By the end of this section, you will be able to do the following:
The learning objectives in this section will help your students master the following standards:
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:
[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.
[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.
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 a c is called centripetal acceleration.
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
The equation expressed in the form ac = rω2 is useful for solving problems where you know the angular velocity rather than the tangential velocity.
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.
Click to view content 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?
[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 constant acceleration (by changing direction), we know from Newton’s second law of motion that there must be a constant net external force acting on the object. 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. Any net force causing uniform circular motion is called a centripetal force. 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 Fc. 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.
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 ′ .
This video explains why a centripetal force creates centripetal acceleration and uniform circular motion. It also covers the difference between speed and velocity and shows examples of uniform circular motion.
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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.
Click to view content 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?
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.
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 .
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. Was it more useful to use the equation a c = v 2 r a c = v 2 r or a c =r ω 2 a c =r ω 2 in this activity? Why?
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).
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 ac gives a c = v 2 r = ( 25.0m/s ) 2 500m = 1.25 m/s 2 . a c = v 2 r = ( 25.0m/s ) 2 500m = 1.25 m/s 2 .
To compare this with the acceleration due to gravity (g = 9.80 m/s2), 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 =0.128g a c =0.128g, which means that the centripetal acceleration is about one tenth the acceleration due to gravity.
We know that F c =m v 2 r
F c =m v 2 r . Therefore, F c = m v 2 r = ( 900kg ) ( 25.0m/s ) 2 600m = 938N.
F c = m v 2 r = ( 900kg ) ( 25.0m/s ) 2 600m = 938N.
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 =938N
f= F c =938N
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 mg f= μ s N= μ s mg .
9.
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?
10.
Calculate the centripetal acceleration of an object following a path with a radius of a curvature of 0.2 m and at an angular velocity of 5 rad/s.
11.
What is uniform circular motion?
12.
What is centripetal acceleration?
13.
Is there a net force acting on an object in uniform circular motion?
14.
Identify two examples of forces that can cause centripetal acceleration.
Use the Check Your Understanding questions to assess whether students master the learning objectives of this section. If students are struggling with a specific objective, the formative assessment will help identify which objective is causing the problem and direct students to the relevant content. |
What happens to the centripetal acceleration if the radius stays the same but the speed gets larger?
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