1. 11/18 do now – on a new sheet Consider the position-time plots below. Sketch the shape of the corresponding velocity-time graphs. Hint: 1.Slope in p-t graph is velocity 2.The lines in v-t graph are straight, not curved

2. Due today –worksheet 2.1.2B – Forces on Angles Packet –Castle Learning Test – Due tonight Due tomorrow –Homework: castle learning Due Friday –Projectile test correction –Project correction –Worksheet 1.2.3 ground launched projectile packet correction Homework is posted on google classroom - code: nsdnlt

3. Circular Motion Chapter Outline Lesson 1: Motion Characteristics for Circular Motion Lesson 2: Applications of Circular Motion

4. Objective - Lesson 1: Motion Characteristics for Circular Motion 1.Speed and VelocitySpeed and Velocity 2.AccelerationAcceleration 3.The Centripetal Force RequirementThe Centripetal Force Requirement 4.Mathematics of Circular MotionMathematics of Circular Motion

5. 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.

6. R is radius of the circle 2πR is the circumference of the circle T is time to make one lap around the circle, also known as period. Speed is constant The average speed and the radius of the circle are directly proportional.

7. Velocity is changing The Direction of the Velocity Vector The direction of the velocity vector at every instant is in a direction tangent to the circle.

8. 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, which is tangent to the circle.. Since the velocity is changing. The object is accelerating. v i represents the initial velocity v f represents the final velocity t represents the time interval

9. Direction of the Acceleration Vector The velocity change is directed towards the center of the circle. The acceleration is in the same direction as this velocity change. The acceleration is directed towards the center of the circle.

10. The Centripetal Force According to Newton's second law of motion, an object which experiences an acceleration requires a net force.Newton's second law of motion The direction of the net force is in the same direction as the acceleration. Since acceleration is directed toward the center, the net force must be toward the center. This net force is referred to as the centripetal force. The word centripetal means center seeking. In uniform circular motion, the Centripetal Force is the NET force.

11. Inertia, Force and Acceleration Centrifugal force is a fictitious force.

12. Centripetal force is the reason objects moves in a circle As a car makes a turn in a circle, centripetal force is provided by friction. As a bucket of water tied to a string and spun in a circle, centripetal force is provided by the tension acting upon the bucket. As the moon orbits the Earth in a circle, centripetal force is provided by gravity acting upon the moon.

13. An object in uniform circular motion moves at ____________ speed. Its velocity is ___________ to the circle and its acceleration is directed toward the ___________ of the circle. The object experiences ____________ which is directed in the same direction as the acceleration, toward the _________ of the circle. This net force is called ______________ force. 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. Summary

14. 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.

15. 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. b. c. No, no change in velocity yes, change in velocity, right yes, change in velocity, left d. yes, change in velocity, left

16. 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.

17. example 1.Which vector below represents the direction of the velocity vector when the object is located at point B on the circle? 2.Which vector below represents the direction of the acceleration vector when the object is located at point C on the circle? 3.Which vector below represents the direction of the force vector when the object is located at point A on the circle? An object is moving in a clockwise direction around a circle at constant speed. d b d

18. Example The centripetal force acting on the space shuttle as it orbits Earth is equal to the shuttle’s A.inertia B.momentum C.velocity D.weight

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

20. Relationships between quantities v avg and R are direct: R double, v double v avg and T are inverse: T double, v halves a c and v are direct squared: v double, a c quadruple. F c and v are direct squared: v double, F c quadruple. a c and R are inverse: R double, a c halves. F c and R are inverse: R double, F c halves. F c and m are direct: m double, F c doubles. a c and m are not related: m double, a c unchanged.

21. 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 1.F 2.2F 3.F/2 4.4F F = m v 2 / r F in directly proportional to v 2 F in increased by 2 2

22. 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? It will require more force to accelerate a half-full bucket of water compared to a quarter-full bucket. According to the equation F net = mv 2 / R, force and mass are directly proportional. So the greater the mass, the greater the force.

23. 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 1.doubled 2.halved 3.unchanged 4.quadrupled a c = v 2 / R

24. 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? 1.F 2.2F 3.½ F 4.¼ F F f = µF Norm F Norm = F net = mv 2 /R The child is half as massive as the adult, therefore F Norm of the child is half of that of the adult and its F f is half of that of friction of the adult also.

25. 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 1.the same 2.twice as great 3.one-half as great 4.four times as great F = m v 2 / r

26. 11/19 do now Identify the resultant in the following vector addition diagram. Finally, indicate which two vectors were added to achieve this resultant (express as an equation such as X + Y = Z).

27. Homework: castle learning Due Friday –Projectile test correction –Project correction –Worksheet 1.2.3 ground launched projectile packet correction Due Monday –Force and Vector Application Packet correction –Castle learning test correction –Circular and satellite motion packet Homework is posted on google classroom - code: nsdnlt

28. Solving problems using Equations Identify the given and unknown. Choose an equation Substitute number with units Answer with units

29. Example 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 = v 2 / R a = (10.0 m/s) 2 / (25.0 m) a = (100 m 2 /s 2 ) / (25.0 m) a = 4 m/s 2 F net = m a F net = (900 kg) (4 m/s 2 ) F net = 3600 N

30. 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 = ?F net = ? v = d / t v = (1/42π12.0m) /(2.1s) v = 8.97 m/s a = v 2 / R a = (8.97 m/s) 2 /(12.0 m) a = 6.71 m/s 2 F net = m*a F net = (95.0 kg)*(6.71 m/s 2 ) F net = 637 N

31. 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: F net Step 1: find speed: v = (2 π R) / T = 6.22 m/s. Step 2: find the acceleration: a = v 2 / R = (6.22 m/s) 2 /(2.90 m) = 13.3 m/s 2 Step 3: find net force: F net = m a = (40 kg)(13.3 m/s/s) = 532 N.

32. example A go-cart travels around a flat, horizontal, circular track with a radius of 25 meters. The mass of the go-cart with the rider is 200. kilograms. The magnitude of the maximum centripetal force exerted by the track on the go-cart is 1200. newtons. Which change would increase the maximum speed at which the go-cart could travel without sliding off this track? A.Decrease the coefficient of friction between the go-cart and the track. B.Decrease the radius of the track. C.Increase the radius of the track. D.Increase the mass of the go-cart.

33. Lesson 2: Applications of Circular Motion 1.Newton's Second law - RevisitedNewton's Second law - Revisited 2.Amusement Park PhysicsAmusement Park Physics

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

35. Steps in solving problems involving forces 1.Drawing Free-Body Diagrams 2.Determining the Net Force from Knowledge of Individual Force Values 3.Determining Acceleration from Net Force Or Determining Individual Force Values from Knowledge of the Acceleration

36. 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. Case 1: car driven in a circle

37. 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. Given: m = 945 kg; v = 10.0 m/s; R = 25.0 m Find: F frict = ? μ = ? F f = μF Norm ; F Norm = mg; F f = μmg 3780 N = μ(9270 N); μ = 0.41 F f = F net = m*v 2 /R F f = (945 kg)*(10m/s) 2 /25m F f = 3780 N FfFf F Norm F grav

38. 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) F f = μF N = μF N ; F f = (0.85)(9270N) = 7880 N 7880 N = (945 kg)(v 2 ) / (35.0 m); v = 17.1 m/s F f = F net F net = m*v 2 /R FfFf F Norm F grav

39. Case 2: a swing bucket In a swing bucket, the net force (F c ) is the combinations of gravity and tension. At top: F net = F tens + F g F net = mv 2 /R At bottom: F net = F tens - F g F net = mv 2 /R

41. 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. m = 1.5 kg a = ________ m/s/s F net = _________ N F grav = mg = 14.7 N a = v 2 / R = 16 m/s 2 F net = ma = 24 N, down F net = F grav + F tens 24 N = 14.7 N + F ten F tens = 24 N - 14.7 N = 9.3 N Example

42. 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. m = 1.5 kg a = ________ m/s/s F net = _________ N F grav = m g = 14.7 N a = v 2 / R = 36 m/s 2 F net = ma = 54 N, up F net = F ten - F grav 54 N = F ten – 14.7 N F ten = 54 N +14.7 N = 68.7 N Example

43. 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.

44. example Anna Litical is riding on The Demon at Great America. Anna experiences a downwards acceleration of 15.6 m/s 2 at the top of the loop and an upwards acceleration of 26.3 m/s 2 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 a top = 15.6 m/s 2, down a bottom = 26.3 m/s 2, up F grav = mg = 8476 N. Find: F norm at top and bottom

45. top of Loop F net = m * a F net = (864 kg)∙(15.6 m/s 2 ) = 13 478 N, downward F net = F norm + F grav 13478 N = F norm + 8475 N F norm = 5002 N

46. Bottom of Loop F net = m * a F net = (864 kg) * (26.3 m/s 2 ) = 22 723 N, upward F net = F norm + (-F grav ) 22723 N = F norm + (-8476 N) F norm = 31199 N