* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Lecture-08-09
Survey
Document related concepts
Hooke's law wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Jerk (physics) wikipedia , lookup
Coriolis force wikipedia , lookup
Hunting oscillation wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Nuclear force wikipedia , lookup
Fictitious force wikipedia , lookup
Seismometer wikipedia , lookup
Centrifugal force wikipedia , lookup
Work (thermodynamics) wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Transcript
Chapter 6 Applications of Newton’s Laws Midterm # 1 Average = 16.35 Standard Deviation = 3.10 Well done! 2 Who Wins? Kevin Parks (5’6”, 90kg) takes a handoff from Michael Rocco, but the hole closes as he hits the line. He is pinned between to O-linemen behind him, and 3 D-linemen ahead... and everyone is pushing, lifting, grabbing, and kicking. His acceleration is 1 m/s2 upfield. What is the net force on KP? a) I need to know a lot more about angles, masses of the line men, etc. b) I could answer this if only I had the initial velocity c) I would also need to know what planet this happened on. d) 90 N, upfield. e) This isn’t the Tech game, is it? 3 Who Wins? Kevin Parks (5’6”, 90kg) takes a handoff from Michael Rocco, but the hole closes as he hits the line. He is pinned between to O-linemen behind him, and 3 D-linemen ahead... and everyone is pushing, lifting, grabbing, and kicking. His acceleration is 1 m/s2 upfield. What is the net force on KP? a) I need to know a lot more about angles, masses of the line men, etc. b) I could answer this if only I had the initial velocity c) I would also need to know what planet this happened on. d) 90 N, upfield. e) This isn’t the Tech game, is it? If you know the acceleration of a body, you know the direction of the net force. If you also know the mass, then you know the magnitude. 4 Will It Budge? A box of weight 100 N is at rest on a floor where μs = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move? a) moves to the left, because the force of static friction is larger than the applied force b) moves to the right, because the applied force is larger than the static friction force c) the box does not move, because the static friction force is larger than the applied force d) the box does not move, because the static friction force is exactly equal the applied force e) The answer depends on the value for μk. Static friction (s = 0.4 ) m T 5 Will It Budge? A box of weight 100 N is at rest on a floor where μs = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move? a) moves to the left, because the force of static friction is larger than the applied force b) moves to the right, because the applied force is larger than the static friction force c) the box does not move, because the static friction force is larger than the applied force d) the box does not move, because the static friction force is exactly equal the applied force e) The answer depends on the value for μk. The static friction force has a maximum of sN = 40 N. The tension in the rope is only 30 N. So Static friction (s = 0.4 ) m T the pulling force is not big enough to overcome friction. Follow-up: What happens if the tension is 35 N? What about 45 N? 6 Springs Hooke’s law for springs states that the force increases with the amount the spring is stretched or compressed: The constant k is called the spring constant. 7 Springs Note: we are discussing the force of the spring on the mass. The force of the spring on the wall are equal, and opposite. 8 Springs and Tension S1 S2 A mass M hangs on spring 1, stretching it length L1 Mass M hangs on spring 2, stretching it length L2 Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch? a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller c) L1 or L2, whichever is bigger d) depends on which order the springs are attached e) L1 + L2 9 Springs and Tension A mass M hangs on spring 1, stretching it length L1 Mass M hangs on spring 2, stretching it length L2 Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch? S1 S2 Fs=T W a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller c) L1 or L2, whichever is bigger d) depends on which order the springs are attached e) L1 + L2 10 Springs and Tension A mass M hangs on spring 1, stretching it length L1 Mass M hangs on spring 2, stretching it length L2 Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch? S1 S2 Fs=T W a) (L1 + L2) / 2 b) L1 or L2, whichever is smaller c) L1 or L2, whichever is bigger d) depends on which order the springs are attached e) L1 + L2 Spring 1 supports the weight. Spring 2 supports the weight. Both feel the same force, and stretch the same distance as before. 11 Recall -- Instantaneous acceleration Velocity vector is always in the direction of motion; acceleration vector can points in the direction velocity is changing: 12 Circular Motion and Centripetal Force An object moving in a circle must have a force acting on it; otherwise it would move in a straight line! The net force must have a component centripetal pointing to the center of the circle force This force may be provided by the tension in a string, the normal force, or friction, or.... 13 Circular Motion and Centripetal Acceleration An object moving in a circle must have a force acting on it; otherwise it would move in a straight line. a a If the speed is constant, the direction of the acceleration (which is due to the net force) is towards the center of the circle. The magnitude of this centripetal component of the force is given by: For circular motion problems, it is often convenient to choose coordinate axes with one pointing along the direction of this centripetal force 14 Circular Motion An object may be changing its speed as it moves in a circle; in that case, there is a tangential acceleration as well: 15 Missing Link A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the Ping-Pong ball leaves the track, which path will it follow? d e a b c 16 Missing Link A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the Ping-Pong ball leaves the track, which path will it follow? d e a b c Once the ball leaves the tube, there is no longer a force to keep it going in a circle. Therefore, it simply continues in a straight line, as Newton’s First Law requires! Follow-up: What physical force provides the centripetal force? 17 Tetherball a) toward the top of the pole In the game of tetherball, b) toward the ground the struck ball whirls c) along the horizontal component of the tension force around a pole. In what direction does the net d) along the vertical component of the tension force force on the ball point? e) tangential to the circle T W 18 Tetherball In the game of tetherball, a) toward the top of the pole the struck ball whirls b) toward the ground around a pole. In what c) along the horizontal component of the tension force direction does the net force on the ball point? d) along the vertical component of the tension force e) tangential to the circle The vertical component of the tension balances the weight. The horizontal component of tension provides the W T T necessary centripetal force that accelerates the ball to towards the pole and keeps it moving in a circle. W 19 20 Examples of centripetal force when no friction is needed to hold the track! 21 A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest? 22 22 A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest? necessary centripetal force: Only force on puck is tension in the string! To support mass M, the necessary tension is: 23 Circular motion and apparent weight Car at the bottom of a dip Note: at any specific point, any curve can be approximated by a portion of a circle This normal force is the apparent, or perceived, weight 24 Going in Circles I You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by a) N remains equal to mg b) N is smaller than mg your seat is equal to your weight mg. c) N is larger than mg How does N change at the top of the d) none of the above Ferris wheel when you are in motion? 25 Going in Circles I You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by a) N remains equal to mg b) N is smaller than mg your seat is equal to your weight mg. c) N is larger than mg How does N change at the top of the d) none of the above Ferris wheel when you are in motion? You are in circular motion, so there has to be a centripetal force pointing inward. At the top, the only two forces are mg (down) and N (up), so N must be smaller than mg. Follow-up: Where is N larger than mg? 26 26 Vertical circular motion Centripetal acceleration must be A vertical (up) B horizontal C vertical (down) Condition for falling: N=0 at C: So, as long as: at the top, then N>0 and pointing down. (note: before falling, apparent weight is 27 in the opposite direction to true weight!)27 Chapter 7 Work and Kinetic Energy http://people.virginia.edu/~kdp2c/downloads/WorkEnergySelections.html Work Done by a Constant Force The definition of work, when the force is parallel to the displacement: SI unit: newton-meter (N·m) = joule, J 29 29 Working Hard... or Hardly Working Atlas holds up the world. Sisyphus pushes his rock up a hill. (b) (a) Who does more work? 30 Working Hard... or Hardly Working Atlas holds up the world. Sisyphus pushes his rock up a hill. (b) (a) With no displacement, Atlas does no work 31 Forces not along displacement If the force is at an angle to the displacement: 32 32 Friction and Work I A box is being pulled a) friction does no work at all across a rough floor b) friction does negative work at a constant speed. What can you say c) friction does positive work about the work done by friction? 33 Friction and Work I A box is being pulled a) friction does no work at all across a rough floor b) friction does negative work at a constant speed. What can you say c) friction does positive work about the work done by friction? N Displacement Friction acts in the opposite direction to the displacement, so the work is negative. Pull f Or using the definition of work (W = F (Δr)cos ), because = 180º, then W < 0. mg 34 Friction and Work II Can friction ever do positive work? a) yes b) no 35 Friction and Work II Can friction ever do positive work? a) yes b) no Consider the case of a box on the back of a pickup truck. If the box moves along with the truck, then it is actually the force of friction that is making the box move. ... but .... the friction isn’t really doing the work, it is only transmitting the forces to the box, while work is done by the truck engine. My view: the language is confusing so I’m not interested in arguing the point. 36 Convenient notation: the dot product The work can also be written as the dot product of the force and the displacement: vector “dot” operation: project one vector onto the other 37 Force and displacement The work done may be positive, zero, or negative, depending on the angle between the force and the displacement: 38 38 Sum of work by forces = work by sum of forces If there is more than one force acting on an object, the work done by each force is the same as the work done by the net force: 39 Units of Work Lifting 0.5 L H2O up 20 cm = 1 J 40 Play Ball! In a baseball game, the catcher stops a 90-mph a) catcher has done positive work pitch. What can you say b) catcher has done negative work about the work done by the c) catcher has done zero work catcher on the ball? 41 Play Ball! In a baseball game, the catcher stops a 90-mph a) catcher has done positive work pitch. What can you say b) catcher has done negative work about the work done by the c) catcher has done zero work catcher on the ball? The force exerted by the catcher is opposite in direction to the displacement of the ball, so the work is negative. Or using the definition of work (W = F (Δr)cos ), because = 180º, then W < 0. Note that the work done on the ball is negative, and its speed decreases. Follow-up: What about the work done by the ball on the catcher? 42 Tension and Work A ball tied to a string is a) tension does no work at all being whirled around in b) tension does negative work a circle with constant c) tension does positive work speed. What can you say about the work done by tension? 43 Tension and Work A ball tied to a string is a) tension does no work at all being whirled around in b) tension does negative work a circle with constant c) tension does positive work speed. What can you say about the work done by tension? No work is done because the force acts in a perpendicular direction to the displacement. Or using the definition of work (W = F (Δr)cos ), T = 90º, then W = 0. v Follow-up: Is there a force in the direction of the velocity? 44 Work by gravity Fg A ball of mass m drops a distance h. What is the total work done on the ball by gravity? a W = Fd = Fgx h h W = mgh Path doesn’t matter when asking “how much work did gravity do?” Only the change in height! A ball of mass m rolls down a ramp of height h at an angle of 45o. What is the total work done on the ball by gravity? a Fgx = Fg sinθ N h = L sinθ Fg W = Fd = Fgx L = (Fg sinθ) (h / sinθ) h W = Fg h = mgh θ 45 Motion and energy When positive work is done on an object, its speed increases; when negative work is done, its speed decreases. 46 Kinetic Energy As a useful word for the quantity of work we have done on an object, thereby giving it motion, we define the kinetic energy: 47 Work-Energy Theorem Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy. (True for rigid bodies that remain intact) 48 Lifting a Book You lift a book with your hand in a) mg r such a way that it moves up at b) FHAND r constant speed. While it is c) (FHAND + mg) r moving, what is the total work d) zero done on the book? e) none of the above r FHAND v = const a=0 mg 49 Lifting a Book You lift a book with your hand in a) mg r such a way that it moves up at b) FHAND r constant speed. While it is c) (FHAND + mg) r moving, what is the total work d) zero done on the book? e) none of the above The total work is zero because the net force acting on the book is zero. The work done by r FHAND the hand is positive, and the work done by gravity is negative. The sum of the two is v = const a=0 zero. Note that the kinetic energy of the book does not change either! mg Follow-up: What would happen if FHAND were greater than mg? 50 Kinetic Energy I By what factor does the a) no change at all kinetic energy of a car b) factor of 3 change when its speed is c) factor of 6 tripled? d) factor of 9 e) factor of 12 51 Kinetic Energy I By what factor does the a) no change at all kinetic energy of a car b) factor of 3 change when its speed is c) factor of 6 tripled? d) factor of 9 e) factor of 12 Because the kinetic energy is mv2, if the speed increases by a factor of 3, then the KE will increase by a factor of 9. 52 Slowing Down If a car traveling 60 km/hr can a) 20 m brake to a stop within 20 m, what b) 30 m is its stopping distance if it is c) 40 m traveling 120 km/hr? Assume d) 60 m that the braking force is the same e) 80 m in both cases. 53 53 Slowing Down If a car traveling 60 km/hr can a) 20 m brake to a stop within 20 m, what b) 30 m is its stopping distance if it is c) 40 m traveling 120 km/hr? Assume d) 60 m that the braking force is the same e) 80 m in both cases. F d = Wnet = KE = 0 – and thus, |F| d = mv2, mv2. Therefore, if the speed doubles, the stopping distance gets four times larger. 54 54 Work Done by a Variable Force We can interpret the work done graphically: 55 Work Done by a Variable Force If the force takes on several successive constant values: 56 Work Done by a Variable Force We can then approximate a continuously varying force by a succession of constant values. 57 Work Done by a Variable Force The force needed to stretch a spring an amount x is F = kx. Therefore, the work done in stretching the spring is 58