Chapter 4 Motion in two and three dimensions. Two principles for 2D and 3D motions: Two principles for 2D and 3D motions: 1) The principle of independence.

ChapterChapter 4 4

Motion in two and three Motion in two and three dimensionsdimensions

Two principlesTwo principles for 2D and 3D motions: for 2D and 3D motions:

1) The principle of independence of force1) The principle of independence of force

2) The principle of superposition of motion2) The principle of superposition of motion

F1F1

F2F2 F3F3

The particle starts at t=0 with initial positionThe particle starts at t=0 with initial position

and an initial velocity .and an initial velocity .

Section 4-1 Motion in three dimensions Section 4-1 Motion in three dimensions with constant accelerationwith constant acceleration

Now we consider a particle move in Now we consider a particle move in three dimensionsthree dimensions with with constant accelerationconstant acceleration. We can represent the . We can represent the acceleration as a vector: acceleration as a vector:

kajaiaa zyx

kzjyixr o000

kvjvivv zyx 0000

In a similar way: In a similar way:

tavv xxx

0

tavv yyy

0

tavv zzz

0

tavv

0

a constant a constant ,a

, , ,x y za a a all constants all constants

tavx xxtx

2

00 2

1

tavz zztz

2

00 2

1

tavy yyty

2

00 2

1 2

00 2

1tatvrr

（（ 4-14-1 ））

（（ 4-24-2 ））

Section 4-2 Newton’s laws in three Section 4-2 Newton’s laws in three dimensional vector form dimensional vector form

(4-3)(4-3)

Which includes the three component Which includes the three component equations equations

(4-4)(4-4)

amF

xx maF

yy maF zz maF

Sample problemSample problem

1. A crate of mass 1. A crate of mass m=62 kgm=62 kg is sliding without is sliding without friction with an initial velocity of friction with an initial velocity of v0=6.4 m/sv0=6.4 m/s a along the floor. In an attempt to move it in a dilong the floor. In an attempt to move it in a different direction, fferent direction, Tom pushes opposite to its Tom pushes opposite to its initial motion with a constant force of a magninitial motion with a constant force of a magnitude F1=81Nitude F1=81N, while , while Jane pushes in a perpeJane pushes in a perpendicular direction with a constant force of mndicular direction with a constant force of magnitude F2=105Nagnitude F2=105N.. If they each push for 3.0 If they each push for 3.0s, in what direction is the crate moving when s, in what direction is the crate moving when they stop pushing?they stop pushing?

Section 4-3 Projectile motionSection 4-3 Projectile motion

Figure 4-4 shows the initial Figure 4-4 shows the initial motion of a projectile at the motion of a projectile at the instant of launch. instant of launch. Its initial Its initial velocity is , directed at an velocity is , directed at an angle from the horizontal.angle from the horizontal.

ov

0

00 x 00 y

ovy

x

0

Fig 4-4 A particle Fig 4-4 A particle is launched with is launched with initial velocity initial velocity

mg

We choose suitable coordinate We choose suitable coordinate system to make: system to make:

o

The components of the initial velocity are The components of the initial velocity are

(4-6)(4-6)

Gravity is the only force acting on the particle, so the Gravity is the only force acting on the particle, so the

components of the net force arecomponents of the net force are

(4-7)(4-7) (4-8) (4-8)

(4-9)(4-9) Position components:Position components: (4-10)(4-10)

00 cosvvox 00 sinvvoy

0 xF mgFy 0xa ga y

xx vv 0 gtvv yy 0

tvx x0 20 2

1gttvy y

From Eqs. (4-10), we canFrom Eqs. (4-10), we can eliminate t and obtain theliminate t and obtain the relationship between x and y (after considering Ee relationship between x and y (after considering Eqs. (4-6)): qs. (4-6)):

22

000 )cos(2

tan xv

gxy

(4-13)(4-13)

which is the equation of a trajectory (which is the equation of a trajectory ( 轨线轨线 ) of ) of the projectile, the projectile, the equation of a parabolathe equation of a parabola. .

Hence the trajectory of a projectile is parabolic.Hence the trajectory of a projectile is parabolic.

Fig 4-5 trajectory of a projectileFig 4-5 trajectory of a projectile

v x0

v y0

v x0

vy

R

0vv x0

v x0

v x0

v y0

x

y

o

The The “horizontal range R”“horizontal range R” of the projectile is of the projectile is defined as the distance along the horizontal defined as the distance along the horizontal where the projectile return to the level from where the projectile return to the level from which it was launched.which it was launched.

Let Let y=0y=0 in Eq(4-13), we obtain the range in Eq(4-13), we obtain the range RR::

(4-14)(4-14)

0

20

00

20 2sincossin

2

g

v

g

vR

当子弹从枪口射出时，椰子刚好从树上由静止自由下落 . 试说明为什么子弹总可以射中椰子 ?

Sample problemSample problem

4-3. In a contest to drop a package on a targ4-3. In a contest to drop a package on a target, one contestant’s plane is flying at a constet, one contestant’s plane is flying at a constant horizontal velocity of 155km/h at an elevant horizontal velocity of 155km/h at an elevation (ation ( 海拔海拔 ) of 225m toward a point directly ) of 225m toward a point directly above the target. At what angle of sight shoabove the target. At what angle of sight should the package be released to strike the taruld the package be released to strike the target?get?

Section 4-4 Drag forces and the effects on Section 4-4 Drag forces and the effects on motionsmotions

Drag forceDrag force is a frictional force whichis a frictional force which

experienced by any object that moves experienced by any object that moves

through a fluid medium, such as air or water. through a fluid medium, such as air or water. Drag forceDrag force must be taken into account in the must be taken into account in the

design of aircraft and seacraft. design of aircraft and seacraft. Drag forcesDrag forces prevent the velocity from prevent the velocity from

increasing without limit in the nature. increasing without limit in the nature.

Falling motion with drag forceFalling motion with drag force

We assume that the We assume that the magnitude of the dragmagnitude of the drag forceforce DD depends linearly on the speed: depends linearly on the speed: (4-17)(4-17)

vbD

We choose the We choose the y axisy axis to be vertical and the positive to be vertical and the positive direction to be downward.direction to be downward.

yy bvmgF yy mabvmg

yyy v

m

bga

dt

dv

(4-18)(4-18)

(4-19)(4-19)

dt

m

bvg

dv

y

y (4-20)(4-20)

andand

With at time t=0, we integrate two With at time t=0, we integrate two sides of Eq.(4-20) sides of Eq.(4-20)

then we obtain then we obtain

(4-21)(4-21) (4-22) (4-22)

0yv

tv

y

y dt

m

bvg

dvy

00

( )ymg bv b

ln tmg m

)1(t

m

b

y eb

mgv

0 tm

b

e• For large For large tt,,

b

mgvy

The magnitude of the terminal speed The magnitude of the terminal speed approaches a approaches a constant valueconstant value, not increasing without limit., not increasing without limit.

• For small For small tt,, 1tm

bt

m

be

tm

b

1SoSo

gttm

b

b

mgtvy )]1(1[)( (small t) (4-23)(small t) (4-23)

At the beginning of the motion, it is nearly a freely At the beginning of the motion, it is nearly a freely falling motion. falling motion.

mgmg mgmg mgmg

DDDD

How is D changing with tHow is D changing with time in general?ime in general?

It is found that a cat is muIt is found that a cat is much safer when it falls from ch safer when it falls from higher place. WHY? higher place. WHY?

mgmg

DD

(for cats)(for cats)

Projectile motion with drag forceProjectile motion with drag force

X (m)

Y(m)

Without drag force

With drag force

060060

-

79

vbD

When the drag forceWhen the drag force is considered,is considered,the range Rthe range R and and the the maximum heightmaximum height H will H will be reduced.be reduced.

Ro

The trajectory is also The trajectory is also no longer symmetricno longer symmetric about about the maximum; the descending motion is much the maximum; the descending motion is much steeper than the ascending motion.steeper than the ascending motion.

4-5 Uniform Circular Motion4-5 Uniform Circular Motion

In uniform circular motion, the particle In uniform circular motion, the particle moves at moves at constant speedconstant speed in ain a circular pathcircular path. .

Since the direction of velocity Since the direction of velocity changes in the motion, it is changes in the motion, it is an an acceleration motionacceleration motion..

How to find How to find acceleration acceleration from the constant speedfrom the constant speed for for uniform circular motion?uniform circular motion?

Fig 4-16Fig 4-16

O

v

r

Fig 4-16Fig 4-16

O

v1p 2p

v

y

Find acceleration for the motionFind acceleration for the motion

x

yv1

xv1xv2

yv2

cos1 vv x sin1 vv y

cos2 vv x sin2 vv y

(4-25)(4-25) As the particle moves along As the particle moves along the arc from to ,it the arc from to ,it covers a distance of , covers a distance of , and a time interval .and a time interval .

1p 2p2r

v

rt

2

Acceleration:Acceleration:

012

t

vva xxx

(4-27)(4-27)

sin

)/2(

sinsin 212

r

v

vr

vv

t

vva yyy

(4-28)(4-28)

r

p

In order to find the In order to find the instantaneous accelerationinstantaneous acceleration, we , we take approaches zero, (then angle goes to take approaches zero, (then angle goes to zero) so that and both approach zero) so that and both approach pp , which , which givesgives

1p 2pt

r

v

r

v

r

vay

2

0

22

0

sinlim)]

sin([lim

(4-29)(4-29)

• The The minusminus sign indicating that the acceleration sign indicating that the acceleration at at p points toward the center of the circle. points toward the center of the circle.

• Point Point p is an arbitrary point on the circle, so is an arbitrary point on the circle, so Eq.(4-29) is a general result for the motion.Eq.(4-29) is a general result for the motion.

• The acceleration is called The acceleration is called centripetal accelerationcentripetal acceleration or or radial accelerationradial acceleration. The corresponding force is . The corresponding force is called called centripetal forcecentripetal force oror seeking center forceseeking center force..

ya

4-6 Relative Motion4-6 Relative Motion

Fig 4-17Fig 4-17

ssr '

psr

'psr

P

sspsps rrr ''

(4-31)(4-31)

s,s, ss’: inertial frame’: inertial frame

If we have two If we have two inertial frames inertial frames ss and and s’,s’, how about how about the relationship between the motions in them?the relationship between the motions in them?

0

ss

x

y

x’

S’S’ y’

0’

Take the derivative with respect to time of Take the derivative with respect to time of Eq.(4-31), we have Eq.(4-31), we have

(4-32)(4-32) Eq.(4-32) is a law of the transformation of Eq.(4-32) is a law of the transformation of

velocitiesvelocities. It is often called the . It is often called the Galilean formGalilean form of the law of transformation of velocities. of the law of transformation of velocities.

sspsps vvv ''

It permits us to transform a velocity of It permits us to transform a velocity of one frame of reference (s’) to another one frame of reference (s’) to another frame of reference.frame of reference.

Differentiate Eq.(4-32), we have Differentiate Eq.(4-32), we have

(4-33)(4-33)

dt

vd

dt

vd

dt

vdsspsps

''

'psps aadt

vd

dt

vd psps

'

The accelerations of The accelerations of P P measured by two measured by two observers are identical.observers are identical.

Eq.(4-34) indicates directly Eq.(4-34) indicates directly Newton’s second lawNewton’s second law can be equally well can be equally well appliedapplied in in any inertial framesany inertial frames..

oror (4-34)(4-34)

The last term of eq.(4-33) vanishes, because The last term of eq.(4-33) vanishes, because the relative velocity of two reference frames the relative velocity of two reference frames must be a constant. Thusmust be a constant. Thus

Sample problemSample problem

1. The compass in an airplane indicates that 1. The compass in an airplane indicates that it is headed due east; its it is headed due east; its air speed indicatorair speed indicator reads 215 km/h. A reads 215 km/h. A steady wind of 65 km/hsteady wind of 65 km/h is is blowing due north. (a) What is the velocity of blowing due north. (a) What is the velocity of the plane with respect to the ground? (b) If the plane with respect to the ground? (b) If the pilot wishes to fly due east, what must the pilot wishes to fly due east, what must be the heading? That is, what must the be the heading? That is, what must the compass read?compass read?