acceleration of a tennis racket and then, in our first Worked Example, compare the centripetal acceleration of a car rounding a curve to gravitational acceleration. For the second Worked Example, we’ll calculate the force required to make a car round a curve.

    Snap Lab

    Estimating Centripetal Acceleration

    In this activity, you will measure the swing of a golf club or tennis racket to estimate the centripetal acceleration of the end of the club or racket. You may choose to do this in slow motion. Recall that the equation for centripetal acceleration is a c = v 2 r a c = v 2 r or a c = r ω 2 a c = r ω 2 .

    • One tennis racket or golf club
    • One timer
    • One ruler or tape measure
    1. Work with a partner. Stand a safe distance away from your partner as he or she swings the golf club or tennis racket.
    2. Describe the motion of the swing—is this uniform circular motion? Why or why not?
    3. Try to get the swing as close to uniform circular motion as possible. What adjustments did your partner need to make?
    4. Measure the radius of curvature. What did you physically measure?
    5. By using the timer, find either the linear or angular velocity, depending on which equation you decide to use.
    6. What is the approximate centripetal acceleration based on these measurements? How accurate do you think they are? Why? How might you and your partner make these measurements more accurate?

    Teacher Support

    Teacher Support

    The swing of the golf club or racket can be made very close to uniform circular motion. For this, the person would have to move it at a constant speed, without bending their arm. The length of the arm plus the length of the club or racket is the radius of curvature. Accuracy of measurements of angular velocity and angular acceleration will depend on resolution of the timer used and human observational error. The swing of the golf club or racket can be made very close to uniform circular motion. For this, the person would have to move it at a constant speed, without bending their arm. The length of the arm plus the length of the club or racket is the radius of curvature. Accuracy of measurements of angular velocity and angular acceleration will depend on resolution of the timer used and human observational error.

    Grasp Check

    1. It should be simpler to use a c = r ω 2 a c = r ω 2 because measuring angular velocity through observation would be easier.
    2. It should be simpler to use a c = v 2 r a c = v 2 r because measuring tangential velocity through observation would be easier.
    3. It should be simpler to use a c = r ω 2 a c = r ω 2 because measuring angular velocity through observation would be difficult.
    4. It should be simpler to use a c = v 2 r a c = v 2 r because measuring tangential velocity through observation would be difficult.

    Worked Example

    Comparing Centripetal Acceleration of a Car Rounding a Curve with Acceleration Due to Gravity

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

    Strategy

    Because linear rather than angular speed is given, it is most convenient to use the expression a c = v 2 r a c = v 2 r to find the magnitude of the centripetal acceleration.

    Entering the given values of v = 25.0 m/s and r = 500 m into the expression for a_c gives

    a c = v 2 r = ( 25.0 m/s ) 2 500 m = 1.25 m/s 2 . a c = v 2 r = ( 25.0 m/s ) 2 500 m = 1.25 m/s 2 . Discussion

    To compare this with the acceleration due to gravity (g = 9.80 m/s 2 ), we take the ratio a c / g = ( 1.25 m/s 2 ) / ( 9.80 m/s 2 ) = 0.128 a c / g = ( 1.25 m/s 2 ) / ( 9.80 m/s 2 ) = 0.128 . Therefore, a c = 0.128 g a c = 0.128 g , which means that the centripetal acceleration is about one tenth the acceleration due to gravity.

    Worked Example

    Frictional Force on Car Tires Rounding a Curve

    1. Calculate the centripetal force exerted on a 900 kg car that rounds a 600-m-radius curve on horizontal ground at 25.0 m/s.
    2. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
    3. If the car would slip if it were to be traveling any faster, what is the coefficient of static friction between the tires and the road? Could we conclude anything about the coefficient of static friction if we did not know whether the car could round the curve any faster without slipping?

    Strategy and Solution for (a)

    We know that F c = m v 2 r F c = m v 2 r . Therefore,

    F c = m v 2 r = ( 900 kg ) ( 25.0 m/s ) 2 600 m = 938 N. F c = m v 2 r = ( 900 kg ) ( 25.0 m/s ) 2 600 m = 938 N.

    Strategy and Solution for (b)

    The image above shows the forces acting on the car while rounding the curve. In this diagram, the car is traveling into the page as shown and is turning to the left. Friction acts toward the left, accelerating the car toward the center of the curve. Because friction is the only horizontal force acting on the car, it provides all of the centripetal force in this case. Therefore, the force of friction is the centripetal force in this situation and points toward the center of the curve.

    f = F c = 938 N f = F c = 938 N

    Strategy and Solution for (c)

    If the car is about to slip, the static friction is at its maximum value and f = μ s N = μ s m g f = μ s N = μ s m g . Solving for μ s μ s , we get μ s = 938 900 × 9.8 = 0.11 μ s = 938 900 × 9.8 = 0.11 . Regardless of whether we know the maximum allowable speed for rounding the curve, we can conclude this is a minimum value for the coefficient.

    Discussion

    Since we found the force of friction in part (b), we could also solve for the coefficient of friction, since f = μ s N = μ s m g f = μ s N = μ s m g . The static friction is only equal to μ s N μ s N when it is at the maximum possible value. If the car could go faster, the friction at the given speed would still be the same as we calculated, but the coefficient of static friction would be larger.

    Practice Problems

    What is the centripetal acceleration felt by the passengers of a car moving at 12 m/s along a curve with radius 2.0 m?