# Uniform circular motion: Movement in a circle at a constant speed.

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1 9.0 - Introduction A child riding on a carousel, you riding on a Ferris wheel: Both are examples of uniform circular motion. When the carousel or Ferris wheel reaches a constant rate of rotation, the rider moves in a circle at a constant speed. In physics, this is called uniform circular motion. Developing an understanding of uniform circular motion requires you to recall the distinction between speed and velocity. Speed is the magnitude, or how fast an object moves, while velocity includes both magnitude and direction. For example, consider the car in the graphic on the right. Even as it moves around the curve at a constant speed, its velocity constantly changes as its direction changes. A change in velocity is called acceleration, and the acceleration of a car due to its change in direction as it moves around a curve is called centripetal acceleration. Although the car moves at a constant speed as it moves around the curve, it is accelerating. This is a case where the everyday use of a word acceleration and its use in physics differ. A non-physicist would likely say: If a car moves around a curve at a constant speed, it is not accelerating. But a physicist would say: It most certainly is accelerating because its direction is changing. She could even point out, as we will discuss later, that a net external force is being applied on the car, so the car must be accelerating. Uniform circular motion begins the study of rotational motion. As with linear motion, you begin with concepts such as velocity and acceleration and then move on to topics such as energy and momentum. As you progress, you will discover that much of what you have learned about these topics in earlier lessons will apply to circular motion. In the simulation shown to the right, the car moves around the track at a constant speed. The red velocity vector represents the direction and magnitude of the car s instantaneous velocity. The simulation has gauges for the x and y components of the car's velocity. Note how they change as the car travels around the track. These changes are reflected in the centripetal acceleration of the car. You can also have the car move at different constant speeds, and read the corresponding centripetal acceleration in the appropriate gauge. Is the centripetal acceleration of the car higher when it is moving faster? Note: If you go too fast, you can spin off the track. Happy motoring! Uniform circular motion Uniform circular motion: Movement in a circle at a constant speed. The toy train on the right moves on a circular track in uniform circular motion. The identical lengths of the velocity vectors in the diagram indicate a constant magnitude of velocity a constant speed. When an object is moving in uniform circular motion, its speed is uniform (constant) and its path is circular. The train does not have constant velocity; in fact, its velocity is constantly changing. Why? As you can also see in the diagram to the right, the direction of the velocity vector changes as the train moves around the track. A change in the direction of velocity means a change in velocity. The velocity vector is tangent to the circle at every instant because the train s displacement is tangent to the circle during every small interval of time. Uniform circular motion is important in physics. For instance, a satellite in a circular orbit around the Earth moves in uniform circular motion. Uniform circular motion Motion in a circle with constant speed Velocity changes! Instantaneous velocity always tangent Period Period: The amount of time it takes for an object to return to the same position. The concept of period is useful in analyzing motion that repeats itself. We use the example of the toy train shown in Concept 1 to illustrate a period. The train moves around a circular track at a constant rate, which is to say in uniform circular motion. It returns to the same position on the track after equal intervals of time. The period measures how long it takes the train to complete one revolution. In this example, it takes the train six seconds to make a complete lap around the track. When an object like a train moves in uniform circular motion, that motion is often described in terms of the period. Many other types of motion 242 Copyright Kinetic Books Co. Chapter 9

2 can be discussed using the notion of a period, as well. For example, the Earth follows an elliptical path as it moves around the Sun, and its period is called a year. A metronome is designed to have a constant period that provides musicians with a source of rhythm. The equation on the right enables you to calculate the period of an object moving in uniform circular motion. The period is the circumference of the circle, 2 r, divided by the object s speed. To put it more simply, it is distance divided by speed. Period Time to complete one revolution Period for uniform circular motion T = period r = radius v = speed What is the period of the train? Copyright Kinetic Books Co. Chapter 9 243

3 9.3 - Interactive checkpoint: a spinning CD A CD player spins a CD at 205 revolutions per minute (rpm). The CD has a diameter of 12.0 cm. How fast is a point on the outer edge of the CD moving, in? Answer: v = Centripetal acceleration Centripetal acceleration: The centrally directed acceleration of an object due to its circular motion. An object moving in uniform circular motion constantly accelerates because its direction (and therefore its velocity) constantly changes. This type of acceleration is called centripetal acceleration. Any object moving along a circular path has centripetal acceleration. In Concept 1 at the right is a vector analysis of centripetal acceleration that uses a toy train as an example of an object moving along a circular path. As the drawing indicates, the train s velocity is tangent to the circle. In uniform circular motion, the acceleration vector always points toward the center of the circle, perpendicular to the velocity vector. In other words, the object accelerates toward the center. This can be proven by considering the change in the velocity vector over a short period of time and using a geometric argument (an argument that is not shown here). Centripetal acceleration Acceleration due to change in direction in circular motion In uniform circular motion, acceleration: Has constant magnitude Points toward center The equation for calculating centripetal acceleration is shown in Equation 1 on the right. The magnitude of centripetal acceleration equals the speed squared divided by the radius. Since both the speed and the radius are constant in uniform circular motion, the magnitude of the centripetal acceleration is also constant. With uniform circular motion, the only acceleration is centripetal acceleration, but for circular motion in general, there may be both centripetal acceleration, which changes the object s direction, and acceleration in the direction of the object s motion (tangential acceleration), which changes its speed. If you ride on a Ferris wheel which is starting up, rotating faster and faster, you are experiencing both centripetal and tangential acceleration. For now, we focus on uniform circular motion and centripetal acceleration, leaving tangential acceleration as another topic. Centripetal acceleration a c = centripetal acceleration v = speed r = radius 244 Copyright Kinetic Books Co. Chapter 9

5 It can also be a normal force; for example, the walls of a clothes dryer supply a normal force that keeps the clothes moving in a circle, while the holes in those walls allow water to spin out of the fabric. Or, as in the case of the motorcycle rider in Example 1, the centripetal force can be a combination of forces, such as the normal force from the wall and the force of friction. Sometimes the source of a centripetal force is easily seen, as with a string or the walls of a dryer. Sometimes that force is invisible: The force of gravity cannot be directly seen, but it keeps the Earth in its orbit around the Sun. The centripetal force can also be quite subtle, such as when an airplane tilts or banks; the air passing over the plane s angled wings creates a force inward. In each of these examples, a force causes the object to accelerate toward the center of its circular path. Identifying the force or forces that create the centripetal acceleration is a key step in solving many problems involving circular motion. Forces and centripetal acceleration Force causes circular motion Directed toward center Any force can be centripetal Forces and centripetal acceleration F = net force m = mass v = speed r = radius A daredevil bike rider goes around a circular track. The bike and rider together have the mass shown. What is the centripetal force on them? F = (180 kg)(25 ) 2 /15 m F = (180)(625)/15 F = 7500 N 246 Copyright Kinetic Books Co. Chapter 9

6 Force directed toward center Sample problem: banked curves The car goes around a banked circular track. What bank angle will let it negotiate the curve without any friction on the tires? Curves on roads are often banked, which is to say that they are tilted at an angle instead of flat. The question above asks for the angle at which the car can go around a circular curve without requiring friction. The normal force of the road supplies the force necessary to create the centripetal acceleration. This may at first seem implausible, but the analysis in this section shows how it can be done. Draw a free-body diagram We draw a free-body diagram of the forces on the car, assuming there is no friction. The banking of the curve is designed to keep the car on course. The normal force is perpendicular to the road surface, so it is at the same angle from the vertical as the bank angle. Variables speed radius bank angle v = 10 r = 30 m x component y component normal force F Nx = F N sin F Ny = F N cos weight 0 mg sin 270 = mg What is the strategy? 1. The horizontal acceleration of the car is centripetal acceleration. Apply Newton's second law, setting the horizontal force equal to the car s mass times its centripetal acceleration. 2. Since the car does not accelerate in the vertical direction, the sum of the vertical forces (the vertical component of the normal force and the weight of the car) is zero. State this as an equation. 3. Combine these equations and simplify. Physics principles and equations Newton's second law F = ma The equation for centripetal acceleration a c = v 2 /r Copyright Kinetic Books Co. Chapter 9 247

7 Step-by-step solution The first series of steps uses the diagram above. Step Reason 1. F x = F N sin = mv 2 /r The horizontal component of the normal force is the centripetal force 2. F y = F N cos + ( mg) = 0 Newton's second law; no acceleration in vertical dimension 3. F N cos = mg rewrite 4. divide equation 1 by equation 3 5. simplify 6. trigonometric identity This gives us an equation we can use to solve the problem. Step Reason 7. = arctan (v 2 /rg) solve equation 6 for 8. enter values 9. = arctan (0.34) arithmetic 10. = 19 value of arctan The perhaps surprising and definitely fortunate result is that the bank angle does not depend on the mass of the vehicle. For a given speed and radius, the same angle will work for a tricycle or a truck. This explains why road curves have speed limit signs that announce the maximum safe speed for a vehicle without having to specify its mass Sample problem: centripetal force on a pendulum The yo-yo swings outward at the angle shown as the carousel rotates. What is the radius of the carousel from its center to the yo-yo? Your friend decides to dangle a yo-yo from its string while riding a carousel that rotates at a constant speed. The yo-yo swings outward. Knowing the angle at which the yo-yo hangs, and the period of the carousel s rotation, you can find the radius of the carousel, as the problem asks. To solve this problem, you need to consider the source of the centripetal force on the yo-yo. Since the yo-yo hangs at an angle, there is a horizontal component of the string tension. The horizontal component of tension provides the force for centripetal acceleration of the yo-yo. 248 Copyright Kinetic Books Co. Chapter 9

8 Draw a free-body diagram A free-body diagram of the yo-yo can be used to analyze the forces. We use F for the tension force. Variables period T = 7.3 s angle of yo-yo = 18 radius speed of yo-yo r v x component y component tension F x = F sin F y = F cos weight of yo-yo 0 mg sin 270 = mg What is the strategy? 1. Using Newton's second law, write an equation stating that the horizontal component of the tension creates the centripetal acceleration of the yo-yo. 2. The yo-yo does not accelerate vertically. State this as another equation using Newton's second law. 3. Combine and simplify the equations. 4. Use the relationship of period and speed to eliminate the speed of the yo-yo from the resulting equation. 5. Simplify and solve the equation. Physics principles and equations Newton's second law F = ma The equation for centripetal acceleration a c = v 2 /r An equation for the period of an object in uniform circular motion T = 2 r/v Step-by-step solution We first use Newton's second law to find an equation for the string angle involving the radius and the yo-yo s speed. Step Reason 1. F x = F sin = mv 2 /r The x component of the tension is the centripetal force 2. F y = F cos + ( mg) = 0 Newton's second law; no acceleration in vertical dimension 3. F cos = mg rewrite 4. divide equation 1 by equation 3 5. simplify 6. trigonometric identity Copyright Kinetic Books Co. Chapter 9 249

9 We created an equation that has two terms we do not know: the radius and the speed of the yo-yo. We use the relationship of period and speed to reduce the equation to one unknown term, the radius, and solve. Step Reason 7. T = 2 r/v period of object in uniform circular motion 8. v = 2 r/t solve for v 9. substitute for v in equation solve for r 11. enter values 12. r = 4.3 m evaluate Accelerating reference frames and fictitious forces Fictitious force: A perception of force caused by the acceleration of a reference frame. Imagine that you are riding in the back seat of a car, heading home after a day of tennis. Suddenly, you see a tennis ball on the floor roll toward the back of the car while you feel yourself moving back and the seat pressing harder against you. From experience, you know what has happened: The driver has pressed her foot down on the gas pedal, increasing the car s velocity. She caused the car to accelerate, which in turn caused the ball to roll and created the pushing sensation you experienced. If you were to consider this solely from the reference frame of the car, you would be hard-pressed to explain the origin of the force that accelerated the ball. In this reference frame, no force can be identified to explain why the ball accelerated toward you, because no force is being applied on it. Of course, you could look outside, but consider perhaps that it is a dark night, or you are instead in a spacecraft, where reference points are not as readily viewed. You can explain what is occurring by using a different reference frame, but we are focusing on what you perceive while inside the car, with its interior defining your reference frame. A force diagram using the car as the reference is shown in Concept 1. The normal force of the floor up balances the weight of the ball down so there is no net force in the vertical direction, and the ball is not accelerating up or down. This all fits with your understanding of physics. However, in this frame of reference the ball is accelerating backward with no apparent force on it. This defies Newton s second law: There is acceleration but no net force. So, you may decide to look for an explanation. Accelerating reference frame Objects appear to violate Newton's laws Fictitious forces seem to reconcile violation To explain this, you can do one of two things. First, you can use what physicists call a fictitious force (sometimes called a pseudoforce). You can imagine that some force pushes you and the ball backward. This new force is called a fictitious force because there is no true force you can point to (such as gravity or a normal force) to explain it. Repeat: This is a fictitious force, and by fictitious, we mean one that truly does not exist. If you leave this chapter thinking there is some force pushing the ball backward, we have failed! The ball moves backward because the reference frame (the car) is accelerating, and Newtonian mechanics do not hold in an accelerating reference frame. Non-accelerating reference frame Car's acceleration explains observed motion To explain the ball s acceleration, you could also change the reference frame to one that is not accelerating, such as the one used by an observer standing on the ground, as shown in Concept 2. (This reference frame accelerates slightly due to the Earth s motion, but we ignore this here.) From this person s perspective, the car is accelerating forward, as is the passenger inside. The normal force of the seat causes the person inside to accelerate. The ball is not accelerating, which is why it moves backward relative to the car. The ball moves at a constant velocity, not accelerating, because there is no net force acting on it. Circular motion provides several familiar situations involving fictitious forces. Consider the clothes in a dryer as shown in Concept 3. As the dryer spins, the clothes press against its walls. Does some outward force, a centrifugal force, cause this? No. In fact, the wall of the dryer pushes against the clothes inward, supplying the inward centripetal force. (A force of friction not shown in the diagram also results from this normal force.) The dryer supplies both the mechanism to put the clothes in motion and the force to keep them from exiting. 250 Copyright Kinetic Books Co. Chapter 9

12 m = mass v = speed r = radius g = acceleration due to gravity What is the minimum speed at the top for the toy to complete the loop? v = Interactive checkpoint: maximum loop-the-loop radius A roller coaster has a loop preceded by a ramp. A car is placed on the track so that its speed at the top of the loop will be What is the maximum radius that the loop can have so that the car will not fall off the track? Answer: r = m Interactive summary problem: race curves In the simulation on the right, you are asked to race a truck on an S-shaped track against the computer. This time, the first curve is covered with snow and you are racing against a snowmobile. As you go around the track, the static friction between the tires of your truck and the snow or pavement provides the centripetal force. If you go too fast, you will exceed the maximum force of friction and your truck will leave the track. If you go as fast as you can without sliding, you will beat the snowmobile. The snowmobile runs the entire race at its maximum speed. The blue truck negotiates each curve at a constant speed, but these speeds must be different for you to win the race. You set the speed of the blue truck on each curve. Straightaway sections are located at the start of the race and between the two curves. The simulation will automatically supply the acceleration you need on the straightaway sections. The blue truck has a mass of 1,800 kg. The first curve is icy, and the coefficient of static friction of the truck on this curve is (The snowmobile has a greater coefficient thanks to its snow-happy treads.) On the second curve, the coefficient of static friction is The radius of the first curve is 13 m, and the second curve is 11 m. Set the speed of the blue truck on each curve as fast Copyright Kinetic Books Co. Chapter 9 253

13 as it can go without sliding off the track, and you will win. You set the speed in increments of 0.1 in the simulation. If you need to round a value after your calculations, make sure you round down to the nearest 0.1. (If you round up, you will be exceeding the maximum safe speed.) Press GO to begin the race, and RESET if you need to try again. If you have difficulty with this problem, you may want to review the section on static friction in a previous chapter and the section on centripetal acceleration in this chapter Gotchas A car is moving around a circular track at a constant speed of 20 km/h. This means its velocity is constant, as well. Wrong. The car s velocity changes because its direction changes as it moves. Since an object moving in uniform circular motion is constantly changing direction, it is hard at any point in time to know the direction of its velocity and the direction of its acceleration. This is not true. The velocity vector is always tangent to the circle at the location of the object. Centripetal acceleration always points toward the center of the circle. No force is required for an object to move in uniform circular motion. After all, its speed is constant. Yes, but its velocity is changing due to its change in direction, which means it is accelerating. By Newton s second law, this means there must be a net force causing this acceleration. Centripetal force is another type of force. No, rather it is a way to describe what a force is doing. The normal force, gravity, tension each of these forces can be a centripetal force if it is causing an object to move in uniform circular motion Summary Uniform circular motion is movement in a circle at a constant speed. But while speed is constant in this type of motion, velocity is not. Since instantaneous velocity in uniform circular motion is always tangent to the circle, its direction changes as the object's position changes. The period is the time it takes an object in uniform circular motion to complete one revolution of the circle. Since the velocity of an object moving in uniform circular motion changes, it is accelerating. The acceleration due to its change in direction is called centripetal acceleration. For uniform circular motion, the acceleration vector has a constant magnitude and always points toward the center of the circle. Newton's second law can be applied to an object in uniform circular motion. The net force causing centripetal acceleration is called a centripetal force. Like centripetal acceleration, it is directed toward the center of the circle. A centripetal force is not a new type of force; rather, it describes a role that is played by one or more forces in the situation, since there must be some force that is changing the velocity of the object. For example, the force of gravity keeps the Moon in a roughly circular orbit around the Earth, while the normal force of the road and the force of friction combine to keep a car in circular motion around a banked curve. 254 Copyright Kinetic Books Co. Chapter 9

14 Chapter 9 Problems Conceptual Problems C.1 A string's tension force supplies the centripetal force needed to keep a yo-yo whirling in a circle. (a) What force supplies the centripetal force keeping a satellite in uniform circular motion around the Earth? (b) What kinds of forces keep a roller coaster held to a looping track? C.2 Health professionals use a device called a centrifuge to separate the different components of blood. If you allow a sample to sit long enough, Earth's gravity will cause it to separate on its own. This happens because the liquids and solids in blood have different densities. The denser solids sink to the bottom of a test tube, while less dense liquids rise to the top. To speed up the process of separation, a centrifuge spins blood sample tubes at high speeds in uniform circular motion. How does the spinning of the centrifuge speed up the separation process? C.3 You move a roller coaster with a loop-the-loop from the Earth to your new amusement park on the Moon. (a) How does the minumum speed to complete the loop-the-loop compare between the Earth and the Moon? (b) A roller coaster car starts from rest on a hill that preceeds the loop. On the Earth, if the car is released on the hill from a height h above the bottom of the loop, it will have the minimum speed required to get around the loop. Is the release height required to just get around the same loop on the Moon greater than, less than, or equal to h? (a) i. It is greater on the Moon ii. iii. (b) i. Greater ii. iii. It is less on the Moon There is no difference Less Equal C.4 Does an object moving in uniform circular motion have constant centripetal acceleration? Yes No C.5 If a satellite in a circular orbit is accelerating toward the Earth, then why doesn't the satellite hit the Earth? C.6 A father holds his three-year-old daughter's hands and swings her around in a circle, lifting her off her feet. Why is it harder for him to hold on the the faster he turns? C.7 The hammer throw is a track-and-field event, popular in Scotland, in which a ball on a rope (the "hammer") is whirled around the thrower in a circle before being released. The goal is the send the ball as a projectile as far down the field as possible. At a track meet, the circle in which a ball is whirled by a hammer thrower is at a 45 angle to the ground. To achieve the longest distance, at what point in its tilted circular orbit should the thrower release the ball? i. At its lowest point ii. iii. iv. At its highest point Halfway as it moves from the lowest to the highest point Halfway as it moves from the highest to the lowest point C.8 Two beads are tied to a string at different positions, and you swing the string around your head at a constant rate so that the beads move in uniform circular motion. Bead A is closer to your hand than bead B. Compare (a) the periods of A and B; (b) the speeds of A and B; (c) the centripetal accelerations of A and B. (a) i. A and B have the same period ii. iii. A has a longer period B has a longer period (b) i. A and B have the same speed ii. iii. A has a greater speed B has a greater speed (c) i. A and B have same the acceleration ii. iii. A has a greater acceleration B has a greater acceleration Copyright Kinetic Books Co. Chapter 9 Problems 255

15 C.9 An object is moving at a constant speed around a circle. (a) In which of these cases does the magnitude of the centripetal acceleration of the object increase? (Assume all other factors are kept the same.) (b) In which case does the centripetal acceleration increase the most? (a) The object's speed doubles The object's speed is halved The radius of the circle doubles The radius of the circle is halved (b) i. The object's speed doubles ii. iii. iv. The object's speed is halved The radius of the circle doubles The radius of the circle is halved C.10 Pretend that the Earth is rotating so fast that if you were standing at a fixed point on the equator, your weight would equal the centripetal force required to keep you in uniform circular motion around the Earth's center. If you stood on a scale at the equator, what would it read? Section Problems Section 0 - Introduction 0.1 Use the simulation in the interactive problem in this section to answer the following questions. (a) Is the centripetal acceleration of the car higher when it is moving faster? (b) If the speed of the car remains constant, do the x and y components of the car's velocity change as the car goes around the track? (a) Yes No (b) Yes No Section 2 - Period 2.1 Jupiter's distance from the Sun is meters and it takes seconds to complete one revolution of the Sun in its roughly circular orbit. What is Jupiter's speed? 2.2 Saturn travels at an average speed of around the Sun in a roughly circular orbit. Its distance from the Sun is m. How long (in seconds) is a "year" on Saturn? s 2.3 Mars travels at an average speed of around the Sun, and takes s to complete one revolution. How far is Mars from the Sun? m 2.4 Long-playing vinyl records, still used by club DJs, are 12 inches in diameter and are played at 33 1/3 revolutions per minute. What is the speed (in ) of a point on the edge on such a record? Section 4 - Centripetal acceleration 4.1 A runner rounds a circular curve of radius 24.0 m at a constant speed of What is the magnitude of the runner's centripetal acceleration? In a carnival ride, passengers are rotated at a constant speed in a seat at the end of a long horizontal arm. The arm is 8.30 m long, and the period of rotation is 4.00 s. (a) What is the magnitude of the centripetal acceleration experienced by a rider? (b) State the acceleration in "gee's," that is, as a multiple of the gravitational acceleration constant g. (a) (b) 2 g 4.3 Consider the radius of the Earth to be m. What is the magnitude of the centripetal acceleration experienced by a person (a) at the equator and (b) at the North Pole due to the Earth's rotation? (a) (b) Copyright Kinetic Books Co. Chapter 9 Problems

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