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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
2 Position vector , equation of motion and 2 Position vector , equation of motion and displacementdisplacement
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.
2 Position vector , equation of motion and 2 Position vector , equation of motion and displacementdisplacement
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
2 Position vector , equation of motion and 2 Position vector , equation of motion and displacementdisplacement
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
2 Position vector , equation of motion and 2 Position vector , equation of motion and displacementdisplacement
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
44 AccelerationAcceleration
？
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
44 AccelerationAcceleration
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
33 Tangential and normal acceleration of the Tangential and normal acceleration of the circular motion circular motion
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
33 Tangential and normal acceleration of the Tangential and normal acceleration of the circular motion circular motion
For General curve Motion （ natural coordinate system ）
ntdd ee
ta
2vv
tdd e
tsv
ddswhere is curvature radius.
33 Tangential and normal acceleration of the Tangential and normal acceleration of the circular motion circular motion
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
33 Tangential and normal acceleration of the Tangential and normal acceleration of the circular motion circular motion
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
11 Time and spaceTime and space
'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
11 Time and spaceTime and space
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
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