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Circular Motion and Satellite Motion - Chapter Outline

Circular Motion and Satellite Motion - Chapter Outline. Lesson 1: Motion Characteristics for Circular Motion Lesson 2: Applications of Circular Motion Lesson 3: Universal Gravitation Lesson 4: Satellite Motion. Objective - Lesson 1: Motion Characteristics for Circular Motion.

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Circular Motion and Satellite Motion - Chapter Outline

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  1. Circular Motion and Satellite Motion - Chapter Outline Lesson 1: Motion Characteristics for Circular Motion Lesson 2: Applications of Circular Motion Lesson 3: Universal Gravitation Lesson 4: Satellite Motion

  2. Objective - Lesson 1: Motion Characteristics for Circular Motion • Speed and Velocity • Acceleration • The Centripetal Force Requirement • Mathematics of Circular Motion

  3. Uniform circular motion • Uniform circular motion is the motion of an object in a circle with a constant or uniform speed. The velocity is changing because the direction of motion is changing.

  4. Speed and velocity • Speed is constant: • Since the distance of one complete cycle around the perimeter of a circle is the circumference Circumference = 2·π·R (R is the radius of the circle) • The time (T) to make one cycle around the circle is called one period. • The average speed of an object in uniform circular motion is:

  5. vavg = 2·π·R / T • If objects moving around circles of different radius in the same period, the object traversing the circle of larger radius must be traveling with the greatest speed. In fact, the average speed and the radius of the circle are directly proportional.

  6. The Direction of the Velocity Vector The best word that can be used to describe the direction of the velocity vector is the word tangential. The direction of the velocity vector at any instant is in the direction of a tangent line drawn to the circle at the object's location.

  7. summary • To summarize, an object moving in uniform circular motion is moving around the perimeter of the circle with a __________speed. While the speed of the object is constant, its velocity is changing. Velocity, being a vector, has a constant magnitude but a changing direction. The direction is always directed _________ to the circle and as the object turns the circle, the tangent line is always pointing in a new direction. constant tangent The average speed and the radius of the circle are directly proportional. The larger radius the greatest the speed.

  8. Check Your Understanding • A tube is been placed upon the table and shaped into a three-quarters circle. A golf ball is pushed into the tube at one end at high speed. The ball rolls through the tube and exits at the opposite end. Describe the path of the golf ball as it exits the tube. The ball will move along a path which is tangent to the spiral at the point where it exits the tube. At that point, the ball will no longer curve or spiral, but rather travel in a straight line in the tangential direction.

  9. example • A vehicle travels at a constant speed of 6.0 meters per second around a horizontal circular curve with a radius of 24 meters. The mass of the vehicle is 4.4 × 103 kilograms. An icy patch is located at P on the curve. On the icy patch of pavement, the frictional force of the vehicle is zero.  Which arrow best represents the direction of the vehicle's velocity when it reaches icy patch P? a b c d

  10. Acceleration • An object moving in uniform circular motion is moving in a circle with a uniform or constant speed. The velocity vector is constant in magnitude but changing in direction. • Since the velocity is changing. The object is accelerating. where vi represents the initial velocity and vf represents the final velocity after some time of t

  11. Direction of the Acceleration Vector • The velocity change is directed towards point C - the center of the circle. • The acceleration of the object is dependent upon this velocity change and is in the same direction as this velocity change. The acceleration is directed towards point C as well - the center of the circle.

  12. example • The initial and final speed of a ball at two different points in time is shown below. The direction of the ball is indicated by the arrow. For each case, indicate if there is an acceleration. Explain why or why not. Indicate the direction of the acceleration. a. No, no change in velocity yes, change in velocity, right b. yes, change in velocity, left c. yes, change in velocity, left d.

  13. example • Explain why an object moving in a circle at constant speed can be said to experience an acceleration. An object which experiences either a change in the magnitude or the direction of the velocity vector can be said to be accelerating. This explains why an object moving in a circle at constant speed can be said to accelerate - the direction of the velocity changes.

  14. example • Identify the three controls on an automobile which allow the car to be accelerated. The accelerator allows the car to increase speed. The brake pedal allows the car to decrease the speed. And the steering wheel allows the car to change direction.

  15. example An object is moving in a clockwise direction around a circle at constant speed. • Which vector below represents the direction of the velocity vector when the object is located at point B on the circle? • Which vector below represents the direction of the acceleration vector when the object is located at point C on the circle? • Which vector below represents the direction of the velocity vector when the object is located at point C on the circle? • Which vector below represents the direction of the acceleration vector when the object is located at point A on the circle? d b a d

  16. The Centripetal Force Requirement • According to Newton's second law of motion, an object which experiences an acceleration requires a net force. • The direction of the net force is in the same direction as the acceleration. So for an object moving in a circle, there must be an inward force acting upon it in order to cause its inward acceleration. This is sometimes referred to as the centripetal force requirement. • The word centripetal means center seeking. For object's moving in circular motion, there is a net force acting towards the center which causes the object to seek the center. • The Centripetal Force is the NET force

  17. Inertia, Force and Acceleration

  18. examples of centripetal force As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion. As a car makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion. As a bucket of water is tied to a string and spun in a circle, the tension force acting upon the bucket provides the centripetal force required for circular motion.

  19. Summary inward • An object in uniform circular motion experiences an __________ net force. This inward force is sometimes referred to as a _______________ force, where centripetal describes its direction. Without this centripetal force, an object could never alter its direction. The fact that the centripetal force is directed perpendicular to the tangential velocity means that the force can alter the direction of the object's velocity vector without altering its magnitude. centripetal

  20. Check your understanding • An object is moving in a clockwise direction around a circle at constant speed • Which vector below represents the direction of the force vector when the object is located at point A on the circle? • Which vector below represents the direction of the force vector when the object is located at point C on the circle? • Which vector below represents the direction of the velocity vector when the object is located at point B on the circle? •  Which vector below represents the direction of the velocity vector when the object is located at point C on the circle? • Which vector below represents the direction of the acceleration vector when the object is located at point B on the circle? d b d a c

  21. Mathematics of Circular Motion m: mass v: speed R: radius T: period

  22. v2 a = R Equations as a Guide to Thinking • An equation expresses a mathematical relationship between the quantities present in that equation. This equation shows for a constant mass and radius, both Fnet and a is directly proportional to the v2. F ~ v2 If the speed of the object is doubled, the net force required for that object's circular motion and its acceleration are quadrupled. And if the speed of the object is halved (decreased by a factor of 2), the net force required and its acceleration are decreased by a factor of 4.

  23. example • A car going around a curve is acted upon by a centripetal force, F. If the speed of the car were twice as great, the centripetal force necessary to keep it moving in the same path would be • F • 2F • F/2 • 4F F = m v2 / r F in directly proportional to v2 F in increased by 22

  24. v2 a = R Centripetal force and mass of the object This equation shows for a constant speed and radius, the Fnet is directly proportional to the mass. If the mass of the object is doubled, the net force required for that object's circular motion is doubled. And if the mass of the object is halved (decreased by a factor of 2), the net force required is decreased by a factor of 2. • Centripetal acceleration and mass of the object Centripetal acceleration is not affected by the mass of the object

  25. example • Anna Litical is practicing a centripetal force demonstration at home. She fills a bucket with water, ties it to a strong rope, and spins it in a circle. Anna spins the bucket when it is half-full of water and when it is quarter-full of water. In which case is more force required to spin the bucket in a circle? Explain using an equation as a "guide to thinking.” It will require more force to accelerate a half-full bucket of water compared to a quarter-full bucket. According to the equation Fnet = m•v2 / R, force and mass aredirectly proportional. So the greater the mass, the greater the force. 

  26. example • A Lincoln Continental and a Yugo are making a turn. The Lincoln is four times more massive than the Yugo. If they make the turn at the same speed, then how do the centripetal forces acting upon the two cars compare. Explain. The centripetal force on the Continental is four times greater than that of a Yugo. According to the equation Fnet=(m•v2) / R, force and mass are directly proportional. So 4 times the mass means 4 times the force.

  27. Example • The diagram shows a 5.0-kilogram cart traveling clockwise in a horizontal circle of radius 2.0 meters at a constant speed of 4.0 meters per second. If the mass of the cart was doubled, the magnitude of the centripetal acceleration of the cart would be • doubled • halved • unchanged • quadrupled ac = v2 / R

  28. example • A 60.-kilogram adult and 30.-kilogram child are passengers on a rotor ride at an amusement park. When the rotating hollow cylinder reaches a certain constant speed, v, the floor moves downward. Both passengers stay "pinned" against the wall of the rotor, as shown in the diagram. The magnitude of the frictional force between the adult and the wall of the spinning rotor is F. What is the magnitude of the frictional force between the child and the wall of the spinning rotor? • F • 2F • ½ F • ¼ F         Ff = µFNorm FNorm = Fnet = mv2/R Compare to the adult, the child’s mass is half, therefore FNorm of the child is half of that of the adult and its Ff is half of that of friction of the adult also.

  29. v2 a = R Centripetal Force, acceleration and the radius This equation shows for a constant speed and mass, the Fnet and acceleration a is inversely proportional to the radius If the radius of the object is doubled, the net force required for that object's circular motion and its acceleration are both halved. And if the radius of the object is halved (decreased by a factor of 2), the net force required and its acceleration are both increased by a factor of 2.

  30. example • Two masses, A and B, move in circular paths as shown in the diagram. The centripetal acceleration of mass A, compared to that of mass B, is • the same • twice as great • one-half as great • four times as great F = m v2 / r

  31. Equations as a Recipe for Problem-Solving • A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car. Given: = 900 kg; v = 10.0 m/s R = 25.0 m Find: a = ? Fnet = ? a = v2 / R a = (10.0 m/s)2 / (25.0 m) a = (100 m2/s2) / (25.0 m) a = 4 m/s2 Fnet = m • a Fnet = (900 kg) • (4 m/s2) Fnet = 3600 N

  32. example • A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path which is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback. Given: m = 95.0 kg; R = 12.0 m; Traveled 1/4 of the circumference in 2.1 s Find: v = ? a = ? Fnet = ? a = v2 / R a = (8.97 m/s)2/(12.0 m) a = 6.71 m/s2 v = d / t v = (1/4•2•π•12.0m) /(2.1s) v = 8.97 m/s Fnet = m*a Fnet = (95.0 kg)*(6.71 m/s2) Fnet = 637 N

  33. example • Determine the centripetal force acting upon a 40-kg child who makes 10 revolutions around the Cliffhanger in 29.3 seconds. The radius of the barrel is 2.90 meters. Given: m = 40 kg; R = 2.90 m; T = 2.93 s (since 10 cycles takes 29.3 s). Find: Fnet Step 1: find speed: v = (2 • π • R) / T = 6.22 m/s. Step 2: find the acceleration: a = v2 / R = (6.22 m/s)2/(2.90 m) = 13.3 m/s2 Step 3: find net force: Fnet = m • a = (40 kg)(13.3 m/s/s) = 532 N.

  34. practices • An object with a mass of 0.5 kilogram is swung by a string in a horizontal circle of radius 1 meter at a speed of 2 meters per second. What is the magnitude of the acceleration of the mass? • A 4.0-kilogram model airplane travels in a horizontal circular path of radius 12 meters at a constant speed of 6.0 meters per second. What is the magnitude of the centripetal acceleration of the airplane? • The diagram shows a student seated on a rotating circular platform, holding a 2.0-kilogram block with a spring scale. The block is 1.2 meters from the center of the platform. The block has a constant speed of 8.0 meters per second. [Frictional forces on the block are negligible.] what is the reading on the spring scale? A race car starting from rest accelerates uniformly at a rate of 4.00 meters per second2. What is the car’s speed after it has traveled 200. meters? [show work, including equation, substitution with numbers and units, answer with units]

  35. Lesson 2: Applications of Circular Motion • Newton's Second law - Revisited • Amusement Park Physics

  36. Applications of Circular Motion • Newton's Second Law - Revisited Where Fnet is the sum (the resultant) of all forces acting on the object. Newton's second law was used in combination of circular motion equations to analyze a variety of physical situations. Note: centripetal force is the net force!

  37. Steps in solving problems involving forces • Drawing Free-Body Diagrams • Determining the Net Force from Knowledge of Individual Force Values • Determining Acceleration from Knowledge of Individual Force Values Or Determining Individual Force Values from Knowledge of the Acceleration

  38. Case 1: car driven in a circle Free body diagram FNorm When a car is moving in a horizontal circle on a level surface, there is a net force (centripetal force) acting on it, the net force is FRICTION, which is directed toward the center of the circle. Ff Fgrav • Find individual force and net force • Since there is no vertical acceleration, FN = Fg= mg; Fnet = ∑ F = Ff; friction force is the net force.

  39. example • A 945-kg car makes a 180-degree turn with a speed of 10.0 m/s. The radius of the circle through which the car is turning is 25.0 m. Determine the force of friction and the coefficient of friction acting upon the car. FNorm Given: m = 945 kg; v = 10.0 m/s; R = 25.0 m Find: Ffrict = ? μ = ? Ff Fgrav Ff = Fnet = m*v2/R Ff = (945 kg)*(10m/s)2/25m Ff = 3780 N Ff = μFNorm;FNorm =mg; Ff = μmg 3780 N = μ(9270 N); μ = 0.41

  40. example • The coefficient of friction acting upon a 945-kg car is 0.850. The car is making a 180-degree turn around a curve with a radius of 35.0 m. Determine the maximum speed with which the car can make the turn. Given: m = 945 kg; μ = 0.85; R = 35.0 m Find: v = ? (the minimum speed would be the speed achieved with the given friction coefficient) FNorm Ff = Fnet Ff Ff = μFN =μFN ; Ff= (0.85)(9270N) = 7880 N Fgrav Fnet = m*v2/R 7880 N = (945 kg)(v2) / (35.0 m); v = 17.1 m/s

  41. Case 2: a swing bucket In a swing bucket, the net force (Fc) is the combinations of gravity and tension. At top: Fnet = Ftens + Fg Fnet = mv2/R At bottom: Fnet = Ftens - Fg Fnet = mv2/R

  42. Example • A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the top of the circular loop, the speed of the bucket is 4.00 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the top of the circular loop. Fgrav = mg = 14.7 N a = v2 / R = 16 m/s2 Fnet = ma = 24 N, down Fnet = Fgrav + Ftens 24 N = 14.7 N + Ften Ftens = 24 N - 14.7 N = 9.3 N m = 1.5 kg a = ________ m/s/s Fnet = _________ N

  43. Example • A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the bottom of the circular loop, the speed of the bucket is 6.00 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the bottom of the circular loop. Fgrav = m • g = 14.7 N a = v2 / R = 36 m/s2 Fnet = ma = 54 N, up Fnet = Ften - Fgrav 54 N = Ften – 14.7 N Ften = 54 N +14.7 N = 68.7 N m = 1.5 kg a = ________ m/s/s Fnet = _________ N

  44. Case 3: Roller Coasters and Amusement Park Physics In a roller coaster, the centripetal force is provided by the combination of normal force and gravity.

  45. example • Anna Litical is riding on The Demon at Great America. Anna experiences a downwards acceleration of 15.6 m/s2 at the top of the loop and an upwards acceleration of 26.3 m/s2 at the bottom of the loop. Use Newton's second law to determine the normal force acting upon Anna's 864 kg roller coaster car at top and at bottom of the loop. Given: m = 864 kg atop = 15.6 m/s2 , down abottom = 26.3 m/s2 , up Fgrav = m•g = 8476 N. Find: Fnorm at top and bottom

  46. Bottom of Loop • Fnet = m * a • Fnet = (864 kg) * (26.3 m/s2 ) = 22 723 N, upward Fnet = Fnorm + (-Fgrav) 22723 N = Fnorm + (-8476 N) Fnorm = 31199 N

  47. top of Loop • Fnet = m * a • Fnet = (864 kg)∙(15.6 m/s2 ) = 13 478 N, downward Fnet = Fnorm + Fgrav 13478 N = Fnorm + 8475 N Fnorm = 5002 N

  48. example • Anna Litical is riding on The American Eagle at Great America. Anna is moving at 18.9 m/s over the top of a hill which has a radius of curvature of 12.7 m. Use Newton's second law to determine the magnitude of the applied force of the track pulling down upon Anna's 621 kg roller coaster car. Given: m = 621 kg; v = 18.9 m/s; R=12.7 m Fgrav = m * g = 6086 N a = v2 / R = (18.9 m/s)2 / (12.7 m) = 28.1 m/s2 Find: Fapp at top of hill

  49. Fnet = m • a Fnet = (621 kg)•(28.1 m/s2 ), down Fnet = 17450 N, down Fnet = Fapp + Fgrav 17450 N = Fapp + 6086 N Fapp = 11381 N

  50. Practices • Calculate the magnitude of the centripetal force acting on Earth as it orbits the Sun, assuming a circular orbit and an orbital speed of 3.00 x 104 m/s. [show all work, including the equation and substitution with units.] • An unbalanced force of 40. newtons keeps a 5.0-kilogram object traveling in a circle of radius 2.0 meters. What is the speed of the object? • A student on an amusement park ride moves in a circular path with a radius of 3.5 meters once every 8.9 seconds. What is the student’s average speed? • In an experiment, a 0.028-kilogram rubber stopper is attached to one end of a string. A student whirls the stopper overhead in a horizontal circle with a radius of 1.0 meter. The stopper completes 10 revolutions in 10 seconds. • Determine the speed of the whirling stopper. • Calculate the magnitude of the centripetal force on the whirling stopper. • The combined mass of a race car and its driver is 600. kilograms. Traveling at constant speed, the car completes one lap around a circular track of radius 160. meters in 36.0 seconds. • Calculate the speed of the car, to the nearest tenth of a meter per second. • Calculate the magnitude of the centripetal acceleration of the car. Round your answer to the nearest tenth of a meter per second2.

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