Chapter 8: Dynamics in a plane

Transcription

1 8.1 Dynamics in 2 Dimensions p Chapter 8: Dynamics in a plane 8.2 Velocity and Acceleration in uniform circular motion (a review of sec. 4.6) p Dynamics of Uniform Circular Motion p Circular Orbits p Why does the water stay in the bucket? p Non-uniform Circular Motion p We do not cover 8.5 Fictitious Forces 1

2 Chapter 8: dynamics in a plane 8.1 Dynamics in 2 Dimensions p

3 Newton's FIRST and SECOND laws carried over to 2D motion. This makes it possible to study many different situations. Paticularly important: projectile motion A projectile is an object that moves in two dimensions under the influence of only the gravitational force 3

4 Cliction 8.1 The components of this particle s acceleration are 1. a x > 0, a y > a x = 0, a y > a x < 0, a y > a x > 0, a y < a x < 0, a y < 0. 4

5 8.2 Velocity and Acceleration in Uniform Circular Motion p

6 Velocity of the particle: v is always tangent to the circle Speed: Chapter 8: Dynamics in a plane The speed is constant and has the right unit, m/s a always points in towards the center of the circle and thus is perpendicular to the velocity. This is because the speed is constant Magnitude of acceleration: = a 6

7 Chapter 8: Dynamics in a plane Example: ferris wheel at the Heritage Park Estimates from a visit: radius r = 6m period T = 9 s so ω = 2π/T 0.7 rad/s acceleration: a = ω2 r = 2.9 m/s2 0.3 g In addition, the weight force always pulls you down so that your apparent weight changes. This gives you the funny feeling ^ 7

8 The r-t and r-t-z coordinate systems For circular motion it is more convenient to decompose a vector A into: Radial component A r (positive if pointing inwards) Tangential component A t (positive for counterclockwise) We will sometimes also need a third coordinate axis z perpendicular to the circle: rtz coordinate system 8

9 Components of V and a in the rtz system The three components of the relevant vectors velocity and acceleration are With respect to radial and tangential direction, each of these vectors has only one non-zero component: 9

10 Rank in order, from largest to smallest, the centripetal accelerations (a r ) a to (a r ) e of particles a to e. 1. (a r ) b > (a r ) e > (a r ) a > (a r ) d > (a r ) c 2. (a r ) b = (a r ) e > (a r ) a = (a r ) c > (a r ) d 3. (a r ) b > (a r ) a = (a r ) c = (a r ) e > (a r ) d 4. (a r ) b > (a r ) a = (a r ) a > (a r ) e > (a r ) d 5. (a r ) b > (a r ) e > (a r ) a = (a r ) c > (a r ) d Cliction

11 8.3 Dynamics of Uniform Circular Motion p

12 The relation between speed and magnitude of acceleration is quite often very useful. KNOW IT BY HEART!! Newton's 2 nd law allows to infer the net force from this: Hence, there must always be some force that keeps the object on the circular trajectory 12

13 Example: turning the corner Forces acting on car: Chapter 8: Dynamics in a plane Weight w downwards (w z = -w) Normal force upwards (n z = n) Static friction force f s radially inwards, 0 < (f s ) r < µ s n. The friction force keeps the car in the corner. 13

14 The car continues to corner if it is not too fast, i.e., as long as For an r = 50 m corner radius and µ s = 1 we find 14

15 Cliction 8.3 A block on a string spins in a horizontal circle on a frictionless table. Rank order, from largest to smallest, the tensions T a to T e acting on blocks a to e. 1. T b > T a > T d > T c > T e 2. T d > T b = T e > T c > T a 3. T e > T c = T d > T a = T b 4. T e > T d > T c > T b > T a 5. T d > T b > T e > T c > T a 15

16 End of Week 8 16

17 8.4 Circular Orbits p

18 If the launch speed of the a projectile is sufficiently large, there comes a point where The curve of the trajectory and the curve of the Earth are parallel. In this case the projectile falls but it never gets any closer to the ground! The projectile is said to be in a CIRCULAR ORBIT (c) An object (say a satellite) moving in a circle of radius r around a planet with an acceleration of gravity g will be orbiting at a speed The period T is then T orbit = 2πr Vorbit = 2π r g V = r.g orbit 18

19 8.6 Why does the water stay in the bucket?) p

20 Example: looping in a roller coaster The only forces acting on the car are its weight and the normal force exerted by the track Your apparent weight gives you an intuition about the normal force's magnitude 20

21 The two forces always add up to the acceleration: At the top w points radially inward At the botton its the opposite 21

22 That means: at the bottom your apparent weight (= magnitude of n) is increased, at the top it is decreased, At the top our result for n r becomes negative for small enough velocities. This is unphysical: our assumption that the car stays on a circular orbit is not true anymore 22

23 What really goes on: weight and normal force are added to create the net force that is needed to produce the acceleration a = v 2 / r If v is too small the track had to pull the car rather than to push it, but this is impossible. Instead, the car derails At a minimum critical speed v c, the acceleration is entirely produced by the weight of the car 23

24 The critical speed The critical speed is reached when the normal force at the top is zero (you are not pushed in the chair anymore): v c = ω c r critical angular velocity 24

25 A car is rolling over the top of a hill at speed v. At this instant, Cliction n > w. 2. n = w. 3. n < w. 4. We can t tell about n without knowing v. 25

26 Clicker Question 7.8 A roller coaster car does a loop-the-loop. Which of the free-body diagrams below shows the forces on the car at the top of the loop? Rolling friction can be neglected. A B C D E 26 Slide 7-26

27 Clicker Question 7.8 Answer A roller coaster car does a loop-the-loop. Which of the free-body diagrams below 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. w n A B C D E 27 Slide 7-27

28 8.7 Non-uniform Circular Motion: p

29 NO Tangential acceleration Tangential acceleration is NOT zero The tangential acceleration is what causes the particle to change the speed with which It goes around the circle. 29

30 Equations for non-uniform circular motion in the rtz system Initial angle Initial angular velocity Tangential acceleration For uniform circular motion, simply replace the tangential acceleration by zero, or a t = 0, in the equations. 30

31 Clicker Question 7.3 An ice hockey puck is tied by a string to a stake in the ice. The puck is then swung in a circle. What force or forces does the puck feel? A. A new force: the centripetal force. B. A new force: the centrifugal force. C. One or more of our familiar forces pushing outward. D. One or more of our familiar forces pulling inward. E. I have no clue. 31 Slide 7-31

32 Clicker Question 7.3 Answer An ice hockey puck is tied by a string to a stake in the ice. The puck is then swung in a circle. What force or forces does the puck feel? A. A new force: the centripetal force. B. A new force: the centrifugal force. C. One or more of our familiar forces pushing outward. D. One or more of our familiar forces pulling inward. E. I have no clue. The rules about what is or is not a force haven t changed. 1. Force must be exerted at a point of contact (except for gravity) 2. Force must have an identifiable agent doing the pushing or pulling. 3. The net force must point in the direction of acceleration (Newton s second law) 32 Slide 7-32

33 Clicker Question 7.4 An ice hockey puck is tied by a string to a stake in the ice. The puck is then swung in a circle. What force is producing the centripetal acceleration of the puck? A. Gravity B. Air resistance C. Friction D. Normal force E. Tension in the string Draw a free-body diagram in which you see the puck from ahead or behind, with the z-axis perpendicular to the ice. 33 Slide 7-33

34 Clicker Question 7.4 Answer An ice hockey puck is tied by a string to a stake in the ice. The puck is then swung in a circle. What force is producing the centripetal acceleration of the puck? A. Gravity B. Air resistance C. Friction D. Normal force E. Tension in the string 34 Slide 7-34

35 Clicker Question 7.5 A coin sits on a turntable as the table rotates ccw. The free-body diagrams below show the coin from behind, moving away from you. Which is the correct diagram? z z z z z r r r r r B D A C E 35 Slide 7-35

36 Clicker Question 7.5 Answer A coin sits on a turntable as the table rotates ccw. The free-body diagrams below show the coin from behind, moving away from you. Which is the correct diagram? z Center of circle is to the left. z z z What force is this? z r r r r Static friction! r B D A Net force must point to the center of the circle. C E 36 Slide 7-36

37 Chapter 8 Reading Quiz 37

38 Circular motion is best analyzed in a coordinate system with 1. x- and y-axes. 2. x-, y-, and z-axes. 3. x- and z-axes. 4. r-, t-, and z-axes. 38

39 The quantity with the symbol ω is called 1. the circular weight. 2. the angular velocity. 3. the circular velocity. 4. the centripetal acceleration. 39

40 For uniform circular motion, the net force 1. points toward the center of the circle. 2. points toward the outside of the circle. 3. is tangent to the circle. 4. is zero. 40

41 Selected Problems 41

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46 End of Chapter 8 IMPORTANT: Print a copy of the SUMMARY page (p. 229) and add it here to your lecture notes. It will save you crucial time when trying to recall: Concepts, Symbols, and Strategies 46

47 Chapter 7: motion in a circle 7.5 Fictitious Forces and Apparent Weight p

48 Chapter 7: motion in a circle Why are you pushed to the outside of the corner? The reason, the centrifugal force, is a fictitious force. Such forces are no real forces Fictitious forces appear in noninertial reference frames: If the reference frame is accelerated, uniform straight motion seems to correspond to an accelerated motion. 48

49 Chapter 7: motion in a circle The centrifugal forces simply describes the tendency to continue on a straight line Important: since fictitious forces are related to non-inertial reference frames, they do not appear in a free-body diagram They are related to the apparent weight that we discussed before. Main difference: we now have circular instead of straight motion 49