1、Chapter 4Motion in two and three dimensionsTwo principlesfor 2D and 3D motions:1)The principle of independence of force2)The principle of superposition of motionF1F2F3The particle starts at t=0 with initial positionand an initial velocity .Section 4-1 Motion in three dimensions with constant acceler
2、ationNow we consider a particle move in three dimensions with constant acceleration.We can represent the acceleration as a vector:+=kajaiaazyx+=kzjyixro000+=kvjvivvzyx0000In a similar way:tavvxxx+=0tavvyyy+=0tavvzzz+=0tavv+=0a constant,a,xyza a aall constants tavxxxtx20021+=tavzzztz20021+=tavyyyty20
3、021+=20021tatvrr+=（4-1）（4-2）Section 4-2 Newtons laws in three dimensional vector form(4-3)Which includes the three component equations (4-4)=amFxxmaF=yymaF=zzmaF=Sample problem1.A crate of mass m=62 kg is sliding without friction with an initial velocity of v0=6.4 m/salong the floor.In an attempt to
4、 move it in a different direction,Tom pushes opposite to its initial motion with a constant force of a magnitude F1=81N,while Jane pushes in a perpendicular direction with a constant force of magnitude F2=105N.If they each push for 3.0s,in what direction is the crate moving when they stop pushing?Se
5、ction 4-3 Projectile motionFigure 4-4 shows the initial motion of a projectile at the instant of launch.Its initial velocity is ,directed at an angle from the horizontal.ov000=x00=yovyx0Fig 4-4 A particle is launched with initial velocity mgWe choose suitable coordinate system to make:oThe component
6、s of the initial velocity are (4-6)Gravity is the only force acting on the particle,so the components of the net force are(4-7)(4-8)(4-9)Position components:(4-10)00cosvvox=00sinvvoy=0=xFmgFy=0=xagay=xxvv0=gtvvyy=0tvxx0=2021gttvyy=From Eqs.(4-10),we can eliminate t and obtain the relationship betwee
7、n x and y(after considering Eqs.(4-6):22000)cos(2tanxvgxy=(4-13)which is the equation of a trajectory(轨线)of the projectile,the equation of a parabola.Hence the trajectory of a projectile is parabolic.Fig 4-5 trajectory of a projectilevx0vy0vx0vyR0vvx0vx0vx0vy0 xyoThe“horizontal range R”of the projec
8、tile is defined as the distance along the horizontal where the projectile return to the level from which it was launched.Let y=0 in Eq(4-13),we obtain the range R:(4-14)02000202sincossin2gvgvR=当子弹从枪口射出时，椰子刚好从树上由静止当子弹从枪口射出时，椰子刚好从树上由静止自由下落自由下落.试说明为什么子弹总可以射中椰子试说明为什么子弹总可以射中椰子?Sample problem4-3.In a cont
9、est to drop a package on a target,one contestants plane is flying at a constant horizontal velocity of 155km/h at an elevation(海拔)of 225m toward a point directly above the target.At what angle of sight should the package be released to strike the target?Section 4-4 Drag forces and the effects on mot
10、ionsDrag force is a frictional force whichexperienced by any object that moves through a fluid medium,such as air or water.Drag force must be taken into account in the design of aircraft and seacraft.Drag forces prevent the velocity from increasing without limit in the nature.Falling motion with dra
11、g forceWe assume that the magnitude of the drag forceD depends linearly on the speed:(4-17)vbD=We choose the y axis to be vertical and the positive direction to be downward.yybvmgF=yymabvmg=yyyvmbgadtdv=(4-18)(4-19)dtmbvgdvyy=(4-20)andWith at time t=0,we integrate two sides of Eq.(4-20)then we obtai
12、n(4-21)(4-22)0=yv=tvyydtmbvgdvy00()ymgbvblntmgm=)1(tmbyebmgv=0tmbe For large t,bmgvyThe magnitude of the terminal speed approaches a constant value,not increasing without limit.For small t,1tmbtmbetmb1Sogttmbbmgtvy=)1(1)(small t)(4-23)At the beginning of the motion,it is nearly a freely falling moti
13、on.mgmgmgDDHow is D changing with time in general?It is found that a cat is much safer when it falls from higher place.WHY?mgD(for cats)Projectile motion with drag forceX(m)Y(m)Without drag forceWith drag force060060=-79vbD=When the drag forceis considered,the range R and the maximum height H will b
14、e reduced.RoThe trajectory is also no longer symmetric about the maximum;the descending motion is much steeper than the ascending motion.4-5 Uniform Circular MotionIn uniform circular motion,the particle moves at constant speed in a circular path.Since the direction of velocity changes in the motion
15、,it is an acceleration motion.How to find acceleration from the constant speed for uniform circular motion?Fig 4-16OvrFig 4-16Ov1p2pvyFind acceleration for the motionxyv1xv1xv2yv2cos1vvx=sin1vvy=cos2vvx=sin2vvy=(4-25)As the particle moves along the arc from to ,it covers a distance of ,and a time in
16、terval .1p2p2 rvrt2=Acceleration:012=tvvaxxx(4-27)sin)/2(sinsin212rvvrvvtvvayyy=(4-28)rpIn order to find the instantaneous acceleration,we take approaches zero,(then angle goes to zero)so that and both approach p,which gives1p2ptrvrvrvay20220sinlim)sin(lim=(4-29)The minus sign indicating that the acceleration at p points toward the center of the circle.Point p is an arbitrary point on the circle,so Eq.(4-29)is a general result for the motion.The acceleration is called centripetal accelerationor
