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 í acceleration í 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
andy
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!9.1 - 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 í 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
Instantaneous velocity always tangent
· Velocity changes!
9.2 - 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
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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,
2ʌr
, 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?
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 m/s?
Answer:
v =
m/s
9.4 - 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).
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
Acceleration due to change in direction
in circular motion
In uniform circular motion, acceleration:
· Has constant magnitude
· Points toward center
Centripetal acceleration
a
c= centripetal acceleration
v = speed
r = radius
What is the centripetal
acceleration of the train?
Accelerates toward center
9.5 - Interactive problems: racing in circles
The two simulations in this section let you experience uniform circular motion and centripetal acceleration as you race your car against the computer’s.
In the first simulation, you race a car around a circular track. Both your car and the computer’s move around the loop at constant speeds. You control the speed of the blue car. Halfway around the track, you encounter an oil slick. If the centripetal acceleration of your car is greater than 3.92 m/s2 at this point, it will leave the track
and you will lose. The radius of the circle is 21.0 m.
To win the race, set the centripetal acceleration equal to 3.92 m/s2 in the centripetal
acceleration equation, solve for the velocity, and then round down the velocity to the nearest 0.1 m/s; this is a value that will keep your car on the track and beat the computer car. Enter this value using the controls in the simulation. Press GO to start the simulation and test your calculations.
In the second simulation, the track consists of two half-circle curves connected by a straight section. Your blue car runs the entire race at the speed that you set for it. You want to set this speed to just keep the car on the track. The first curve has a radius of 14.0 meters; the second, 8.50 meters. On either curve, if the centripetal acceleration of your car exceeds 9.95 m/s2, its tires will lose traction on the curve,
causing it to leave the track. If your car moves at the fastest speed possible without leaving the track, it will win. Again, calculate the speed of the blue car on each curve but using a centripetal acceleration value of 9.95 m/s2, and round down to the
nearest 0.1 m/s. Since the car will go the same speed on both curves, you need to decide which curve determines your maximum speed. Enter this value, then press GO.
If you have difficulty solving these interactive problems, review the equation relating centripetal acceleration, circle radius, and speed.
9.6 - Newton's second law and centripetal forces
If you hold onto the string of a yo-yo and twirl it in a circle overhead, as illustrated in
Concept 1, you know you must hold the string firmly or the yo-yo will fly away from you. This is true even when the toy moves at a constant speed. A force must be applied to keep the yo-yo moving in a circle.
A force is required because the yo-yo is accelerating. Its change in direction means its velocity is changing. Using Newton’s second law,
F = ma
, we can calculate the amount of force as the product of the object’s mass and its centripetal acceleration. That equation is shown in Equation 1. It applies to any object moving in uniform circular motion. The force, called a centripetal force, points in the same direction as the acceleration, toward the center of the circle.The term “centripetal” describes any force that causes circular motion. A centripetal force is not a new type of force. It can be the force of tension exerted by a string, as in the yo-yo example, or it can be the force of friction, such as when a car goes around a curve on a level road.
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 m/s)
2/15 m
F = (180)(625)/15
F = 7500 N
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Force directed toward center
9.7 - Sample problem: banked curves
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
ș
from the vertical as the bank angle.Variables
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
ȈF = ma
The equation for centripetal acceleration
a
c= v
2/r
The car goes around a banked
circular track. What bank angle ș will
let it negotiate the curve without any
friction on the tires?
speed
v = 10 m/s
radius
r = 30 m
bank angle
ș
x component
y component
normal force
F
Nx= F
Nsin ș
F
Ny= F
Ncos ș
weight
0
mg sin 270° = ímg
Step-by-step solution
The first series of steps uses the diagram above.
This gives us an equation we can use to solve the problem.
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.
Step
Reason
1. ȈF
x= F
Nsin ș = mv
2/r
The horizontal component of the normal force is the centripetal
force
2. ȈFy
= F
Ncos ș + (–mg) = 0
Newton's second law; no acceleration in vertical dimension
3. FN
cos ș = mg
rewrite
4.
divide equation 1 by equation 3
5.
simplify
6.
trigonometric identity
Step
Reason
7.
ș = arctan (v
2/rg)
solve equation 6 for ș
8.
enter values
9. ș = arctan (0.34)
arithmetic
10. ș = 19°
value of arctan
9.8 - Sample problem: centripetal force on a pendulum
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.
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?
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
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
ȈF = ma
The equation for centripetal acceleration
a
c= v
2/r
An equation for the period of an object in uniform circular motion
T = 2ʌr/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.
period
T = 7.3 s
angle of yo-yo
ș = 18°
radius
r
speed of yo-yo
v
x component
y component
tension
F
x= F sin ș
F
y= F cos ș
weight of yo-yo
0
mg sin 270° = ímg
Step
Reason
1. ȈFx
= F sin ș = mv
2/r
The x component of the tension is the centripetal force
2. ȈFy
= F cos ș + (–mg) = 0
Newton's second law; no acceleration in vertical dimension
3. F cos ș = mg
rewrite
4.
divide equation 1 by equation 3
5.
simplify
6.
trigonometric identity
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 = 2ʌr/v
period of object in uniform circular motion
8. v = 2ʌr/T
solve for v
9.
substitute for v in equation 6
10.
solve for r
11.
enter values
12. r = 4.3 m
evaluate
9.9 - 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. 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.
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.
Accelerating reference frame
Objects appear to violate Newton's laws
Fictitious forces seem to reconcile
violation
Non-accelerating reference
frame
Car's acceleration explains observed
motion
Clothes presumably do not consider the acceleration of their reference frames, but you factor this into your behavior at times. For example, when you are in a car that is traveling around a curve in the road, you may feel as if there is a force pulling you outward. And, if the car is moving fast enough or the curve is particularly tight, you might push against the car with your foot or hand to stabilize yourself. Doing so increases the normal force of the car on your body, causing your body to remain in a circular path.
Despite the efforts of physicists, the concept of centrifugal force is hard to eradicate. Perhaps explaining accelerating reference frames is just too hard. For instance, a U.S. government press release about the safety of a carnival ride states: “The ride is a whirling cylinder which uses centrifugal force to hold the riders to their seats as the seats rise….”
Accelerating reference frame,
circular
Circular motion linked to fictitious forces
Circular motion requires centripetal
force
· Outward centrifugal “force” (fictitious)
9.10 - Artificial gravity
When a spacecraft is far away from the Earth and any other massive body, the force of gravity is near zero. As a result of this lack of gravitational force, the astronauts and their equipment float in space. Although perhaps amusing to watch and experience, floating presents an unusual challenge because humans are accustomed to working in environments with enough gravity to keep them, and their equipment, anchored to the floor. (Note: This same feeling of weightlessness occurs in a spacecraft orbiting the Earth, but the cause of the apparent weightlessness in this case arises from the free-fall motion of the craft and the astronaut, not from a lack of gravitational force.)
A number of science fiction books and films have featured spacecraft that rotate very slowly as they travel through the universe. Arthur C. Clarke’s science fiction novel Rendezvous with Rama is set in a massive rotating spacecraft, as is part of the movie 2001, which is also based on his work. This rotation supplies artificial gravity í the illusion of gravity í to the astronauts and their equipment. Artificial gravity has effects similar to true gravity, and as a result can mislead the people riding in such machines to believe they are experiencing true gravitational force.
Why does the rotation of a spacecraft produce the sensation of gravity? Consider what happens when an airplane takes off from a runway: You feel a force pulling you back into your seat, as if the force of gravity were increasing. The force of gravity has not been significantly altered (in fact, it decreases a bit as you gain elevation). However, while the airplane accelerates upward, you feel a greater normal force pushing up from your seat, and you may interpret this subconsciously as increased gravity.
A roughly analogous situation occurs on a rotating spaceship. The astronauts are rotating in uniform circular motion. The outside wall of the station (the floor, from the astronauts’ perspective) provides the centrally directed normal force that is the centripetal force. This force keeps the astronauts moving in a circle. From the astronauts’ perspective, this force is upwards, and they relate it to the upward normal force of the ground they feel when standing on the Earth. On Earth, the normal force is equal but opposite to the force of gravity. Because they typically associate the normal force with gravity, the astronauts may erroneously perceive this force from the spacecraft floor as being caused by some form of artificial or simulated gravity. Artificial gravity is a pseudo, or fictitious, force. The astronauts assume it exists because of the normal force. The perception of this fictitious force is a function of the
acceleration of the astronauts as they move in a circle. It would disappear if the spacecraft stopped rotating.
Although discussed as the realm of science fiction, real-world carnival rides (like the “Gravitron”) use this effect. Riders are placed next to the wall of a cylinder. The cylinder then is rotated at a high speed and the floor (or seats) below the riders is lowered. The walls of the cylinder supply a normal force and the force of friction keeps the riders from
A rotating space station provides an illusion of gravity.
Artificial gravity
Space station rotates
Floor of craft provides centripetal force
Person (incorrectly) assumes normal
force counters force of gravity
To simulate Earth's gravity, what
should the radius of the space
slipping down.
station be?
9.11 - Loop-the-loop
Above, our toy is successfully completing a circle around the loop, in the process “defying” gravity. This loop-the-loop scenario provides another opportunity to apply the concept of centripetal force.
We ignore friction and consider only the gravitational force and the normal force on the toy from the track. The normal force always points toward the center of the loop, while the gravitational force always points down. Consequently, the net force on the toy and its speed change as it goes around the loop. For example, the normal force is horizontal when the toy is at the 3 o’clock position and vertical at the top of the loop.
We apply Newton’s second law to the toy at the top of the loop, where the gravitational and normal forces both point down toward the center of the loop. At that point, the sum of the normal and gravitational forces equals the toy’s mass times its centripetal acceleration.
At the top of the loop, we will assume the toy is moving at the minimum speed required to keep from falling off the track. At this minimum speed, the normal force from the track is zero. This means the only force acting on the toy at this point is gravity; it provides all the centripetal force.
To derive the equation for the minimum speed required to negotiate the top of the loop, we set the weight equal to the mass times the centripetal acceleration. By re-arranging this equation, we can solve for the speed, as we show in Equation 1 on the right. Note that the toy’s mass is not a factor in this equation. To apply this equation precisely, the radius is that of the circle of motion of the car's center of mass, not the radius of the track itself.
Since in most cases the center of mass of the car is close to the loop’s track, the radius of the loop is often used instead. For the toy on the right, this is a significant
approximation, but for typical roller coasters, it is much more reasonable.
If the toy moves faster at the top of the loop than the speed calculated in the equation, the centripetal acceleration will be greater. The normal force of the track will supply the additional force required for the increased acceleration. If the toy is moving more slowly than the speed calculated on the right, it will lose contact with the track.
As mentioned, the equation on the right shows how to calculate the minimum speed at the top of the loop required for a given radius of the object's motion. The greater the radius, the greater the speed required. To give you an idea of the speed required for navigating a loop built on a larger scale, the minimum speed for a radius of 10 meters is 9.9 m/s (about 36 km/h).
Although roller coasters are lightly regulated, efforts are underway to legislate a maximum G force. This force represents the rider’s perceived weight as a multiple of his normal weight. When experiencing a force of 2 G, the rider feels twice as heavy as usual. The perceived weight is a combination of the rider's actual weight and the forces creating the acceleration of the roller coaster. Suggested maximum G forces range from 2.5 G to 3 G. Here is an example to give you an idea of the accelerations involved. For ease of comparison, we assume the same speed at the top and bottom of the loop, though this would not be true for a real coaster. A person moving in a vertical loop with a radius of 10 meters, at a constant speed of 16 m/s (58 km/h), experiences a 1.6 G force at the top of the loop but a sensational 3.6 G force at the bottom.
A toy car loops the loop.
At top of loop:
Weight plus any normal force are
centripetal force(s)
At minimum speed, normal force is zero
At top of loop:
Minimum speed at top of loop
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 = 1.6 m/s
9.12 - 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 15.0 m/s. What is the
maximum radius that the loop can
have so that the car will not fall off the
track?
Answer:
r =
m
9.13 - 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 0.51. (The snowmobile has a greater coefficient thanks to its snow-happy treads.) On the second curve, the coefficient of
static friction is 0.84. 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
253
as it can go without sliding off the track, and you will win.
You set the speed in increments of 0.1 m/s in the simulation. If you need to round a value after your calculations, make sure you round down to the nearest 0.1 m/s. (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.
9.14 - 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 í each of these forces can be a centripetal force if it is causing an object to move in uniform circular motion.
9.15 - 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.
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. It is less on the Moon iii. There is no difference (b) i. Greater
ii. Less iii. 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. At its highest point
iii. Halfway as it moves from the lowest to the highest point iv. 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. A has a longer period iii. B has a longer period (b) i. A and B have the same speed
ii. A has a greater speed iii. B has a greater speed
(c) i. A and B have same the acceleration ii. A has a greater acceleration iii. B has a greater acceleration
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. The object's speed is halved iii. The radius of the circle doubles iv. 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 7.78×1011 meters and it takes 3.74×108 seconds to complete one revolution of the Sun in its roughly circular orbit. What is Jupiter's speed?
m/s
2.2 Saturn travels at an average speed of 9.66×103 m/s around the Sun in a roughly circular orbit. Its distance from the Sun is 1.43×1012 m. How long (in seconds) is a "year" on Saturn?
s
2.3 Mars travels at an average speed of 2.41×104 m/s around the Sun, and takes 5.94×107 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 m/s) of a point on the edge on such a record?
m/s
Section 4 - Centripetal acceleration
4.1 A runner rounds a circular curve of radius 24.0 m at a constant speed of 5.25 m/s. What is the magnitude of the runner's
centripetal acceleration? m/s2
4.2 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) m/s2
(b) g
4.3 Consider the radius of the Earth to be 6.38×106 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) m/s2
(b) m/s2
4.4 When tires are installed or reinstalled on a car, they are usually first balanced on a device that spins them to see if they wobble. A tire with a radius of 0.380 m is rotated on a tire balancing device at exactly 460 revolutions per minute. A small stone is embedded in the tread of the tire. What is the magnitude of the centripetal acceleration experienced by the stone?
m/s2
4.5 A toy airplane connected by a guideline to the top of a flagpole flies in a circle at a constant speed. If the plane takes 4.5 s to complete one loop, and the radius of the circular path is 11 m, what is the magnitude of the plane's centripetal acceleration?
m/s2
4.6 You tie a string to a rock and twirl it at a constant speed in a horizontal circle with a radius of 1.30 m, 2.10 m above the ground. The rock comes loose and travels as a projectile a horizontal distance of 9.40 m before striking the ground. (a) What was the magnitude of the centripetal acceleration of the rock when it was on the string? (b) What was its speed when it was on the string?
(a) m/s2
(b) m/s
Section 5 - Interactive problems: racing in circles
5.1 Use the information given in the first interactive problem in this section to calculate the initial speed that keeps the blue car on
the track and wins the race. For safety, round your answer down to the nearest 0.1 m/s. Test your answer using the simulation.
m/s
5.2 Use the simulation in the second interactive problem in this section to calculate the initial speed that keeps the blue car on
the track and wins the race. For safety, round your answer down to the nearest 0.1 m/s. m/s
Section 6 - Newton's second law and centripetal forces
6.1 An astronaut in training rides in a seat that is moved in uniform circular motion by a radial arm 5.10 meters long. If her speed
is 15.0 m/s, what is the centripetal force on her in "G's," where one G equals her weight on the Earth? "G's"
6.2 A ball with mass 0.48 kg moves at a constant speed. A centripetal force of 23 N acts on the ball, causing it to move in a circle
with radius 1.7 m. What is the speed of the ball? m/s
6.3 A bee loaded with pollen flies in a circular path at a constant speed of 3.20 m/s. If the mass of the bee is 133 mg and the
radius of its path is 11.0 m, what is the magnitude of the centripetal force? N
6.4 Fifteen clowns are late to a party. They jump into their sporty coupe and start driving. Eventually they come to a level curve, with a radius of 27.5 meters. What is the top speed at which they can drive successfully around the curve? The coefficient of static friction between the car's tires and the road is 0.800.
m/s
6.5 You are playing tetherball with a friend and hit the ball so that it begins to travel in a circular horizontal path. If the ball is 1.2 meters from the pole, has a speed of 3.7 m/s, a mass of 0.42 kilograms, and its (weightless) rope makes a 49° angle with the pole, find the tension force that the rope exerts on the ball just after you hit it.
N
6.6 A car with mass 1600 kg drives around a flat circular track of radius 28.0 m. The coefficient of friction between the car tires and the track is 0.830. How fast can the car go around the track without losing traction?
m/s
Section 7 - Sample problem: banked curves
7.1 A car with mass 2200 kg goes around a banked circular track in a path with radius 23 m. The track makes an angle of 15°
with the horizontal. How fast should the car go around the track so that there will be no "sideways" friction force acting on the tires?
m/s
7.2 A car goes around the circular banked curve
shown in the illustration at a speed of 21 m/s. At what radius does it travel if there is no frictional force between the car tires and the track?
m
7.3 What would the bank angle be for a circular racetrack with radius 120 m so that a car can go around the curve safely at a
maximum of 21 m/s, without the help of frictional force to keep it on the road? °
7.4 A jet of mass 9.7×104 kg makes a horizontal circular turn and "banks" or tilts its wings as it does so. The turn is 1200 m in radius and the jet has a constant speed of 160 m/s. The lift created by the plane's wings is perpendicular to the wing surface. (a) What is the magnitude of the lift force on the wings? (b) What is the angle of the lift force with respect to the horizontal?
(a) N
(b) °
Section 8 - Sample problem: centripetal force on a pendulum
8.1 You are at the carnival and decide to go onthe swing ride. It is a high rotating platform from which swing seats hang like pendulums. As the platform begins to turn, your swing's chain does not stay perpendicular to the ground but angles out from the vertical. If your distance from the center of rotation is 8.50 meters and you go around once every 12.0 seconds, what angle does your swing make with the vertical?
°
8.2 You have lovingly restored a red 1964 Ford Falcon two-door sedan, complete with a pair of red fuzzy dice hanging from the rear-view mirror. When you go down a straight road at 26.0 m/s, the dice hang straight down, but when you enter a flat circular turn with radius 97.0 m at the same speed, they hang at an angle ș from the vertical. What is ș?
°
8.3 A tetherball is suspended on a 3.80 m rope from a tall pole. The ball is hit so that it travels in a horizontal circle around the pole with a constant speed of 5.60 m/s. What angle does the rope make with the pole?
°
Section 9 - Accelerating reference frames and fictitious forces
9.1 You hang a ball on a string from the ceiling inside a parked minivan. Then your friend starts up the van and drives forward at
a constant acceleration while you measure the angle the string makes from its original position. If the angle is 9.00°, what is the magnitude of the van's acceleration?
m/s2
Section 10 - Artificial gravity
10.1 A rotating space station has radius 1.31e+3 m, measured from the center of rotation to the outer deck where the crew lives.
What should the period of rotation be if the crew is to feel that they weigh one-half their Earth weight? s
Section 11 - Loop-the-loop
11.1 You are designing a roller coaster for a new amusement park on the Moon. You envision a loop with a radius of 35 meters in
your track. (a) What minimum speed at the top will a car have to possess in order to successfully negotiate the loop? The gravitational acceleration on the Moon is 1.6 m/s2. (b) What speed at the top of the loop would a car need to have if the same roller coaster was built on Earth?
(a) m/s
(b) m/s
11.2 A car drives over a hill that is shaped as a circular arc with radius 65.0 m. The car has a constant speed of 14.0 m/s and a
mass of 423 kg. What is the magnitude of (a) the centripetal force on the car at the top of the hill and (b) the normal force exerted on the car by the road at this point?
(a) N
(b) N
11.3 A road has a hill with a top in the shape of a circular arc of radius 32.0 m. How fast can a car go over the top of the hill
without losing contact with the ground? m/s
11.4 A roller coaster consists of a hill followed by a loop. A car is released from rest on the hill from a height h above the bottom of the loop. If the car has the minimum speed at the top of the loop to avoid falling off, what is the ratio of h to r, the radius of the loop?
to 1
11.5 A roller coaster car does a vertical loop-the-loop with a radius of 23.0 meters. The centripetal force on a person in the car is 2.50 G's, which is 2.50 times her weight on Earth. How many times faster than the minimum velocity required to get around the loop is she going?
times the minimum velocity
Section 13 - Interactive summary problem: race curves
13.1 Use the simulation in the interactive problem in this section to calculate the speed for (a) the first curve and (b) the second
curve that keeps the blue truck on the track and wins the race. For safety, round your answer down to the nearest 0.1 m/s.
(a) m/s
(b) m/s
Additional Problems
A.1 Rebecca goes on a popular ride at an amusement park that involves a cylinder rotating on a vertical axis, with a radius of
6.20 m. Rebecca stands inside, with her back against the wall of the cylinder. The ride rotates and presses Rebecca against the wall of the cylinder, and when she reaches a speed v, the floor of the ride drops so that she can no longer stand on it. The coefficient of static friction between the wall and Rebecca's clothing is 0.820. What should Rebecca's minimum speed v be when the floor is lowered, so that she does not fall?
m/s
A.2 The solar system moves around the galactic center in a roughly circular path at a radius of about 2.6×1020 m. The system's orbital speed around the center is 2.2×105 m/s. (a) What is the period of this circular motion in years? (b) What is the sun's centripetal acceleration? (c) What is the magnitude of the centripetal force keeping the sun in this path given that the sun's mass is 1.99×1030 kg?
(a) years
(b) m/s2
(c) N
A.3 A soapbox racer rounds a curve banked at 28.0° with a radius of 21.0 meters. The tread on the racer's old tires is nearly gone
so it can't provide much friction to keep the racer on track. How fast can the racer go around the curve without requiring friction to stay on track?
m/s
A.4 What period (in hours) would the Earth have to have for a person standing at the equator to experience a centripetal force
equal to the gravitational force exerted on him by the Earth? When moving about, this person would be literally walking on air. The radius of the Earth is 6.38×106 meters.
hours
A.5 An ant and a beetle are riding a spinning
record on a record player. The ant stands on the outside edge of the record while the beetle stands on the record's label. The ant is four times as far from the center of rotation as the beetle. What is the ratio of the ant's centripetal acceleration to the beetle's?
to 1
A.6 A dirty teddy bear is thrown into a top-loading washing machine. The cylinder in the machine rotates about a vertical axis. At what minimum speed does the wall of the cylinder have to move to stick the teddy to it so that he doesn't slide down? At this point, the frictional force acting upwards on the bear will balance his weight (this is the spin cycle, so the water has been drained and there is no buoyant force). The coefficient of static friction between a wet teddy bear and a washing machine wall is 0.500 and the washing machine cylinder has a radius of 0.310 meters.
m/s
A.7 A car is traveling down a straight road at a constant 17 m/s. A tire on the car has a radius of 0.35 m. Imagine a dot on the edge of the tire. It will rotate in uniform circular motion with respect to the center of the tire, but the whole tire along with the rest of the car is moving linearly forward at the same time. What is the instantaneous velocity of the dot when it is (a) at the lowest point of its motion (on the road) and (b) at the highest point of its motion? (c) What is the average horizontal speed of the dot? (You do not need to do any calcuations to answer this last question.)
(a) m/s
(b) m/s
(c) m/s
A.8 A car drives around two circular curves on two different roads. The two curves have the same radius of curvature. Coincidentally, the maximum speed that the car can drive through either of the turns is the same for both roads. The car drives through both turns at this speed. The first road is frictionless, but it is banked at 14 degrees off the horizontal. The other turn is flat. What is the coefficient of friction between the car tires and the road on the unbanked turn?
A.9 A relay satellite for satellite TV orbits the Earth above the equator at an altitude of 3.579×104 kilometers. This is the altitude required for a geosynchronous orbit í that is, for the satellite to remain above the same point on the Earth's equator as the Earth rotates. What is the speed of the satellite? The Earth has a radius of 6380 kilometers. (Note that a geosynchronous satellite's altitude is several times Earth's radius!)
m/s
A.10 While on vacation you rent a moped and go cruising around a tropical island. If you take a 10.0° banked curve with a radius of 15.0 meters at 12.0 m/s, what is the minimum coefficient of static friction between your tires and the road to keep the tires from slipping as you negotiate the curve?
A.11 You are a traffic safety engineer in charge of determining safe speeds for roads. A particular banked curve has a radius of 11.0 meters and is banked at an angle of 8.00°. The coefficient of static friction between common tires and this road is 0.870. What is the maximum speed that a car can drive this curve? Use both the bank of the curve and the friction on the tires in determining your answer.
m/s
