Gravity and Uniform Circular Motion (UCM)

Name: Gravity and Uniform Circular Motion (UCM)
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Description: Gravity and Uniform Circular Motion (UCM)

Gravity and Uniform Circular Motion (UCM)
© 2015 Pearson Education, Inc.

Section 6.1 Uniform Circular Motion

Review of the Acceleration and ∆ 𝒗 vector

Motion in Two Dimensions: Circular Motion
For circular motion at a constant speed, the acceleration vector a points toward the center of the circle. An acceleration that always points directly toward the center of a circle is called a centripetal acceleration. Centripetal acceleration is just the name for a particular type of motion. It is not a new type of acceleration. © 2015 Pearson Education, Inc.

Motion in Two Dimensions: Circular Motion

Question 3 A toy car moves around a circular track at constant speed. It suddenly doubles its speed — a change of a factor of 2. As a result, the centripetal acceleration changes by a factor of 1/4 1/2 No change since the radius doesn’t change. 2 4 Answer: E © 2015 Pearson Education, Inc. 12

Question 3 A toy car moves around a circular track at constant speed. It suddenly doubles its speed — a change of a factor of 2. As a result, the centripetal acceleration changes by a factor of 1/4 1/2 No change since the radius doesn’t change. 2 4 © 2015 Pearson Education, Inc. 13

Question 4 Acceleration in the turn
World-class female short-track speed skaters can cover the 500 m of a race in 45 s. Turns are very tight, with a radius of approximately 11 m. Estimate the magnitude of the skater’s centripetal acceleration in a turn. © 2015 Pearson Education, Inc.

Question 4 Acceleration in the turn (cont.)
Centripetal acceleration depends on two quantities: The radius of the turn (given as approximately 11 m) The speed. (speed varies during the race, but we get an average speed by using the total distance and time) Therefore: This is a large acceleration—a bit more than g. © 2015 Pearson Education, Inc.

Example Problem Turning a corner at a typical large intersection in a city means driving your car through a circular arc with a radius of about 25 m. If the maximum advisable acceleration of your vehicle through a turn on wet pavement is 0.40 times the free-fall acceleration. What is the maximum speed at which you should drive through this turn? Answer: 9.9 m/s (22 mph) Answer: 9.9 m/s (22 mph)

Question 5 A ball at the end of a string is being swung in a horizontal circle. The ball is accelerating because The speed is changing. The direction is changing. The speed and the direction are changing. The ball is not accelerating. Answer: B © 2015 Pearson Education, Inc.

Question 5 A ball at the end of a string is being swung in a horizontal circle. The ball is accelerating because The speed is changing. The direction is changing. The speed and the direction are changing. The ball is not accelerating. © 2015 Pearson Education, Inc.

Period, Frequency, and Speed
Period (T) = The time interval it takes an object to go around a circle one time. Frequency = The number of revolutions per second or the # of times an object goes around a circle in 1 sec. Note: the SI unit of frequency is inverse seconds, or s–1.

Period, Frequency, and Speed
We can combine this with the expression for centripetal acceleration: 𝑜𝑟 4 𝜋 2 𝑟 𝑇 2 © 2015 Pearson Education, Inc.

Question 6: Spinning some tunes
An audio CD has a diameter of 120 mm and spins at up to 540 rpm. When a CD is spinning at its maximum rate, how much time is required for one revolution? If a speck of dust rides on the outside edge of the disk, how fast is it moving? What is the acceleration? So we need to find: © 2015 Pearson Education, Inc.

Question 6: Spinning some tunes
The diameter of a CD is given as 120 mm, which is m. The radius is m. The frequency is given in rpm; we need to convert this to s−1: Period (T) = © 2015 Pearson Education, Inc.

Question 6 Spinning some tunes (cont.)
The dust speck is moving in a circle of radius m at a frequency of 9.0 s−1. To find the speed: To find the acceleration (centripetal): © 2015 Pearson Education, Inc.

Example 6: Spinning some tunes (cont.)
The speed we calculate for the dust speck is nearly 8 mph, And we’d expect that such a high speed in a small circle would lead to a very large acceleration. © 2015 Pearson Education, Inc.

Section 6.2 Dynamics of Uniform Circular Motion

Dynamics of Uniform Circular Motion
Riders traveling around on a circular carnival ride are accelerating, as we have seen: © 2015 Pearson Education, Inc.

Dynamics of Uniform Circular Motion
not a new kind of force: The is due to a force such as tension, friction, or the normal force. © 2015 Pearson Education, Inc.

Question 8 A ball at the end of a string is being swung in a horizontal circle. What force is producing the centripetal acceleration of the ball? Gravity Air resistance Normal force Tension in the string Answer: D © 2015 Pearson Education, Inc.

Question 9 A ball at the end of a string is being swung in a horizontal circle. What is the direction of the net force on the ball? Tangent to the circle Toward the center of the circle There is no net force. Answer: B © 2015 Pearson Education, Inc.

Question 10 A coin is rotating on a turntable; it moves without sliding. At the instant shown in the figure, which arrow gives the direction of the coin’s velocity? Answer: A © 2015 Pearson Education, Inc.

Question 11 A coin is rotating on a turntable; it moves without sliding. At the instant shown in the figure, which arrow gives the direction of the frictional force on the coin? Answer: D © 2015 Pearson Education, Inc.

Question 12 A coin is rotating on a turntable; it moves without sliding. At the instant shown, suppose the frictional force disappeared. In what direction would the coin move? Answer: A © 2015 Pearson Education, Inc.

Question 8 A ball at the end of a string is being swung in a horizontal circle. What force is producing the centripetal acceleration of the ball? Gravity Air resistance Normal force Tension in the string © 2015 Pearson Education, Inc.

Question 9 A ball at the end of a string is being swung in a horizontal circle. What is the direction of the net force on the ball? Tangent to the circle Toward the center of the circle There is no net force. © 2015 Pearson Education, Inc.

Question 11 A coin is rotating on a turntable; it moves without sliding. At the instant shown in the figure, which arrow gives the direction of the frictional force on the coin? D © 2015 Pearson Education, Inc.

Question 12 A coin is rotating on a turntable; it moves without sliding. At the instant shown, suppose the frictional force disappeared. In what direction would the coin move? A © 2015 Pearson Education, Inc.

Question 13: Forces on a car, part I
Engineers design curves on roads to be segments of circles. They design dips and peaks in roads to be segments of circles with a radius that depends on expected speeds and other factors. A car is moving at a constant speed and goes into a dip in the road. At the very bottom of the dip, is the normal force of the road on the car greater than, less than, or equal to the car’s weight?

Question 13 Forces on a car, part I
The car is accelerating, even though it is moving at a constant speed, because its direction is changing. When the car is at the bottom of the dip, the center of its circular path is directly above it and so its acceleration vector points straight up. The free-body diagram identifies the only two forces acting on the car as the normal force, pointing upward, and its weight, pointing downward. Which is larger: n or w?

Question 13 Forces on a car, part I
points upward, by Newton’s second law Fnet on the car that also points upward. Therefore: the free-body diagram shows that the magnitude of the normal force must be greater than the weight. © 2015 Pearson Education, Inc.

Question 14: Forces on a car, part II
A car is turning a corner at a constant speed. What force provides the necessary centripetal acceleration? We know that the acceleration is directed toward the center of the circle. What force or forces can we identify that provide this acceleration?

Question 14 Forces on a car, part II
Imagine driving a car on a frictionless road, such as a very icy road. You would not be able to turn a corner. Turning the steering wheel would be of no use. The car would slide straight ahead, in accordance with both Newton’s first law and the experience of anyone who has ever driven on ice! So it must be friction that causes the car to turn. It must be a static friction force, not kinetic, because the tires are not skidding: The points where the tires touch the road are not moving relative to the surface. Note: If you skid, your car won’t turn the corner—it will continue in a straight line!

Problem-Solving Strategy 6.1 Circular Dynamics Problems
Draw FBD with the circle viewed edge on and the POSITIVE x-axis pointed toward the center of the circle. Y axis is perpendicular to the plane of the circle.

Question 16 In the track and field event known as the hammer throw, an athlete spins a heavy mass in a circle at the end of a chain. For male athletes, the “hammer” is a mass of 7.3 kg at the end of a 1.2-m chain. A world-class thrower can get the hammer up to a speed of 29 m/s. If an athlete swings the mass in a horizontal circle centered on the handle he uses to hold the chain, what is the tension in the chain? Answer: The tension provides the centripetal force for uniform circular motion with a linear speed v=29 m/s and a radius r=1.2m. The tension is thus 7.3*29^2/1.2 = 5100 N. © 2015 Pearson Education, Inc.

The Unbanked Turn Problem
Question 17 Finding the maximum speed to turn a corner (3 Major Types of UCM problems) The Unbanked Turn Problem What is the maximum speed with which a 1500 kg car can make a turn around a curve of radius 20 m on a level (unbanked) road without sliding? (s = 1.0 for rubber) © 2015 Pearson Education, Inc.

Question 17: Finding the maximum speed to turn a corner (cont.)
n = w = mg. Static friction maximum possible value: © 2015 Pearson Education, Inc.

Question 17 Finding the maximum speed to turn a corner (cont.)
Because the static friction force has a maximum value there will be a maximum speed at which a car can turn without sliding. This speed is reached when the static friction force reaches its maximum value fs max = smg. If the car enters the curve at a speed higher than the maximum, static friction cannot provide the necessary centripetal acceleration and the car will slide. Thus the maximum speed occurs at the maximum value of the force of static friction, or when © 2015 Pearson Education, Inc.

Example 6.7 Finding the maximum speed to turn a corner (cont.)
Using the known value of fs max, we find Rearranging: For rubber tires on pavement s = 1.0.

Question 17: Finding the maximum speed to turn a corner (cont.)
The Facts and the conceptual REAL WORLD Discussion!!!!! 14 m/s ≈ 30 mph, which seems like a reasonable upper limit for the speed at which a car can go around a curve without sliding. WOW!!!!The car’s mass canceled out. The maximum speed does not depend on the mass of the vehicle, though this may seem surprising. AND!!!!The final expression for vmax does depend on the coefficient of friction and the radius of the turn. Yep you have personally witness this concept!!!!You know, from experience, that the speed at which you can take a turn decreases if s is less (the road is wet or icy) or if r is smaller (the turn is tighter).

Question 18: Finding speed on a banked turn
The Banked Turn (NASCAR TRACKS and the Freeways) A curve on a racetrack of radius 70 m is banked at a 15° angle. At what speed can a car take this curve without assistance from friction? © 2015 Pearson Education, Inc.

Example 18 Finding speed on a banked turn (cont.)
The only two forces are the normal force and the car’s weight.(IT’S A NO FRICTION PROBLEM) Even though the car is tilted, it is still moving in a horizontal circle. Choose the x-axis to be horizontal and pointing toward the center of the circle.

nx = n sin  is the only component of force toward the center of the circle. It’s component of the Fn on the car that causes it to turn the corner. where  is the angle at which the road is banked,

From the y-equation, Substituting this into the x-equation and solving for ν give Important Concept: This is ≈ 30 mph. Only at this exact speed can the turn be negotiated without reliance on friction forces. What would we have to do for NASCAR speeds if we did not rely on friction?

Section 6.3 Apparent Forces in Circular Motion

Centrifugal Force? If you are a passenger in a car that turns a corner quickly, it is the force of the car door, pushing inward toward the center of the curve, that causes you to turn the corner. What you feel is your body trying to move ahead in a straight line as outside forces (the door) act to turn you in a circle.(Inertia…baby!!!) A centrifugal force will never appear on a free-body diagram and never be included in Newton’s laws. © 2015 Pearson Education, Inc.

Apparent Weight in Circular Motion

The force you feel, your apparent weight, is the magnitude of the contact force that supports you. When the passenger on the roller coaster is at the bottom of the loop: The net force points upward, so n > w. Her apparent weight is wapp= n, so her apparent weight is greater than her true weight. © 2015 Pearson Education, Inc.

Newton’s second law for the passenger at the bottom of the circle is From this equation, the passenger’s apparent weight is Apparent weight at the bottom is greater than true weight, w. © 2015 Pearson Education, Inc.

Newton’s second law for the passenger at the top of the circle is Note that wx is now positive because the x-axis is directed downward. We can solve for the passenger’s apparent weight: If v is sufficiently large, her apparent weight can exceed the true weight. © 2015 Pearson Education, Inc.

As the car goes slower there comes a point where n becomes zero: The speed for which n = 0 (free fall) is called the critical speed vc. Because for n to be zero we must have , the critical speed is The critical speed is the slowest speed at which the car can complete the circle. © 2015 Pearson Education, Inc.

Question 19 A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a diameter 14 feet, what is the minimum speed the skater must have at the very top of the loop? © 2015 Pearson Education, Inc.

Question 19 The skater’s speed must be sufficiently high to continue in uniform circular motion. Lowest speed is when normal force (n) = 0 MASS…No Effect!!!!!!!!! Think of centripedal force being the push you feel because of inertia when in a turn or in a roller coaster loop. Not enough inertia (mv2/r) to hold you against the track…gravity causes you to fall!!!

Question 20 A physics textbook swings back and forth as a pendulum. Which is the correct free-body diagram when the book is at the bottom and moving to the right? Answer: C © 2015 Pearson Education, Inc. 66

Question 21 A car that’s out of gas coasts over the top of a hill at a steady 20 m/s. Assume air resistance is negligible. Which free-body diagram describes the car at this instant? Answer: A © 2015 Pearson Education, Inc. 67

Question 22 A roller coaster car does a loop-the-loop. Which of the free- body diagrams shows the forces on the car at the top of the loop? Rolling friction can be neglected. Answer: E © 2015 Pearson Education, Inc. 68

Question 24 A coin sits on a turntable as the table steadily rotates counterclockwise. The free-body diagrams below show the coin from behind, moving away from you. Which is the correct diagram? Answer: C © 2015 Pearson Education, Inc. 70

Question 20 A physics textbook swings back and forth as a pendulum. Which is the correct free-body diagram when the book is at the bottom and moving to the right? Centripetal acceleration requires an upward force. C. © 2015 Pearson Education, Inc. 72

Question 21 A car that’s out of gas coasts over the top of a hill at a steady 20 m/s. Assume air resistance is negligible. Which free-body diagram describes the car at this instant? Now the centripetal acceleration points down. A. © 2015 Pearson Education, Inc. 73

Question 22 A roller coaster car does a loop-the-loop. Which of the free- body diagrams shows the forces on the car at the top of the loop? Rolling friction can be neglected. The track is above the car, so the normal force of the track pushes down. E. © 2015 Pearson Education, Inc. 74

Question 24 A coin sits on a turntable as the table steadily rotates counterclockwise. The free-body diagrams below show the coin from behind, moving away from you. Which is the correct diagram? C. © 2015 Pearson Education, Inc. 76

Question 26 A car of mass 1500 kg goes over a hill at a speed of 20 m/s. The shape of the hill is approximately circular, with a radius of 60 m, as in the figure. When the car is at the highest point of the hill, What is the force of gravity on the car? What is the normal force of the road on the car at this point? © 2015 Pearson Education, Inc.

Question 26 Answer: A: the force of gravity on the car is independent of any motion and is simply 𝑚×𝑔 = 1500*9.8 = 1.5x104 N. B: the car is going through uniform circular motion at the top of the hill, and the net force on the car thus equals m*v2/r. The net force on the car = 1500*202/60 = 10,000 N downward. Gravity supplies 15,000 N downward (part (a)), so the normal force must be 5000 N upward for a net force of 10,000 N down. © 2015 Pearson Education, Inc.

Question 1 For uniform circular motion, the acceleration
Is parallel to the velocity. Is directed toward the center of the circle. Is larger for a larger orbit at the same speed. Is always due to gravity. Is always negative. Answer: B © 2015 Pearson Education, Inc.

Question 2 When a car turns a corner on a level road, which force provides the necessary centripetal acceleration? Friction Normal force Gravity Tension Air resistance Answer: A © 2015 Pearson Education, Inc.

Question 3 A passenger on a carnival ride rides in a car that spins in a horizontal circle as shown at right. At the instant shown, which arrow gives the direction of the net force on one of the riders? Answer: D © 2015 Pearson Education, Inc.

Question 4 Newton’s law of gravity describes the gravitational force between The earth and the moon. The earth and the sun. The sun and the planets. A person and the earth. All of the above. Answer: E © 2015 Pearson Education, Inc.

Question 1 For uniform circular motion, the acceleration
Is parallel to the velocity. Is directed toward the center of the circle. Is larger for a larger orbit at the same speed. Is always due to gravity. Is always negative. © 2015 Pearson Education, Inc.

Question 3 A passenger on a carnival ride rides in a car that spins in a horizontal circle as shown at right. At the instant shown, which arrow gives the direction of the net force on one of the riders? D © 2015 Pearson Education, Inc.

Question 4 Newton’s law of gravity describes the gravitational force between The earth and the moon. The earth and the sun. The sun and the planets. A person and the earth. All of the above. © 2015 Pearson Education, Inc.

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