RostockDeltaKinematics(中译全文)
kinematics(动力学)1动力学课件
The subjects that occupied physical scientists through the end of the nineteenth century----mechanics, light, heat, sound, electricity and magnetism-------are usually referred to as classical physics.
Velocity is the rate at which the position changes. • Average Velocity The average velocity of a particle during the time interval Δt is defined as the ratio of the displacement to the time interval.
It is classical physics we must master to understand the macroscopic world we live in, and classical physics is the main subject of the first part of the course.
K Δr and Δr
K Δr ≠ Δr
#mech001
Which of the following statements is (are) true ? (a) Δr is the magnitude of the displacement .
G (b) Δr is the magnitude of the displacement.
z
P1
G v1
Kinematics
Kenematics1. An impatient motorist considers speeding as he travels between two cities. If the trip normally takes2.8 h at an averagespeed of 100.0 km/h, how much time will be saved if he exceeds the speed limit by 10.0 km/h?2. Two friends plan to meet at a cottage for a weekend retreat. One person must drive a distance of 1.5 ⨯ 102 km at anaverage speed of 85 km/h. The other person has only 90.0 km to travel and averages a speed of 1.0 ⨯ 102 km/h. If theyboth depart at the same time, how much earlier does the one friend arrive than the other. (Give your answer in minutes.)3. A driver is travelling at 25 m/s when she spots a sign that reads "BRIDGE OUT AHEAD." It takes her 1.0 s to react andbegin braking. The car slows down at a rate of 3.0 m/s2. Luckily, she stops 5.0 m short of the washed-out bridge.(a) How much time was required to stop the car once the brakes were applied?(b) How far was the driver from the bridge when she first noticed the sign?4. A jogger runs 6.0 km [E], then 4.0 km [N], and finally 2.0 km [W]. The entire trip takes 2.0 h to complete. Calculate thejogger's(a) average speed(b) average velocity5. A car leaves Toronto and drives west at 80.0 km/h for the next 1.5 h. The vehicle then proceeds north at 50.0 km/h forthe next 2.0 h.(a) What is the car's average speed for the entire trip?(b) What is the car's average velocity for the entire trip?6. A group of hikers sets out from point A, proceeds to B, then to C, and finally to D. The entire trip takes 6.0 h.(a) Determine the hikers' average speed for the trip.(b) What is the hikers' final displacement relative to their initial position?(c) If the hikers release a homing pigeon upon their arrival at point D and the bird returns to point A 30 min later, whatis the bird's average velocity during the flight?7. A plane leaves Toronto and flies with an airspeed of 2.20 ⨯ 102 km/h always pointing due east. A wind is blowing fromthe north at 8.0 ⨯ 101 km/h.(a) What is the plane's velocity relative to the ground?(b) What is the plane's displacement from Toronto after flying for 2.5 h?8. A boat sets out from the north shore of a 200 m wide east-flowing river. The boat always faces due south but the currentcarries it 300 m downstream while crossing. The trip takes 2.0 min. Assume three significant digits.(a) What is the boat's displacement during the trip?(b) What is the boat's average velocity during the trip?(c) If the boat's velocity relative to the water is 1.7 m/s [S], what is the velocity of the current?9. The position-time graph below represents the motion of a police car, P, in pursuit of a motorcycle, M.(a) How far ahead of the police car was the motorcycle when the pursuit began?(b) What was the motorcycle's velocity during most of the pursuit?(c) How far did the motorcycle travel before stopping?(d) What was the average velocity of the police car during the pursuit?(e) What was the police car's velocity at t = 8.0 s?(f) If the police car started from rest, what was its acceleration (assumed constant) during this pursuit?10. Study the position-time graph pictured below and answer the questions that follow.(a) What is the object's final displacement?(b) What is the object's average velocity during the interval from t = 3.0 s to t = 7.0 s?(c) What is the object's velocity during the time its motion is uniform?(d) What is the object's average speed during the entire interval?(e) What is the object's average velocity during the entire interval?(f) What is the object's instantaneous velocity at t = 1.5 s?(g) What is the length of time during which the object does not move at all?11. A car pulls away from an intersection when the light turns green. After uniformly accelerating for the next 4.0 s, the carhas travelled a distance of 50 m. The car then proceeds at constant speed.(a) What was the car's acceleration?(b) How fast was the car travelling when it finished accelerating?(c) How long will it take for the car to travel another 50 m at this constant speed?(d) Plot the corresponding position-time graph for the car over its 100-m drive.(e) Plot the corresponding speed-time graph for the car over its 100-m drive.12. A sprinter who is competing in a 100-m race accelerates from rest to a top speed of 10.0 m/s over a distance of 15 m.The remainder of the race is run at a constant speed.(a) What length of time is required for the sprinter toreach top speed?(b) What is the sprinter's acceleration?(c) What is the sprinter's time for the entire race?13. A box accidentally falls from the back of a truck and hits the ground with a speed of 15 m/s. It slides along the groundfor a distance of 45 m before coming to rest. Determine(a) the length of time the box slides before stopping(b) the average acceleration of the box while it'ssliding(c) the time it takes to slide the last 10 m14. A car is travelling along a highway with a speed of 25 m/s when the driver sees an obstruction 1.80 ⨯ 102 m directly ahead. It takes thedriver 0.80 s to react and begin braking.(a) How far does the car travel before it begins to slow down?(b) How long will it take the car to stop once the brakes are applied, provided the car stops just before the obstruction?(c) What is the value of the acceleration of the car if it just misses hitting the obstruction? Assume the acceleration is uniform.15. Within 4.0 s of lift-off, a space shuttle reaches an altitude of 4.50 ⨯ 102 m, uniformly accelerating during the entire time.(a) What is its acceleration?(b) At what speed is the shuttle travelling when itreaches this altitude? (c) How long would it take the shuttle to travel thenext 450 m if it stops accelerating after 4.0 s?16. A cyclist is travelling with a speed of 12.0 m/s when she applies the brakes. After slowing for 3.0 s, her speed has beenreduced to 4.0 m/s.(a) What distance does she travel during this time?(b) What is her acceleration?(c) If she continues braking, how much longer will ittake her to stop? (d) If she continues to travel with her new speed, howfar will she travel during the next 3.0 s?17. A car is travelling at 20 m/s when it pulls out to pass a truck that is travelling at only 18 m/s. The car accelerates at 2.0m/s2 for 4.0 s and then maintains this new velocity. (Assume 2 significant digits.)(a) What distance does the car travel during the period of acceleration?(b) What is the car's speed at the end of the period of acceleration?(c) If the car was originally 8.0 m behind the truck when it pulled out to pass, how far in front of the truck is the car10.0 s later?(d) On the same set of axes, sketch what the position-time graph would look like for both the car and the truck.18. An object is pushed from rest across a sheet of ice, accelerating at 5.0 m/s2 over a distance of 80 cm. The object thenslides at a constant speed for 4.0 s until it reaches a rough section that causes the object to stop in 2.5 s. (Assume 2 significant digits.)(a) What is the speed of the object when it reaches therough section? (b) At what rate does it slow down once it reaches therough section?(c) What is the total distance that the object slides?19. A ball is thrown vertically upward from a window that is 3.6 m above the ground. The ball's initial speed is 2.8 m/s andthe acceleration due to gravity is 9.8 m/s2.(a) What is the ball's speed when it hits the ground?(b) How long after the first ball is thrown should a second ball be simply dropped from the same window so that bothballs hit the ground at the same time?20. An object is pushed along a rough surface and released. It slides for 10.0 s before coming to rest and travels a distance of20.0 cm during the last 1.0 s of the slide. Assuming the acceleration is uniform throughout,(a) How fast was the object travelling upon release?(b) How fast was the object travelling when it reached the halfway point in its slide?21. An object is pushed from rest across a sheet of ice, accelerating at 5.0 m/s2 over a distance of 80.0 cm. The object thenslides with a constant speed for 4.0 s until it reaches a rough section which causes it to stop in 2.5 s.(a) What is the speed of the object when it reaches the rough section?(b) At what rate does the object slow down once it reaches the rough section?(c) What total distance does the object slide throughout its entire trip?22. A ball is thrown vertically upward from a window that is 3.6 m above the ground. Its initial speed is 2.8 m/s.(a) With what speed does the ball hit the ground?(b) How long after the first ball is thrown should a second ball be simply dropped from the same window so that bothballs hit the ground at the same time?23. An object is pushed along a rough horizontal surface and released. It slides for 10.0 s before coming to rest and travels adistance of 20.0 cm during the last 1.0 s of its slide. Assuming the acceleration to be uniform throughout(a) How fast was the object travelling upon release?(b) How fast was the object travelling when it reached the halfway position in its slide?24. An arrow is shot vertically upward with an initial speed of 25 m/s. When it’s exactly halfway to the top of its flight, asecond arrow is launched vertically upward from the same spot. The second arrow reaches the first arrow just as the first arrow reaches its highest point.(a) What is the launch speed of the second arrow?(b) What maximum height does the second arrow reach?25. A pedestrian is running at his maximum speed of 6.0 m/s trying to catch a bus that is stopped at a traffic light. When heis 16 m from the bus, the light changes and the bus pulls away from the pedestrian with an acceleration of 1.0 m/s2.(a) Does the pedestrian catch the bus and, if so, how far does he have to run? (If not, what is the pedestrian’s distance ofclosest approach?)(b) How fast is the bus moving when the pedestrian catches it? (or at the distance of closest approach)(c) On a single set of axes, plot the corresponding position-time graphs of both the bus and pedestrian to confirm youranswer in (a).26. A truck travels at a constant speed of 28.0 m/s in the fast lane of a two-lane highway. It approaches a stationary carstopped at the side of the road. When the truck is still 1.2 ⨯ 102 m behind the car, the car pulls out into the slow lane with an acceleration of 2.6 m/s2.(a) How long will it take the truck to pass the car?(b) How far will the car have travelled when the truck passes it?(c) If the car were to maintain this acceleration, how fast would it be travelling when it overtakes the truck?27. A car leaves point A and drives at 80.0 km/h [E] for 1.50 h. It then heads north at 60.0 km/h for 1.00 h and finally [30.0︒W of N] at 100.0 km/h for 0.50 h, arriving at point B.(a) Determine the displacement of point B from point A.(b) A plane flies directly from point A to point B, leaving 2.00 h after the car has departed from point A. It arrives atpoint B at the same time the car arrives. There is a wind blowing at 60.0 km/h due south for the entire trip. What is the airplane’s airspeed?(c ) What direction must the plane head in order to arrive at point B?(d) How long would the plane’s trip be if there was no wind?28. Two planes fly from Toronto to Philadelphia. Plane A flies via Pittsburgh whereas passengers on plane B have a directflight. Pittsburgh is 350 km due south of Toronto and 390 km due west of Philadelphia. The airspeed of both planes is 400.0 km/h and a steady wind is blowing from the east at 60.0 km/h.(a) What direction must the pilot point the plane flying from Toronto to Pittsburgh? Include a vector diagram ofvelocities.(b) How long will the entire flight take for plane A assuming a 0.50-h layover in Pittsburgh?(c) How much time must the pilot of plane B wait before leaving Toronto if she is to arrive in Philadelphia at the sametime plane A arrives?29. Two canoeists, A and B, live on opposite shores of a 300.0 m wide river that flows east at 0.80 m/s. A lives on the northshore and B lives on the south shore. They both set out to visit a mutual friend X who lives on the north shore at a point 200.0 m upstream from A and 200.0 m downstream from B. Both canoeists can propel their canoes at 2.4 m/s through the water. How much time must canoeist A wait after canoeist B sets out so that they both arrive at X at the same time?Both canoeists make their respective trips by the most direct routes.30. The graph below represents the motion of an object over a recorded time interval. Using methods of graphical analysiswherever possible, determine(a) the object’s displacement relative to its starting position at t = 6.0 s.(b) the object’s average velocity between t = 0.0 s and t = 6.0 s.(c) the object’s average speed between t = 0.0 s and t = 6.0 s.(d) Including t = 0.0 s, how many times during the entire recorded time interval is the object at its starting position?(e) During which interval is the object’s acceleration the greatest? What is the value of the acceleration during thisinterval?(f) Plot the corresponding position-time graph.(g) Plot the corresponding acceleration-time graph.31. A shell is fired from a cliff that is 36 m above a horizontal plane. The muzzle speed of the shell is 80.0 m/s and it is firedat an elevation of 25︒ above the horizontal.(a) Determine the horizontal range of the shell.(b) Determine the velocity of the shell as it strikes the ground.32. A football quarterback attempts a pass to one of the receivers. As the ball is snapped, the receiver leaves the line ofscrimmage and runs directly down field. The quarterback releases the ball 2.0 s later and from a position 3.0 m behind the line of scrimmage. He throws the ball with a speed of 26 m/s at an elevation of 60︒ above the horizontal. The receiver makes a diving reception, catching the ball just as it reaches the ground. See the diagram below.(a) What is the time of flight of the football?(b) What is the average speed of the receiver?33. A circus clown is fired from a cannon into a net that is situated 2.0 m above the cannon and some distance from it. Thecannon is elevated at 50.0︒to the horizontal and the clown’s speed at launch is 15 m/s. See the diagram below.(a) Find the horizontal distance from the cannon where the net needs to placed in order for the clown to land in it.(b) Calculate the clown’s velocity as he lands in t he net.34. A boat is 50.0 m from the base of a cliff, fleeing at 5.0 m/s. A gun, mounted on the edge of the cliff fires a shell at 40.0m/s and hits the boat when it has fled another 50.0 m. See the diagram below.(a) At what angle above the horizontal must the gun be aimed so that the shell will hit the target?(b) How high is the cliff?(c) With what velocity does the shell hit the boat?35. A baseball is hit by a bat and given a velocity of 40.0 m/s at an angle of 30.0︒ above the horizontal. The height of the ballabove the ground upon impact with the bat is 1.0 m.(a) What maximum height above the ground does the ball reach?(b) A fielder is 110.0 m from home plate when the ball is hit and the ball’s trajectory is dire ctly at him. If he beginsrunning at the moment the ball is hit and catches the ball when it is still 3.0 m above the ground, how long does he run before catching the ball?(c) How fast (average speed) does he have to run in order to catch the ball?Kenematics: Answer Section1.distance to traveltime to travel if speeding (v = 110km/h)The time saved is2.time to arrive for person Atime to arrive for person Btime differencePerson B arrives 52 min earlier than person A.3.(a)time to stopThe time required to stop is 8.3 s.(b) distance travelled whilereactingdistance travelled while brakingdistance to bridge when stopped: ∆d = 5.0 mtotal distance: 25 m + 104 m + 5.0 m = 1.3 ⨯ 102 mThe driver was 1.3 ⨯ 102 m from the bridge when she first noticed the sign.4.(a)distance ranaverage speedThe jogger's average speed is 6.0 km/h.(b) jogger's displacementaverage velocityThe jogger's average velocity is 2.8 km/h [NE]. 5.(a) distance travelledtotal distance travelled: 120 km + 100 km = 220 km average speedThe car's average speed is 63 km/h.(b) displacementaverage velocityThe car's average velocity is 45 km/h [50ºW of N].6. (a) distance travelled: 2.0 km + 8.0 km + 8.0 km = 18.0 kmaverage speedThe hikers' average speed is 3.0 km/h.(b)hiker's displacementThe final displacement is 10.0 km [53ºE of S].(c) displacement of homing pigeon on return trip10.0 km [53ºW of N] (opposite displacement of hikers)average velocityThe bird's average velocity is 20 km/h [53ºW of N].7. (a)plane's velocity relative to the groundThe plane's velocity relative to the ground is 2.3 ⨯ 102km/h [70ºE of S].(b)plane's displacement from Toronto after 2.5 hThe plane's displacement from Toronto is 5.9 ⨯ 102 km[70︒ E of S].8.(a) boat's displacementThe boat's displacement is 360 m [56︒ E of S].(b)average velocityThe boat's average velocity is 3.00 m/s [56︒ E of S].(c)velocity of currentThe velocity of the current is 2.5 m/s [E].9.(a) 150 m(b)(c) 300 m(d)(e) instantaneous velocity = slope of tangent to graphat t = 8.0 s (approximately 12 m/s [N]) (f)acceleration(a) 16 m [N](b) average velocity = slope of line joining endpointsat t = 3.0 s and t = 7.0 s (3.5 m/s [N])(c) from t = 3.0 s to t = 4.5 s, velocity = 6.7 m/s [N](d)(e)(f) instantaneous velocity = slope of tangent at t = 1.5s (approximately 1.6 m/s [N]) (g) 1.0 s (from t = 6.0 s to t = 7.0 s)11.(a)(b)(c)(d)(e)(a)The time required is 3.0 s.(b)The acceleration is 3.3 m/s2.(c)time to run 85 m:total time: ∆t = 3.0 s + 8.5 s =11.5 sThe sprinter's time for the raceis 11.5 s.13.(a)The length of time the box slides is 6.0 s.(b)The acceleration of the box is–2.5 m/s2.(c)The time it takes to slide the last10 m is 2.8 s.14.(a)The car travels 20 m.(b) distance to stop: 180 m – 20m = 160 mThe time required to stop is 13s.(c)The acceleration is –2.0 m/s2.15.(a)The acceleration is 56 m/s2.(b)The shuttle is travelling at 2.2 ⨯102 m/s.(c)It would take another 2.0 s totravel the next 450 m.(a)She travels 24 m.(b)Her acceleration is –2.7 m/s2.(c)The time required is 1.5 s.(d)She will travel 12 m.17.(a)The car travels 96 m.(b)The car's speed is 28 m/s.(c) car: ∆d = v∆t = 28 m/s(6.0 s)= 168 mtotal: ∆d = 96 m + 168 m =264 mtruck: ∆d = v∆t = 18 m/s(10.0s) = 180 mCar's lead after 10.0 s: ∆d = 264m – 180 m – 8.0 m = 76 m(d)(a)The object's speed is 2.8 m/s.(b)The rate of acceleration is –1.1 m/s 2.(c) accelerating: ∆d = 0.80 muniform motion:decelerating:total distance: ∆d = 0.80 m + 11.3 m + 3.5 m = 16 m19.(a) Let [up] be the "negative" direction and [down] be the"positive" direction.The ball's speed is 8.9 m/s.(b)first ball:second ball:Time to wait: ∆t = 1.19 s – 0.85 s = 0.3 s20.(a)The object is travelling at 4.0 m/s.(b)total distance:halfway point: ∆d = 10 mspeed at that point:The object is travelling at 2.8 m/s.(a)v 1 = 0.0 m/s a = 5.0 m/s 2∆d = 80.0 cm = 0.800 m v 2 = ?The speed of the object upon reaching the rough section is 2.8 m/s. (b)v 1 = 2.83 m/s v 2 = 0.0 m/s ∆t = 2.5 s a = ?The object’s acceleration is 1.1 m/s 2 and slowing.(c)During the period of acceleration: ∆d = 0.800 mDuring the period of uniform motion: v = 2.83 m/s ∆t = 4.0 s∆d = v ∆t = 2.83 m/s(4.0 s) = 11.32 mDuring the period of deceleration: v 1 = 2.83 m/s v 2 = 0.0 m/s ∆t = 2.5 sTotal distance the object slides: 0.800 m + 11.32 m + 3.54 m = 16 mThe object slides a total distance of 16 m.22. (a)Using a sign convention of “down” as (+) and “up” as (–).The speed of the object when it hits the ground is 8.9 m/s.(b)Time of flight for the first ball:Time of flight for the second ball:The difference in flight times is 1.19 s – 0.86 s = 0.33 s.The second ball should be dropped 0.33 s after the first one is thrown so that both hit the ground at the same time.23. (a)The object’s acceleration during the last 1.0 s: ∆t = 1.0 sv 1 = 40 cm/sv 2 = 0.0 cm/s a = ?This is also the acceleration for the entire trip.The speed upon release:The object was travelling at 4.0 m/s upon release.(b)The distance travelled: ∆t = 10.0 s v 2 = 0.0 cm/s a = –40 cm/s 2 ∆d = ?At the halfway position: ∆d = 1.0 ⨯ 103 cm v 2 = 0.0 cm/s a = –40 cm/s 2 v 1 = ?The object is travelling at 2.8 m/s at the halfway position in its slide.is (+):v1 = –25 m/sv2 = 0.0 m/sa = 9.8 m/s2∆d = ?The arrow travels 31.9 m upward to its highest point. The halfway position is 15.9 m. The time to travel the last half of its flight: Array∆d = –15.9 mv2 = 0.0 m/sa = 9.8 m/s2∆t = ?For the second arrow: Array∆d = -31.9 ma = 9.8 m/s2∆t = 1.80 sv1 = ?The speed of the second arrow at launch is 27 m/s [upward].(b)Finding the maximum height of the second arrow: Array v1 = –26.5 m/sv2 = 0.0 m/sa = 9.8 m/s2∆d = ?The second arrow reaches a maximum height of 36 m [upward].Bus: v 1B = 0.0 m/s, a B = 1.0 m/s 2 Pedestrian: v P = 6.0 m/sBus:Pedestrian: ∆d P = v P ∆t∆d P = 6.0 ∆t∆d P = ∆d B + 16 m 6.0 ∆t = 0.5(∆t )2 + 16solving the quadratic: ∆t = 4.0 s, 8.0 sThe pedestrian does catch the bus after running for 4.0 s.∆d = 6.0 m/s(∆t ) = 6.0 m/s(4.0 s) ∆d = 24 m The pedestrian runs 24 m to catch the bus. (b)v 1B = 0.0 m/s a B = 1.0 m/s 2 ∆t = 4.0 sv 2B = ?v 2B = v 1B + a B ∆t= (1.0 m/s 2)(4.0 s) v 2B = 4.0 m/sThe bus is travelling at 4.0 m/s when the pedestrian catches it.(c) The position-time graph of the motion of the pedestrian is a straight diagonal line that begins at the origin. The line that represents the motion of the bus is a curved line that begins at the 8.0 m mark and crosses that pedestrian’s line twice, at 4.0 s and at 8.0 s.26. (a)Car: v 1C = 0.0 m/s, a C = 2.6 m/s 2 Truck: v T = 28.0 m/sCar:Truck: ∆d T = v T ∆t ∆d T = 28.0 ∆t∆d T = ∆d C + 1.2 ⨯ 102 m28.0 ∆t = 1.3(∆t )2 + 1.2 ⨯ 102 solving the quadratic: ∆t = 5.9 s, 16 sThe truck passes the car after 5.9 s.(b)v 1C = 0.0 m/s a C = 2.6 m/s 2 ∆t = 5.9 s ∆d C = ?The car travels 45 m by the time the truck passes it.(c)v 1C = 0.0 m/s a C = 2.6 m/s 2∆t = 15.6 s (the other root of the quadratic) v 2C = v 1C + a C ∆t = 2.6 m/s 2(15.6 s) v 2C = 41 m/sThe car will be travelling at 41 m/s when it passes the truck if it maintains its acceleration.The car drives the followingdisplacements:= 80.0 km/h [E](1.50 h) = 120 km [E]= 60.0 km/h [N](1.00 h) =60.0 km [N]= 100.0 km/h [30︒W of N](0.50 h) = 50 km [30︒W of N] Using the component method for the displacement of A to B: north-south components: 60 km [N] + 50 km(cos 30︒) [N] = 103.3 km [N]east-west components: 120 km [E] + 50 km(sin 30︒) [W] = 95 km [E]Using Pythagoras Theorem, the magnitude of the displacement is 140 kmUsing trigonometry, the direction is: [43︒ E of N]The displacement of point B from point A is 1.4 ⨯ 102 km [43︒E of N].(b)A vector diagram showing the relationship among the vectors is drawn:v PG = velocity of plane with respect to the ground v PA = velocity of plane with respect to the airv AG = velocity of the air with respect to the groundSince the car takes a total of 3.00 h to reach point B and the plane leaves 2.00 h later but arrives at the same time, the time it takes the plane to make the flight is 1.00 h.= 1.4 ⨯ 102 km/h[43︒ E of N]Using cosine law:The plane’s speed is 1.9 ⨯ 102 km/h. (c)Using sine law:, θ = 13︒The plane must head in a direction of [30ºE of N].(90︒ – 13︒– 47︒)(d)If there is no wind:The plane would take 0.74 h to fly from A to B if there was no wind.28. ANS:(a) The triangle of velocity vectors appears as:The plane must point [8.6︒ E of S].(b) For the first leg of the trip of plane A:time to fly from Toronto to Pittsburgh:Layover time in Pittsburgh is 0.5 h.Pittsburgh to Philadelphia:Time to fly:Total time Toronto to Philadelphia: 0.885 h + 0.5 h + 1.15 h = 2.5 hThe total time for plane A is 2.5 h.(c) Distance from Toronto to Philadelphia:Vector triangle of velocities:Using sine law: β = 5.8︒, then θ = 180︒ – 138︒ – 5.8︒ = 36︒Using cosine law: 353 km/hTime for plane B to fly from Toronto to Philadelphia:Plane B must wait 2.5 h – 1.48 h = 1.0 h.29.Canoeist B:Using sine law: .The component of acrossthe river is: 2.4sin(56︒ + 16︒) =2.28 m/s.The time for B to cross to pointX:Canoeist A:The time for A to reach point X:Canoe A must wait 131.6 s – 125s = 6.6 s.30.(a) displacement = area under graph= 23.75 m [S] + 18.75 m [N] displacement = 5.0 m [S](b)The object’s average velocity during the first 6.0 s is 0.83 m/s [S].(c)The object’s average speed during the first 6.0 s is7.1 m/s.(d) The object is at its starting location 3 times throughout the motion.(e) The object’s acceleration is greatest between t = 6.5 s and 7.0 s. (the greatest slope) acceleration = slope of graph = 30 m/s2 [N](f)(g)31.(a)Time of flight: let “up” be (–) and “down” be (+)v1 = –80.0 m/s(sin 25︒) = –33.8 m/sa = 9.8 m/s2∆d = 36 m∆t = ?36 = (–33.8)∆t + 4.9(∆t)2Solving the quadratic: ∆t = 7.84 sHorizontal range: ∆d = v∆t = 80.0 m/s(cos 25︒)(7.84 s) = 5.7 ⨯ 102 mThe horizontal range of the shell is 5.7 ⨯ 102 m.(b) Horizontal component of final velocity: 80.0m/s(cos 25︒) = 72.5 m/sVertical component of final velocity: v2 = v1 + a∆t = –33.8 m/s + 9.8 m/s2(7.84 s)v2 = 43.0m/sUsing Pythagoras:θ=The shell lands with a velocity of 84 m/s at an angle of 31︒ below the horizontal.32.(a)Time of flight: let “up” be (–) and “down” be (+)v1 = –26 m/s(sin 60º) = –22.5 m/s a = 9.8 m/s2∆d = 2.0 m∆t = ?2.0 = (–22.5)∆t +4.9(∆t)2Solving the quadratic: ∆t = 4.68 sThe time of flight is 4.7 s.(b)Horizontal range: ∆d = v∆t = 26m/s(cos 60︒)(4.68 s) = 60.8 mThe receiver must run: 60.8 m –3.0 m = 57.8 m.The time the receiver has to reachthe football: 4.68 s + 2.0 s = 6.68s.The average speed of the receiver:The receiver must run with anaverage speed of 8.7 m/s.33.(a)Time of flight: let “up” be (–) and “down” be (+)v1 = –15 m/s(sin 50︒) = –11.5 m/sa = 9.8 m/s2∆d = –2.0 m∆t = ?–2.0 = (–11.5)∆t + 4.9(∆t)2Solving the quadratic: ∆t = 0.19 s (way up) and 2.16 s (way down)Horizontal range: ∆d = v∆t = 15 m/s(cos 50º)(2.16 s) = 21 mThe net must be placed 21 m away from the cannon. (b) Horizontal component of final velocity: 15 m/s(cos 50︒) = 9.64 m/sVertical component of final velocity: v2 = v1 + a∆t = –11.5 m/s + 9.8 m/s2(2.16 s)v2 = 9.67m/sUsing Pythagoras:θ=The shell lands with a velocity of 14 m/s at an angle of 45︒ below the horizontal.34. (a)Time of flight of shell: Horizontal range of shell: 100 mHorizontal component of shell’s velocity: Angle of projection:10 m/s = 40.0 m/s(cos θ)θ= 76ºThe gun must be aimed at an angle of 76︒ to the horizontal.(b)Vertical component of shell’s velocity: 40.0 m/s(sin75.5︒) = 38.8 m/s [up]let “up” be (–) and “down” be (+)v1 = –38.8 m/sa = 9.8 m/s2∆t = 10 s∆d = ? The cliff is 1.0 ⨯102 m high.。
Chapter 1 Kinematics(终版
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机器人学运动学分析_Kinematics
Dynamic Analysis of Mechanical Systems
Kinematics
Joints can be classified by the degrees of freedom allowed or restricted. The commonly used joints are listed in Table 1. 3.2.3 Mobility Analysis One of the basic steps in the kinematic analysis of mechanical systems is to determine the number of degrees of freedom or independent coordinates required to determine the configuration of the system. It can be shown that the number of constraints provided by a joint is equal to the number of degrees of freedom eliminated from the unconstraint system as a result of using this joint. In the planar system, the configuration of a rigid body undergoing unconstrained motion can be described by three independent coordinates due to the common constraints. Therefore, a planar system with n unconstrained movable bodies has 3 n DOF. If the system has p joints, each of which provides two constraints (note the common constraints), then the number of degrees of the system can be evaluated by the mobility (Glubler) criterion
帕金森的帕金森定律
第三，这位“领导者"对他的工作来说是不称职的，如果称职就不必寻找助手。
帕金森定律——发生作用的条件
案例分析：
01
产生根源：权力的危机感——产生了可怕的机构人员膨胀的帕金森现象。
02
解决办法：必须将用人权放在公平、公开、平等、科学、合理的用人制度上，不受人为因素的干扰。
03
最需要注意的一点：不要将用人权放在一个被招聘者的直接上司手里。
04
帕金森定律——解决之道
造成组织机构效率低下的原因： （1）领导者宁愿增加可以指挥得动的下属，而不愿要有真正才能的“竞争对手”。 （2）官员数量增加，自然彼此间会创造出许多工作，以便大家都在“做事”。 （3）机构年代越久，人员质量越低，因为当官的不断选择了各种“庸才”，越堆越多。 （4）领导层的会议长短、与讨论问题的重要性成反比。 （5）一个机构成立越久，其委员会规模必然愈大，效能自然较低。 （6）一个机构内部趋于腐败之时，也是他的外部建筑、设施日趋富丽堂皇之时，所以从机构的富丽堂皇的程度可以推测其腐化的程度。 （7）一个机构有多少预算，就必然开支多少钱，因为不用完，下届预算就可能削减。
帕金森为证明他这一定律，引证了一个和行政部门相似的例子，从而认为：工作的时间要求是有弹性的，在所完成的工作与所分配的人员数目之间显然很少或者根本就没有什么联系。
帕金森定律的两条法则：
其一是增加部属法则。 一个行政官员想增加的是下级而不是对手，其目的是为了减少组织中的竞争对手，同时增加下属也可以提高该官员自己的地位。
1958年，英国历史学家、政治学家诺斯科特·帕金森(C.Northcote Parkinson)出版了《帕金森定律》Parkinson‘s Law)一书。帕金森经过多年调查研究，发现一个人做一件事所耗费的时间差别如此之大：他可以在10分钟内看完一份报纸，也可以看半天；一个忙人20分钟可以寄出一叠明信片，但一个无所事事的老太太为了给远方的外甥女寄张明信片，可以足足花一整天：找明信片一个钟头，寻眼镜一个钟头，查地址半个钟头，写问候的话一个钟头零一刻钟……
Kinematic optimization for isotropic stiffness of redundantly actuated parallel manipulators
Kinematic Optimization for Isotropic Stiffness of Redundantly Actuated Parallel Manipulators
I. INTRODUCTION actuated parallel manipulators (or over-actuated parallel manipulators) have many advantages such as enlarged dexterous workspace and higher stiffness compared to the non-redundant analogues. Kinematic singularity region inside the workspace can mostly be eliminated in redundant actuation and the workspace can be enlarged [1]. In addition, internal preload control can enhance the antagonistic stiffness of the mechanism [2,3,4,5]. In order to design the redundantly actuated parallel mechanism, the determination of kinematic parameters of the mechanism and number of excessive actuators, that is force redundancy, should be carefully considered in the kinematic design process. In previous works, the kinematic parameter optimization of the workspace and stiffness for non-redundantly actuated parallel mechanisms has been examined. Liu and Wang [6] introduced performance indexes such as the global
机器人运动学Kinematics_final
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Johann C. Rocholl (Rostock) StyleDelta 机器人运动学by Steve Graves前言首先我要感谢Johann Rocholl所做出的贡献，虽然我们未曾谋面，但本文所写出的大部分内容都要归功于他。
例如，本文所有的Rostock图都是通过他的OpenSCAD模型的修改版本生成的，他的Marlin代码也是我验证此概念的依据。
然后，我要感谢他创造了现在最好的DIY 3D打印机。
顺便说一句，我自己并没有delta 3D打印机，我只有一台型号非常老的cartesian打印机。
这个月，我已经在迫不及待地计划为Kossel Clear（delta 3D打印机项目，由Blue Eagle Labs出资）搭建Kickstarter众筹平台。
引言这是我对Rostock打印机中应用delta机器人的形态的分析报告，已经出版的此类分析报告我还没有看到过，但有一个叫Clavel的人在研究原始delta机器人。
我意识到对Rostock进行协调转化的固件已经唾手可得，并且我在网上查找的时间如果够久，我也能够找到此种转化的一些描述，可是，我喜欢独自理解事物。
我的哲学是，如果一个问题是我能够自己解决的，那么在读现成答案之前我会自己寻找答案。
这种方式让我有更多启发，使我不会受已经被接受的观念所影响而产生偏见，我可以得出看待问题的不同方法而不是现有解。
例如，我使用外心研究正向运动学的方法可能很独特，因为那是我自己想出来的。
一旦你解析了它，这就是一个相当简单的几何问题。
那么，描述几何并定义几何公式就显得很重要了，部分的几何描述也就是建立命名规则。
Delta机器人有三个导轨（Rostock打印机使用一对杆，如图1灰色部分），我们叫这些导轨A、B和C。
每个列都有一个滑座（图1黄色部分）可以沿导轨上下移动，每个滑座有2个平行臂（图1蓝色部分）与运动平台（图1绿色部分）连接，每对臂长度相同，每个滑座与机器人底面的连接都是准确平行的。
底面也被称为“床”（图1红色部分）。
为了让臂之间平行，每个滑座和运动平台连接点的间距是相等的。
图1首先我发现它的一部分几何设计遵循了比例绘图器的原理。
平行臂并不仅仅彼此平行，他们也使得运动平台的边缘与滑座的连接点平行，我们可以从图2机器人的俯视视角得知。
因为3个滑座的连接点都是与“床”平行的，所以运动平台所在平面也肯定与“床”平面是平行的，换而言之，连接滑座与运动平台的平行臂使得平台平面与“床”平面平行了。
我曾看见有人质疑平行臂，这下你知道它们为什么存在了吧。
我认为，平行臂最大的好处就是使运动平台与“床”平行了。
而且因为运动平台边缘与滑座连接点平行，运动平台与滑座上相应点的间距也是一样的，所以，为方便数学处理，我们可以选择一条连接平台边缘中点和相应滑座连接中点的线，我们可以使用这条线完成大部分的计算。
我们把它叫做我们的运动线，线的一端正好处于平台枢轴点中点，另一端正好处于滑座枢轴点中点，当我们提到滑座和平台，这些点就是我们所提及的。
例如，滑座离“床”的高度就是“床”到相应点的距离，滑座离平台的高度就是平台与滑座相应点的垂直距离。
总之，数学处理可以仅仅基于这条连接滑座与平台的运动线，因为平行臂实现了围绕这条线的几何方向。
图2有一些简化的假设要提出。
首先，我们假设这些列构成了一个等边三角形，这不是必需的，也可以是一个不等边三角形，但是这种假设简化了数学运算，使我们可以用单一变量代表列间距，列的间距两两相等，我们叫这个变量为间距S （三角形的边）。
这个假设也使我们能选择一个所有列距原点相等的坐标系，注意，这些列可以被点表示。
在Rostock模型中每个列表示成的点处于2个杆的中点。
接下来，我们将假设3对臂都是相同长度，我们叫它长度L。
我并不认为假设列间距相同或者臂长相同是必需的，即使列距不同或臂长不同，通过数学计算也能解决。
这些不同改变了工作区域的覆盖形状，而这也许能成为机器人打印专门对象（一座房屋？）的一种优势。
下个一假设是，喷头被精确安装在平台正中心。
我们把运动线与平台边缘交点到平台中心的距离叫做E，喷头底部挤压点的坐标叫做X、Y和Z，平台中心点正好位于挤压点上方，它与喷头有着相同的X、Y坐标。
正如以上讨论那样，我们将测量每个滑座的高度，即运动线终止于连接点的长度。
我们把每个滑座离平台的高度叫做Acz、Bcz和Ccz，把滑座离“床”的高度叫做Az、Bz和Cz，把喷头从载体底部延伸的距离叫Hcz，喷头离“床”的高度叫Z。
那么有Az=Z+Acz+Hcz，Bz=Z+Hcz，Cz=Z+Ccz+Hcz，这就使我们简化了问题。
我们能从已知的X、Y平面推导出公式去掉方程里的Z，然后将平面转换到任意Z坐标。
我们只需要关注Acz、Bcz和Ccz，因为它们全部与平台相关，所以我们能计算平台平面中的X、Y坐标。
Az = Z + Acz + HczBz = Z + Bcz + HczCz = Z + Ccz + HczSolving for ZZ = Az - Acz - HczZ = Bz - Bcz - HczZ = Cz - Ccz - Hcz以上结论促使我们讨论Z轴。
如果有人将所有滑座向上移动同样距离，那么他会发现在几何上只改变了头部的Z坐标，所以Z坐标直接与滑座高度（Az、Bz和Cz）相关，这些同样由以上方程表现出来。
对Z坐标的观察可以得知，Acz、Bcz和Ccz只取决于X、Y。
接下来得讨论一下逆运动学。
逆运动学逆运动学是软件控制打印机的方程基础。
它使用给定的X、Y、Z坐标，通过计算滑座位置（Az、Bz、Cz）来使机器人运动到这些坐标上。
图3现在我们需要定义坐标系。
我们要选择一个原点位于“床”中心的笛卡尔坐标系。
我们把列A放在距原点R + Aey + Acoy的Y轴正向上，其他列在以同样距离为半径的圆上，相互成120°角，列B从上方看呈顺时针120°角，列C从列B顺时针旋转120°角或者从列A逆时针旋转120°。
臂与垂直线形成的一个直角三角形和第三条腿都位于平台平面内，观察到这些可以推导出一组重要的方程。
现在我们需要理解第三条腿是如何与平台中心的X、Y点联系起来的。
我的第一想法是运动线穿过平台中央，但实际上并不是，运动线是垂直于平台边缘的。
因为平台边缘是平行于滑座连接点的，平台上的线在X、Y平面上处于同一方向，这是固定的矢量，我们叫它们为Ae、Be和Ce。
通过我们的定义，它们全部有相同的长度而且方向由对应列的方位确定。
由于矢量加法的交换性，我们可以将这些矢量从弧的末端移动到列的位置上，这使枢纽点改变到了虚拟列位置，这些兴趣点形成一个圆弧，简化了运动算法的计算量。
所以从平台边缘到中心的距离将成为软件初始化时的一个常量。
图4现在确定我们可以从通过各个机械臂所形成的直角三角形在X-Y平面垂直上升（虚拟列）和行走中获得何种公式。
我们将三角形在XY平面行走的距离（效应器平台的边缘到支架轴心底下的点）称为Ad，Bd和Cd。
通过使用Pythagorian 定理我们得到了一下的公式：Ad^2 + Acz^2 = L^2Bd^2 + Bcz^2 = L^2Cd^2 + Ccz^2 = L^2现在让我们将列与X,Y平面联系起来。
首要的事情是要描述A，B和C列在X,Y平面的基础位置。
我们将称这些物理位置为Ax，Ay；Bx，By；Cx，Cy。
通过观察我们可以发现，实际轴心点均在支架上每条终止了的作用线之下。
我们可以通过减去一个来自每个表示支架弥补的列向量来计算出这些位置。
我们将称这些轴心的位置为Apx，Apy；Bpx，Bpy；Cpx，Cpy。
我们现在将使用这些值，但在之后，我们可以通过移动这些位置到虚拟列位置来减少位置数字。
对于这种分析，我们保持Z不变，这样我们就可以在一个给定的XY平面上限制住效应器平台。
每一列轴心的作用线围绕着支架上的点。
当支架在效应器平台以上的一个给定高度时，这个在XY平面上可能的位置即效应器平台的边缘，满足在给定作用线上是一个围绕着轴心点的弧形。
如果向量Ae，Be和Ce为零，并且作用线在效应器平台的中心，我们的XY坐标将成为这些弧形的交点。
但当Ae，Be和Ce不为零，这些问题会更加复杂。
让我们做一些等式将Ae，Be和Ce与我们的XY坐标联系起来。
让我们称这些作用线满足效应器平台的坐标为Acx，Acy；Bcx，Bcy；Ccx，Ccy。
我们知道向量可以通过delta X和delta Y来表示，所以我们将这些向量分解为Aex，Aey；Bex，Bey；Cex，Cey。
我们将定义这些向量从效应器平台的边缘指向其中心。
因此我们有一下公式：X = Acx - Aex = Bcx - Bex = Ccx - CexY = Acy - Aey = Bcy - Bey = Ccy - Cey基于我们上面讨论的在弧形上移动的作用线。
一个圆的公式是(X - CX)^2 + (Y - CY)^2 = CR^2。
在这里，CX、CY是圆心而CR是半径。
我们应该能够解决这一列的等式，并且通过相似点确定另外两个。
所以看列A。
(Acx - Apx)^2 + (Acy - Apy)^2 = Ad^2解出上面的Acx和Acy并替代，我们得出：Acx = X + AexAcy = Y + Aey(X + Aex - Apx)^2 + (Y + Aey - Apy)^2 = Ad^2我们看见一个简式，我们可以将这个圆的轴心点视为Apx - Aex, Apy - Aey。
我们将称这个点为Avx，Avy。
我们将作出如下定义：Avx = Apx - AexAvy = Apy - AeyBvx = Bpx - BexBvy = Bpy - BeyCvx = Cpx - CexCvy = Cpy - Cey这些都是我们“虚拟”列的位置。
这些位置可以在初始化时在软件里被计算出来，然后用到之后所有的计算中。
我们在图5和6展示了这些虚拟列。
我们也展示了在作用线上的单一机械臂。
所有图六是我们的数学模型的一个物理表示。
这些机械臂在X,Y坐标上并在虚拟列上一个支架弥补加效应器弥补远离物理列位置的位置。
这就是我们将运用在硬件中的模型。
图5图6现在用我们简化的模型，我们有一下的公式：(X - Avx)^2 + (Y - Avy)^2 = Ad^2 = L^2 - Acz^2解出AczAcz^2 = L^2 - (X - Avx)^2 - (Y - Avy)^2Acz = sqrt(L^2 - (X - Avx)^2 - (Y - Avy)^2)通过求效应器平台以上的支架高度的相似度公式，得出Bcz和Ccz为：Bcz = sqrt(L^2 - (X - Bvx)^2 - (Y - Bvy)^2)Ccz = sqrt(L^2 - (X - Cvx)^2 - (Y – Cvy)^2)我们想要的真实值是基底上每个支架的距离。