11/26 do now – on a new sheet of paper

1. 11/26 do now – on a new sheet of paper • A rightward force of 302 N is applied to a 28.6-kg crate to accelerate it across the floor. The coefficient of friction between the crate and the floor is 0.750. Determine the acceleration of the crate. • Hint (draw a FBD, find all forces first) • Show work • equations, • substitute number and units, • answer with units

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

3. Circular Velocity Objects traveling in circular motion have constant speed and constantly CHANGING velocity – changing in direction but not magnitude How do we define VELOCITY? Velocity is TANGENT to the circle at all points What ‘d’ are we talking about? What ‘t’ are we talking about? PERIOD (T) Time for one revolution CIRCUMFERENCE C = 2πr = πd If this is true, why does ANYTHING move in a circle?

4. 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 • Speed: constant • Direction of motion: tangent of the path • Velocity: • Speed: constant • Direction: changing

5. Speed and velocity • Calculation of the Average Speed vavg = ∆d / ∆t • The distance of one complete cycle around the perimeter of a circle is known as 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: vavg = 2·π·R / T

6. 2·π·R vavg = T • The average speed and the Radius of the circle are directly proportional. • The average speed and the Period of the circle are inversely proportional.

7. 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.

8. 2·π·R vavg = T constant • 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. tangent The average speed is directly proportional to the radius and inversely proportional to the period.

9. 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.

10. 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

11. 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

12. 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. +

13. 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.

14. 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.

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. Class work Circular and satellite motion packet – pages 1 & 2

17. 11/27 do now • A falling skydiver is accelerating in the downward direction at 3.0 m/s/s. The mass of the skydiver (including parachute gear) is 50 kg. Determine the air resistance force on the skydiver (and accompanying parachute). • Show work

18. Objectives • The Centripetal Force Requirement • Mathematics of Circular Motion • Homework – castle learning

19. 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.

20. Centripetal Force Inertia causes objects to travel STRAIGHT Paths can be bent by FORCES CENTRIPETAL FORCE bends an object’s path into a circle - pulling toward the CENTER

22. Inertia, Force and Acceleration for an Automobile Passenger • Observe that the passenger (in blue) continues in a straight-line motion until he strikes the shoulder of the driver (in red). Once striking the driver, a force is applied to the passenger to force the passenger to the right and thus complete the turn. This force is the cause for turning. Without this force, the passenger would still travel in a straight line at constant speed (inertia).

23. Inertia, Force and Acceleration

24. Without a centripetal force, an object in motion continues along a straight-line path. With a centripetal force, an object in motion will be accelerated and change its direction.

25. Misconception The doors to the “Gravitron” close and it starts to spin. You are pushed against the outside edge of the ride and pinned there, You must be experiencing “centrifugal force” throwing you outward from the ride! Right?

26. What is really happening? As the Gravitron starts to spin, friction between your body and the ride start you moving vc Fc Once you are moving, your body wants to go STRAIGHT … but you can’t… The walls keep pushing you back in toward the center of the ride!

27. What is it you feel? centrifugal (center fleeing) force A ‘fictitious’ or ‘inertial’ force that is experienced from INSIDE a circular motion system centripetal (center seeking) force A true force that pushes or pulls an object toward the center of a circular path

28. The Centripetal Force and Direction Change • Any object moving in a circle (or along a circular path) experiences a centripetal force. This is the centripetal force requirement. • The word centripetal is merely an adjective used to describe the direction of the force. We are not introducing a new type of force but rather describing the direction of the net force acting upon the object that moves in the circle.

29. The Centripetal Force is the NET force

30. 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.

31. inward • To summarize, 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

32. 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

33. Class work • Practice packet pp. 3-4

34. 2·π·R vavg = T 4π2R a = v2 T2 a = R Mathematics of Circular Motion

35. v2 a = R Relationship between quantities This equation shows for a constant mass and radius, both Fnet and a is directly proportional to the v2. F ~ v2 a ~ 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.

36. 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

37. 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

38. 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. 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. 

39. 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

40. 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.

41. 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

42. objectives • Circular motion – using equations for Problem-Solving • Applications of Circular Motion

43. 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

44. 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

45. 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.

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

47. 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!

48. 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

49. 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 Ff = Fnet = m*v2/R Ff = (945 kg)*(10m/s)2/25m Ff = 3780 N Fgrav Ff = μFNorm;FNorm =mg; Ff = μmg 3780 N = μ(9270 N); μ = 0.41

50. 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