Forces

1. Forces

2. Unit 2: Forces Chapter 6: Systems in Motion • 6.1 Motion in Two Dimension • 6.2 Circular Motion • 6.3 Centripetal Force, Gravitation, and Satellites • 6.4 Center of Mass

3. Key Question: Which launch angle will give a marble the best range? 6.1 Investigation: Launch Angle and Range Objectives: • Use the Marble Launcher to find the launch angle that produces the maximum range for a projectile.

4. Motion in two dimensions • Real objects do not move in straight lines alone; their motion includes turns and curves. • To describe a curve you need at least two dimensions (x and y).

5. Displacement • Distance is scalar, but displacement is a vector. • A displacement vector shows a change in position.

6. Displacement • A displacement vector’s direction is often given using words. • Directional words include left, right, up, down, and compass directions.

7. Solving displacement problems • Displacement vectors can be added just like force vectors. • To add displacements graphically, draw them to scale with each subsequent vector drawn at the end of the previous vector. • The resultant vectorrepresents the displacement for the entire trip.

8. Adding vectors A mouse walks 5 meters north and 12 meters west. Use a scaled drawing and a protractor to find the mouse’s displacement. Use the Pythagorean theorem to check your work. • Looking for:… the displacement. • Given:… distances and direction (5 m, N) and (12 m, W) • Relationships:Pythagorean theorem a2+ b2= c2 • Solution:Make a drawing with a scale of 1 cm = 2 meters. Measure the angle from the x axis. Pythagorean theorem: (5)2 + (12)2 = c2 169 = c2 13 = c The mouse walks 13 meters at 157°.

9. Velocity vectors • Velocity is speed with direction, so velocity is a vector. • As objects move in curved paths, their velocity vectors change because the direction of motion changes. • The symbol v is used to represent a velocity vector.

10. Velocity vectors • Suppose a ball is launched at 5 m/s at an angle of 37° • At the moment after launch, the velocity vector for the ball written as a magnitude‑angle pair is v = (5 m/s, 37°).

11. Velocity vectors • In x-y components, the same velocity vector is written as v = (4, 3) m/s. • Both representations tell you exactly how fast and in what direction the ball is moving at that moment.

12. Using velocity vectors A train moves at a speed of 100 km/h heading east. What is its velocity vector in x-y form? • Looking for:… the velocity vector. • Given:… speed (100 km/h) and direction (east). • Relationships: x-velocity is east and y-velocity is north. • Solution:v = (100,0) km/h

13. Projectile motion • Any object moving through air and affected only by the force of gravity is called a projectile. • Flying objects such as airplanes and birds are not projectiles, because they are affected by forces generated from their own power.

14. Projectile motion • The path a projectile follows is called its trajectory. • The trajectory of a projectile is a special type of arch- or bowl-shaped curve called a parabola.

15. Trajectory and range • The rangeof a projectile is the horizontal distance it travels in the air before touching the ground. • A projectile’s range depends on the speed and angle at which it is launched.

16. Two dimensional motion • Projectile motion is two-dimensional because there is both horizontal and vertical motion. • Both speed and direction change as a projectile moves through the air.

17. A ball rolling off a table • The horizontal and vertical components of a projectile’s velocity are independent of each other.

18. Horizontal velocity • The ball’s horizontal motion looks exactly like the its motion if it was it rolling along the ground at 5 m/s.

19. Vertical velocity • The vertical (y) velocity increases due to the acceleration of gravity.

20. NOTE: These equations are suitable only for situations where the projectile starts with zero vertical velocity, such as a ball rolling off a table.

21. Range of a Projectile The range, or horizontal distance, traveled by a projectile depends on the launch speed and the launch angle.

22. Range of a Projectile The range of a projectile is calculated from the horizontal velocity and the time of flight. • The air time and height are greatest when a ball is hit at an angle of 90°, but air time and height are zero when a ball is hit at an angle of 0°.

23. Projectile motion A stunt driver steers a car off a cliff at a speed of 20.0 m/s. The car lands in a lake below 2.00 s later. Find the horizontal distance the car travels and the height of the cliff. • Looking for:… vertical and horizontal distances. • Given:… the time (2.00 s) and initial horizontal speed (20.0 m/s). • Relationships:Use : dx= vxt dy = 4.9t2 • Solution:dx= (20 m/s)(2 s) = 40 m. dy= (4.9 m/s2)(2 s)2 = (4.9 m/s2)(4 s2) = 19.6 m.

24. Unit 2: Forces Chapter 6: Systems in Motion • 6.1 Motion in Two Dimension • 6.2 Circular Motion • 6.3 Centripetal Force, Gravitation, and Satellites • 6.4 Center of Mass

25. Key Question: How does launch speed affect the range of a projectile? 6.2 Investigation: Launch Speed and Range Objectives: • Use the Marble Launcher to explore if a relationship exists between the launch speed and the range of a projectile.

26. Motion in Circles We say an object rotates about its axis when the axis is part of the moving object. A child revolves on a merry-go-round because he is external to the merry-go-round's axis.

27. Motion in Circles Angular speed is the rate at which an object rotates or revolves. There are two ways to measure angular speed number of turns per unit of time (rotations/minute) change in angle per unit of time (degrees/s or radians/s)

28. Angular speed • There are 360 degrees in a full rotation, so one rotation per minute is the same angular speed as 360 degrees per minute

30. Calculating angular speed A merry-go-round makes 10 rotations in 2 minutes. What is its angular speed in rpm? • Looking for:… the angular speed in rotations per minute. • Given:… number of rotations (10) and the time (2 min.) • Relationships:Use:angular speed = rotation time • Solution:angular speed = 10 rotations= 5 rpm 2 minutes

31. The relationship between linear and angular speed • Each point on a rotating object has the same angular speed. • The linear speedof each child is not the same because they travel different distances.

32. The relationship between linear and angular speed • The distance traveled during one revolution equals the circumferenceof the circle.

33. The relationship between linear and angular speed • The linear speed (v) of a point at the edge of a turning circle is the circumference divided by the time it takes to make one full turn. • The linear speed of a point on a wheel depends on the radius, r, which is the distance from the center of rotation.

34. Calculating linear speed The blades on a ceiling fan spin at 60 rpm. The fan has a radius of 0.5 m. Calculate the linear speed of a point at the outer edge of a blade in m/s. • Looking for:… the linear speed in m/s. • Given:… angular speed (60 rpm) and the radius (0.5 m) • Relationships:Use:v = 2  r t • Solution: The blades spin at 60 rotations per minute, so they make 60 rotations in 60 seconds. Therefore it takes 1 second to make one rotation. v = 2 (3.14) (0.5 m) = 3.14 m/s (1 s)

35. Linear, rotational and rolling motion • Rolling is a combination of linear motion and rotational motion. • Holding a bicycle wheel up in the air and moving it to the right is linear motion. • If you lift a bicycle’s front wheel off the ground and make it spin, the spinning wheel is rotational motion.

36. Linear distance equals circumference • The distance the bicycle moves depends on the wheel’s size and angular speed. • When the wheel makes one full rotation, the bicycle goes forward one circumference of the wheel.

37. Unit 2: Forces Chapter 6: Systems in Motion • 6.1 Motion in Two Dimension • 6.2 Circular Motion • 6.3 Centripetal Force, Gravitation, and Satellites • 6.4 Center of Mass

38. Key Question: How do levers work? 6.3 Investigation: Levers and Rotational Equilibrium Objectives: • Explain the meaning of torque and describe its relationship to how levers work. • Use different combinations of weights to balance a lever. • Apply an understanding of rotational equilibrium to determine an unknown mass.

39. Centripetal Force Any force that causes an object to move in a circle is called a centripetal force. A centripetal force is always perpendicular to an object’s motion, toward the center of the circle.

40. Centripetal force and direction • Whether a force makes an object accelerate by changing its speed or by changing its direction or both depends on the direction of the force.

41. Centripetal force and direction • Imagine tying a ball to the end of a string an twirling it in a circle over your head. • The string exerts the centripetal force on the ball to move it in a circle. • The direction of the centripetal force changes as the object moves around you.

42. Centripetal force and inertia • If you give a ball on a string an initial velocity to the left at point A, it will try to keep moving straight to the left. • But the centripetal force pulls the ball to the side.

43. Centripetal force and inertia • A short time later, the ball is at point B and its velocity is 90 degrees from what it was. • But now the centripetal force pulls to the right. • The ball’s inertia makes it move straight, but the centripetal force always pulls it towards the center.

44. Centripetal force and inertia • Notice that the velocity is always perpendicular to the string and therefore to the centripetal force. • The centripetal force and velocity are perpendicular for any object moving in a circle.

45. Newton’s 2nd law and circular motion • An object moving in a circle at a constant speed accelerates because its direction changes. • How quickly an object changes direction depends on its speed and the radius of the circle. • Centripetal accelerationincreases with speed and decreases as the radius gets larger.

46. Centripetal Acceleration Acceleration is the rate at which an object’s velocity changes as the result of a force. Centripetal acceleration is the acceleration of an object whose direction and velocity changes.

47. Newton’s 2nd law and circular motion • Newton’s second law relates force, mass, and acceleration. • The strength of the centripetal force needed to move an object in a circle depends on its mass, speed, and the radius of the circle.

48. Newton’s 2nd law and circular motion • Newton’s second law relates force, mass, and acceleration. • The strength of the centripetal force needed to move an object in a circle depends on its mass, speed, and the radius of the circle. 1. Centripetal force is directly proportional to the mass. A 2-kg object needs twice the force to have the same circular motion as a 1-kg object.

49. Newton’s 2nd law and circular motion 2. Centripetal force is inversely proportional to the radius of its circle. The smaller the circle’s radius, the greater the force. An object moving in a 1 m circle needs twice the force it does when it moves in a 2 m circle at the same speed.

50. Centrifugal force? • Have you ever noticed that when a car makes a sharp turn, you are pushed toward the outside edge of the car? • This apparent outward force is called centrifugal force. • While it feels like there is a force acting on you, centrifugal force is not a true force.