Physics （ Fifth Edition · Part I ） Shanghai Normal university. Department of Physics.

PhysicsPhysics（（ Fifth EditionFifth Edition· · Part IPart I ））

Shanghai Normal university. Department of Physics

ChapterChapter 11 The Kinematics of Mass PointsThe Kinematics of Mass Points

§1.1 The Description of the Motion of Mass points

§1.2 Motion in a Circle

§1.3 Relative Motion

Summary

§1.The description of the Motion of §1.The description of the Motion of Mass PointsMass Points

1. Refernce frame, mass point

2. Position vector, equation of motion and displacement

3. Velocity

4. Acceleration

11 Reference frames, mass pointReference frames, mass point

To determine the location of a body at the reference object quantitatively, a coordinate system is built on it.

1 Reference frames To describe the position of an object, the other object

referred to (Reference Frame ） should be chosen. It is arbitrary.

2 Mass point If we may ignore object’s size and shape, we may

regard the object as a single point with a mass, such a point is usually called a mass point.

坐标系坐标系r

φ

θ plante

The normal unite vector

The tangential unite vector

A moving mass point

τn

Natural coordinate system

Spherical coordinate system

Cartesian coordinate system

2 Position vector , equation of motion and 2 Position vector , equation of motion and displacementdisplacement

1 position vector

kzjyixr

2 2 2r r x y z The absolute value of r

The location of a particle P

relative to the origin of a coordinate

system, represented by the position

vector

.

r

rxcos rzcosrycos

Direction cosines of r

随时间变化

ktjtyitxtr

)()()()( z

)(txx )(tyy

)(tzz

Component equations

2 　 Equation of Motion

0),,( zyxf

Trajectory Equation: by eliminating the parameter t


x

y

o

B

Br

Ar A r

Ar

B

Br

A r

x

y

o BxAx

AB xx

By

Ay AB yy

rrr AB AB rrr

The displacement vector extends from the head of the initial position vector to the head of the later position vector.


3.Displacement—A particle is changing in its position:

222 zyxr the length of this vector, i.e.

Ar

B

Br

A r

x

y

o BxAx

AB xx

By

Ay AB yy

jyixr AAA jyixr BBB

jyyixxr ABAB )()(

AB rrr Consquently

the displacement can be described with three components in Cartesian coordinate O-xyz

kzzjyyixxr ABABAB

)()()(

and


4 Path （） : the actual trajectory of the mass point.s

222 zyxr

rr

21

21

21 zyx

2

2

2

2

2

2 zyx r

Physical meaning of displacement

A) Displacement depends only on the initial and final position the object, it is independent of the real path

.r xi yj zk

B ） showing the properties of vector and superposition of movement

s

),,( 1111 zyxP

),,( 2222 zyxP

)( 1tr

1P

)( 2tr

2Pr

notesThe changing of the radial length of displacement

x

y

Oz

r


The displacement and the path

（ B ） in general case, the

displacement is not equal to the

distance. r s

（ D ） displace is a vector, path is a scalar

.

s

)( 1tr

1p

)( 2tr

2pr

x

y

Oz

's

(C) under what case ？sr

Move in straight line without changing direction; In the limit as . 0t sr

discuss

（ A ）　 the path between P1P2

are not only ，　 e.g. or ,

but the displacement are

unique.

r

s 's

The and have the same direction

rv

Magnitude is 22 )()(

t

y

t

x

v

r

)( ttr

B

)(tr

A

x

y

o

s

1 Average velocity

)()( trttrr

in the time interval ，the mass point moving from A to B ， the displacement is

t

The Average velocity is

jty

itx

tr

v

ji yx vvv Or

33 velocityvelocity

2 Instantaneous velocities

at every point along the trajectory, the instantaneous velocity vector is tangent to the trajectory at that point.

as ,the limit of the average velocity is called the instantaneous velocity ，

0t

jt

yi

t

xtt

00limlimv

t

r

t

rt d

dlim

0

v

sr dd as ，0t

td

de

t

s v

33 velocityvelocity

x

y

o

v

2 2 2d d d( ) ( ) ( )d d d

x y z

t t t

v v

instantaneous speed ： the magnitude of instantaneous velocity

v

yv

xv

ji yx

vvv

jt

yi

t

x d

d

d

d v

The velocity of a mass point in 3-D Cartesian coordinate system is

kt

zj

t

yi

t

x d

d

d

d

d

dv

d

d

s

tv

td

de

t

s v

33 velocityvelocity

Average Speed:t

s

v

r

)( ttr

B

)(tr A

x

y

o

s

d

d

s

tvInstantaneous:

discuss

A mass point is at the endpoint of the position vector at instantaneous time t , the speed is ( )

),( yxr

t

r

d

dt

r

d

d（ A ）（ B ）

t

r

d

d

22 )d

d()

d

d(

t

y

t

x

（ C ）

（ D ）td

sd（ E ）

33 velocityvelocity

Example 1 let the equation of motion be Where,

（ 1 ） what is the velocity at . ( 2 ) plot the trajectory of the mass .

Example 1 let the equation of motion be Where,

（ 1 ） what is the velocity at . ( 2 ) plot the trajectory of the mass .

( ) ( ) ( ) ,r t x t i y t j ��

1( ) (1m s ) 2m,x t t 2 214( ) ( m s ) 2m.y t t

3 st

solution （ 1 ） the following velocity components

1 2d d 11m s , ( m s )

d d 2x y

x yt

t t v v

1 1(1m s ) (1.5m s )i j

v3 st The velocity vector is

The angle with the respect to the position x-axis is

3.561

5.1arctan

（ 2 ） equation motion

1( ) (1m s ) 2mx t t 2 21

4( ) ( m s ) 2my t t

The trajectory equation can be got by eliminating t1 21

4( m ) 3m,y x x

/ mx

/ my

0

轨迹图

2 4 6- 6 - 4 - 2

2

4

6

Example 2 as illustrated in following figure ， A 、 B are connected by a thin rod with length ， A and B may slide along smooth tracks. If A slide at a constant speed toward left, when , what is the velocity of object B60

Solution Based on the coordinate system of figure , the velocity of object A is

OAB is a rectangle triangle ， the length l of the rigid thin rod is a constant

x

y

oA

Bl

v

iit

xixA

vvv d

d

the velocity of object B is

jt

yjyB

d

dvv

y2 2 2x =l

x

y

oA

Bl

v

Furthermore

0d

d2

d

d2

t

yy

t

xx

yield t

x

y

x

t

y

d

d

d

d jt

x

y

xB

d

dv

y

x

t

x tan,

d

dv tanvv B

BvThe direction of points to the positive direction

of the y axis ， when , 1.73Bv v60

1 ） average acceleration

B

v

B

A

v

B

v

v

and have the same direction.

va

x

y

O

at

v

the velocity increment per

unit time

2) (instantaneous) acceleration

0

dlim

dta

t t

v v

44 AccelerationAcceleration

A

v

A

x y za a i a j a k

2

2

2

2

2

2

d d

d dd d

d d

d d

d d

xx

yy

xa

t t

ya

t t

at t

zz

v

v

v zMagnitude is

2 2 2x y za a a a

2

2

d d

d d

ra

t t

v

acceleration jt

it

yx d

d

d

d vv

Magnitude of ā

22

0lim yxt

aat

a

v

The acceleration in 3-D coordination is


？

v v

( ) ( )t t t v v v

ac cb v

讨论

)()( ttt vvv

oaoc intercepted in Ob

Yield cbv

tn vv The change of direction of the velocity

ac nv

The change of the magnitude of the velocity cb tv

( )tv ( )t t

v

v

O

ab

c


O

d

da a

t v

If ？

dv

( )tv

( d )t tv

discussion

( ) ( d )t t t v vgiven

d0

dt

vSo, we can get

Example circular motion with constant speed

sot

ad

dv0a a and ,

dt

da

Geting integral constant according with initial condition

Two kinds of questions in Kinematics:Two kinds of questions in Kinematics:

1d( 1.0s )

da

t

vvSolution ： according the

definition of acceleration

Example 3 a ball drops vertically in a liquid ,the initial velocity of the ball is ， its acceleration is . Question: 1 ） after how long can the ball be considered as no longer moving? And (2) how long has the ball gone through before stopping ？

10 (10m s ) j

v

1( 1.0s )a j v

0

v

y

o

t

0d)1s0.1(

d0

tv

v vv

t

t

y )s0.1(0

1

ed

d vv tyt ty

ded0

)(-1.0s00

-1

v

t)s0.1(0

1

evv

m]e1[10 )s0.1( 1 ty

0

/my

/st

10

-1/m sv0v

0 /st

9.2s, 0, 10mt y v

2.3 4.6 6.9 9.2

8.9974 9.8995 9.9899 9.9990

v 0/10v

/st/my

0/100v 0/1000v 0/10000v

t)s0.1(0

1

evv m]e1[10 )0.1( 1 tsy

p.50 / 1- 8, 9, 13, 17

HomeworkHomework

§1.2 Motion in a Circle§1.2 Motion in a Circle

1. Planar polar coordinate system

2. Angular velocity of the circular motion

3. Tangential and normal acceleration of

circular motion, angular acceleration

4. Circular motion with constant speed and circular motion with constant variation of spped

11 Planar polar coordinate systemPlanar polar coordinate system

A

r

x

y

o

Assume a mass point moves on the ， at some moment it is at point A .the angle between the directional line segment pointing from the coordinate origin O to point A and the axis is . The point A can be determined by ),( rA

Oxy

r

x

is called the planar polar coordination system.

),( r

sin

cos

ry

rx

The transformation relationship between these two coordinate systems

22 Angular velocity and angular acceleration Angular velocity and angular acceleration of the circular motion of the circular motion

t

tt

d

)(d)(

Angular velocity

angular coordinate )(t

Angular accelerationtd

d

Speed

tr

tts

t

0lim

0lim

v x

y

o

r

)()(,dd trt

ts vv

A

B

1v

ro

tee

t dd

dd t

t

vv

2vtttd

deree

t

s vv

ndd e

t

ta

ddv

rt

rt

a dd

dd

tv

The circular motion

1te

2te

Tangential acceleration

1te2te

te

te

tt

0lim

The variation rate of normal unite vector with respect to time

te

dd t

Normal unit vector

33 Tangential and normal acceleration of the Tangential and normal acceleration of the circular motion circular motion

ntdd ee

ta vv

Tangential acceleration （ caused by magnitude variation of the velocity ）

22

t dd

dd

tsr

ta v

Normal acceleration （ originated from the direction change of the velocity ）

rra

22

nv

v

nntt eaeaa 22nt aaa

1v

2v

v

1v

ro

2v

1te

2te


v

Tangential acceleration

rt

a dd

tv

tav,π

2π,0 decreasing

increasingv,2π0,0

tconstan,2π,0 v

te

nea

a

a

tn1tan a

a

π00n a

x

y

o

nntt eaeaa


For General curve Motion （ natural coordinate system ）

ntdd ee

ta

2vv

tdd e

tsv

ddswhere is curvature radius.


4 Circular motion with constant speed and circular motion with constant variation rate of speed

n2

nn ereaa

1 circular motion with constant speed ： speed and angular velocity are both constants .

v0t a

2 circular motion with constant variation rate of speed

t 02

00 21 tt

)(2 0

2

0

2 when ,0t 00 ,

constantAngular acceleration


For a object with a curved shape motion ， which one is the correct among the following statement ：

（ A ） Tangential acceleration definitely does not zero ；

（ B ）　 Normal acceleration definitely does not zero （ except for the point of extremum ）；

（ C ） For the direction of the velocity is same the tangential direction, and normal speed definitely does zero ， so normal acceleration is definitely zero ；

（ D ） A mass point is in a line motion with constant speed ， the acceleration is definitely zero;

（ E ） If the acceleration is a invariable vector ， it must be carry out a motion with constant variation rate of speed .

a

discussion

o

A

B

Av

Bv

r

Example the horizontal speed of supersonic fighter jet at a high altitude point A is 1940 km/h , drives along a curve similar to a circular are to the point B , its speed at B is 2192 km/h , the time lapse is 3s , assume the radius of the arc AB is approximately 3.5km , and the process of driving from A to B can be considered as circular motion with constant variation rate , if the gravitational acceleration can be ignored, what are (1) the acceleration of the fighter jet at point B and ,(2)the real distance the fighter jet goes through from point A to point B?.

ata

na

Solution （ 1 ） the tangential acceleration and are the constant.

ta

ta

d

dt

v

t

0ta dd t

B

A

v

vv

o

A

B

Av

Bv

r

ata

na

2t sm3.23

ta AB vv

1hkm1940 Av1hkm2192 Bv

s3t km5.3AB

Given ：

The normal acceleration at B point 22

n sm106 r

a Bv

The magnitude of acceleration is

22n

2t sm109 aaa

4.12arctann

t a

a the angle is

t

ta0 tdd

B

A

v

vv

1hkm1940 Av1hkm2192 Bv

s3t km5.3ABGiven ：

（ 2 ） in time , the angle swept out by ist r2

2

1ttA

The distance the jet goes through is

2t2

1tatrs A v

Putting the date, we have

m1722s

o

A

B

Av

Bv

r

ata

na

§1.3 Relative motion§1.3 Relative motion

一 . Time and space

二 . Relative motion

Example

11 Time and spaceTime and space

In the two reference frames moving relative to each other, the measurement of length is absolute, regardless of the reference frame. The absoluteness of the time and the length is the foundation of classical or Newton’s mechanics

AB

v

a wagon moves with a relatively slow velocity along the horizontal tracks and goes through point A and B . The time the wagon takes to go from point A to B is the same measured by passenger standing on the wagon and by a ground observer, respectively.

v

相对运动

Relative motion

The ball is in a curved shape motion

The ball is in a line motion at perpendicular direction

How to

transformation ？The motion status of the object depends on the selection of the reference frame


'zz

*

'yy

'xx

u

'oo

0tp 'p

u 'vv

Transformation relationship of velocity

utr

tr

'

Drr '

Relationship of the displacement

'rP

tu

u

'xx

y 'y

z 'z tt o

'o

r

Q 'Q

S frame

frame

)''''( zyxO

)(Oxyz

'S

D

'p


Galileo velocity transformation formula

Notes If the velocity of the mass point approaches to the speed of the light, Galileo velocity transformation can no longer be applied for

The absolute velocity（ observed in S frame）

t

r

d

dv

The relative velocity（ observed in S’ frame ） t

rd

'd 'v

Convected velocity（ the speed of moving frame S’ relative to the basic reference frame S）

u u

v 'v

u 'vv 'r

tu

u

'xx

y 'y

z 'z tt o

'o

r

Q 'Q

D

'p

2 Relative motion2 Relative motion

Example As shown in figure, an experimentalist A controls a bullet launcher on a flat car that moves with a constant speed of 10 m/s along a horizontal track. The launcher fires a bullet with a projectile angle in the opposite direction the cart moving. Another experimentalist B on the ground observes that the bullet moves vertically up, what is the height that the bullet can reach?

60

u

'v v

'x

y 'y

u

o x

60'o

A B

'v

x

y

v'

v'tan

Using velocity transformation formula

xx u v'v yy v'v

Solution the ground reference frame is S with the coordinate system Oxy, and the cart reference frame is S’ frame

xv'v ux yy v'v 1sm10 ux xv'0v

tanxv'v'v yy

1sm3.17 yv

The height is m3.15

2

2

g

y yv

xv'

v'ytan

u

'v v

'x

y 'y

u

o x

60'o

A B

'v

p.51 / 1- 22, 24, 25

HomeworkHomework

Chapter 1 Chapter 1 SummarySummary

Review

Geting integral constant according with initial condition

Two kinds of questions in Kinematics:Two kinds of questions in Kinematics:

The　　　end　　

Physics （ Fifth Edition · Part I ） Shanghai Normal university. Department of Physics.

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Transcript of Physics （ Fifth Edition · Part I ） Shanghai Normal university. Department of Physics.