Part 1 Mechanics
Space shuttle
MECHANICS
Kinematics
Dynamics
How does the
matter move?
Why does the
matter move?
1. Kinematics: description of motion
1.1 Frame of reference and coordinate system
1.2 Physical quantities
1.3 Ideal model and motion
Particle Motion
Rotation
Oscillation
MECHANICS
Kinematics
Dynamics
How does the
matter move?
Why does the
matter move?
2. Dynamics: relation of motion to its causes
2.1 Newton’s laws of motion
2.2 Work and energy
2.3 Momentum and impulse
Particle motion
Reference of frame
quantities to describe motion
linear quantities
method to describe motion
Angular quantities
calculate method
Project
motion
circular
motion
curve
motion
Chapter1-3
Particle Motion
Key word:
scalar
vector
unit vector
magnitude
direction
length
coordinate axis
displacement
distance
vector addition
component
vectors
components
positive
negative
scalar product
vector product
time interval
instant
curved line
line-segment
arrow
origin point
parallel
perpendicular
Key word:
particle
frame of reference
position
displacement
average (/instantaneous) velocity
average (/instantaneous ) acceleration
speed
free fall
acceleration due to gravity
projectile
trajectory
derivative
normal component
tangential component
1. Basic Concepts
1.1 Ideal Model

Particle: It is the body that has only the mass,
but not its shape and size.
a pingpong
the earth
Which one is a particle?
Ideal Models: Simple pendulum, rigid body,
point charge, harmonical oscillator…
1.2 Frame of Reference and Coordinate Axis

Frame of Reference: relative, usually refer to earth
o
x

The Coordinate System: math conception
attached to the real-word bodies
Cartesian
natural
Other coordinates: polar, spherical, cylindrical,
elliptical…
1.3 Scalars and Vectors


Scalar: described by a single number with a unit,
such as 1kg(mass), 103kg/m3(density),
1A(electrical current).
Vector: has both magnitude and a direction, such




as
r , v , a, E
Represe
nt by:
 : Unit vector
 
ˆ
A  A Aˆ  AA

A

B  N
A 
C
Â
M

  
B  A, C   A
1) Components of a vector:
 

A  Ax  Ay


Ax  Ax iˆ , Ay  Ay ˆj

A  Ax iˆ  Ay ˆj
y
iˆ，ˆj :represent unit vectors
in direction of +x-axis or
+y-axis
 
Ay A


O
Ax
x

2
2
A  A  Ax  Ay
tg  Ay / Ax
 
A B  ?
2) Vector Addition
(1) adding with components;
(2) adding by geometrical way.


A  Axiˆ  Ay ˆj, B  Bxiˆ  By ˆj
  
C  A  B  ( Ax  Bx )î  ( Ay  B y ) ĵ
  
C  A B

A

B

A

B
  
C  A B

P3

P2

P4


P
P
1  
P  P1  P2  ...
3) Scalar Product (Dot Product)
 
 
C  A  B  A  B  cos 

A

 AB cos 
 
Suppose: A // B
Then:
 
A B  A B


Suppose: A //  B
 
 
A  A  A2
A B   A B
 
 
Example:W  F  S;
dW  F  dr
Then:
B

A

A

B

B
4) Vector Product (Cross Product)

 
C  A B
 
C  A  B  AB sin
  
C  A B

B
Direction: determined by
right-hand rule
 


Suppose: A // B or A //  B
 
 
Then:
A B  0 A A  0

 
 

Example: M  r  F ; L  r  mv

A
c
2. Physics quantities to describe the particle motion
2.1 Position Vector , Displacement and Motional Equation
1) position vectors

z

r  OP  OA AB BP



P(x,y,z)

r  xiˆ  yˆj  zkˆ

Magnitude is determined by:


r  r  x2  y2  z2
Direction is determined by:
x
x
A
cos   x / r; cos   y / r; cos   z / r
cos 2   cos 2   cos 2   1
r
z

C
y
o
y
B
2) displacement vectors
Displacement Vector:
y
A

rA
O

r
  
r  rB  rA
B

rB
Cautio
n!
x

r

r  AB

r  AB
s  AB
Displacement is different from distance.
Discussion: A very small displacement during a small
time interval
A very small displacement:

dr  AB
y

rA
A
rd
d
O
Cautio
n!
ds

dr

rB
A very small distance:
ds  AB
B
When time interval
approaches to 0:
dr
C

dr  ds
Let: OA  OC
x

dr  dr
AC  rd CB  dr
 AB  AC  CB

 dr  AC  CB  rdˆ  drr̂
Example: a radar station detects an airplane approaching
directly from the east. At first observation, the range to
plane is 360m at 400 above the horizon. The plane is
tracked for another 1230 in the vertical east-west plane, the
range at final contact being 790m. Find displacement of the
airplanes during the period of observation.
Solution:
3) Motional equation

r  xiˆ  yˆj  zkˆ
 
r  r ( t )  x( t )iˆ  y( t ) ˆj  z( t )kˆ
Motional
equation
x  x( t ), y  y(t ), z  z(t )
Example:  x  6 cos 2t

 y  6 sin 2t
Path
graph
Path
equation
y
x2+y2=62
x
2.2 Velocity and Speed
1) Average Velocity
 

r r ( t  t )  r ( t )

vav 

t
t
z

x ˆ y ˆ z ˆ
v av 
i
j
k
t
t
t

r( t )
v av

vav  vav

r  s
A

rr
S
B

r ( t  t )
2) Average Speed
s

t
C
o
x
y
 

r  r ( t  t )  r ( t )
3) Instantaneous Velocity
Caution!


r dr

v  lim

t  0 t
dt

r  s

dr  ds
 dx ˆ dy ˆ dz ˆ
v
i
j k
dt
dt
dt
z
4) Instantaneous Speed
C

r( t )
s
ds
v  lim

t  0  t
dt
A

r
S
B

r ( t  t )

v v
o
x
y
Example: Chose the correct equation

dr
(1)v 
dt

dr

( 3)v 
dt
( 2)v 

dr
dt
dr
( 4)v 
dt
Example: How to determine the direction of V in the
curved-line motion?
y
x
Example: How to find Velocity on an x-t graph?
Vp=?
x
1 Q
P
VQ=?
B
VA=?
VB=?
vp= tg1
tangent
 2  0 vQ=tg2=0
O
A
t1
t2
t
Slope of tangent = instantaneous speed
2.3 Acceleration

v ( t  t )
1) acceleration in Cartesian coordinates

Average acceleration
 

v v ( t  t )  v ( t )

aav 

t
t

Instantaneous acceleration

v (t )
B
A

v


2
v dv d r

a  lim

 2
t  0  t
dt
dt

v (t )
 dv x ˆ dv y ˆ dv z ˆ
a
i
j
k
dt
dt
dt
d 2 x ˆ d 2 y ˆ d 2v ˆ
 2 i  2 j 2 k
dt
dt
dt
 

v  v ( t  t )  v ( t )

v ( t  t )

Components of velocity and acceleration

r  xiˆ  yˆj  zkˆ

 dr dx ˆ dy ˆ dz ˆ
v

i
j
k
dt
dt
dt
dt
dx
dy
dz
vx 
,v y 
,vz 
dt
dt
dt

v  v  v x2  v 2y  v z2

 dv dv x ˆ dv y ˆ dv z ˆ d 2 x ˆ d 2 y ˆ d 2 z ˆ
a

i
j
k
i  2 j 2 k
2
dt
dt
dt
dt
dt
dt
dt
dv y d 2 y
dv z d 2 z
dv x d 2 x
ax 
 2 ,a y 
 2 ,az 
 2
dt
dt
dt
dt
dt
dt

a  a  a x2  a 2y  a z2
Example : direction of acceleration

a

a
x
D
C
B

a
Concave side of the path
E
F
G
O
t
A
Inflection point
Example : Chose the correct equation:
dv
(1)a 
dt

d r

( 3)a 
dt 2
2

dv
( 2)a 
dt

dv
( 4)a 
dt
Example: The position of a particle is given by
 2ˆ
r  t i  (1  2t ) ˆj
(1) calculate:
 
v,a
when t=2s.
(2)when is the velocity perpendicular to acceleration.



2
Solution: (1) r  t i  (1  2t ) j



 dr
v
 2ti  2 j
dt


 dv
a
 2i
dt
(2)
t 0
Example: The motion of a particle is described by the

function
2
2
r  ( 3  2 t )î  ( 2 t  1 ) ĵ
What kind of motion does it undergo?
Tow kinds of problems in kinematics
derivative

r  x(t )iˆ  y(t ) ˆj  z(t )kˆ

 dr

dt
integral
Calculus-based-physics!
 d
a
dt
Example: deduce the following equation if particle move
in straight line with a=c, and t=0, v=v0, x=x0 .
v  v 0  at
1 2
s  v 0 t  at
2
v 2  v 02  2as
Example: Suppose the position of an object is given by
x = t3-9t2+15t+1(SI).
a) Find the initial velocity. When does the object turn
around?
b) Find the displacement and the distance traveled for
the time interval t=0 to t=2s.
Solution: a ) v 
dx
 3t 2  18 t  15  3( t  1 )( t  5 )
dt
 t  0 , v  15 m / s
Because condition for turning around is:
v=0, the object turns around at t=1,t=5
b )x  x( 2 )  x( 0 )  ( 3  1 )m  2 m
S  x( 1 )  x( 0 )  x( 2 )  x( 1 )  12m
Example: The position of a particle is given by

r  R cos tî  R sin tĵ .
a) What kind of motion does it undergo?
b) Find the displacement and the distance traveled for
the time interval t=/ to t= 2/.
Solution: a ) x
 R cos t , y  R sint  x 2  y 2  R 2

 dr
v 
  R sin tî  R cos tĵ ,
dt
 dv

2
2
2
a
  R cos tî  R sin tĵ   r
dt
v
v x2  v 2y  R , a   2 R
The particle moves along a circle with constant speed
 

b )r  r ( 2 /  )  r (  /  )  2 Rî
S  R
Example: A radio-controlled model car is moving on a
plane (xy-plane). The car has x- and y-coordinates
that vary with time according to
x=2t, y=19-2t2(SI).
a) Find the car’s coordinates at time t=1s and t=2s,
then find the displacement and average velocity
during the time interval.
b) Find the instantaneous velocity and acceleration at
t=1s.
c) Find the path equation of the car.
d) When the car is nearest to the origin point of xyplane?
e) What is the distance for t=0s to t=1s.
Solution:
Solution:

a ) r  2 tî  ( 19  2 t 2 ) ĵ


 t  1s , r1  ( 2 î  17 ĵ )m , t  2 s , r2  ( 4 î  11 ĵ )m

r
  

 r  r2  r1  ( 2 î  6 ĵ )m , vav 
 ( 2 î  6 ĵ )m / s
t


 dr
 dv
b ) v 
 2 î  4 tĵ , a 
 4 ĵ
dt
dt


2
 t  1s , v  ( 2 î  4 ĵ )m / s , a  4 ĵm / s
c ) y  19  x 2 / 2 This is a parabola
d )r  x 2  y 2  4t 2  (19  2t 2 )2
dr
when
 0 , r  rmin
dt
 t  3 s , r  rmin  6.08m
ds

e ) v  2 î  4 tĵ  v   2 2  ( 4 t )2
dt
t2
t2
t1
t1
 s   vdt  
1 2
s  v0 t  at
2
 ds  vdt
1  ( 2t ) 2 d ( 2t )
1
 [ 1  ( 2t ) 2  ln( 2t  1  ( 2t ) 2 )]10  1.34m
2
Example:The motion of an object falling from rest in a
resisting medium is described by the equation
dv/dt=A-Bv,
Where A and B are constants. In terms of A and B, find
a) The initial acceleration.
b) The velocity at which the acceleration becomes zero
(the terminal velocity).
c) Show that the velocity at any given t is given by
A
v  ( 1  e  Bt )
B
Solution: a ) a  dv  A  Bv , v  0  a  A
0
dt
dv
A
b ) a 
 A  Bv  0 , v 
dt
B
0
c ) dv  ( A  Bv )dt

v
0
t
A
dv
 Bt
  dt  v 
(
1

e
)
0
A  Bv
B
Example: The man on the bank drag the boat with constant
velocity. Try to find the velocity and acceleration of the
boat, When the distance between the boat and the bank is x.
Solution: Set up coordinate axis in the picture, then draw
the position vector of the boat.

 r  xî  hĵ , x 

 dr dx
v 

î
dt
dt
v x h
  î
x
2

y

o
x

r
2
v h
 d
ˆ
a
 i 3
dt
x
2
r 2  h2
h
2
x
2) Tangential and Normal Acceleration

dv y
dv y
d2x
d2y
d 2z
 dv dv x
a

î 
ĵ 
k̂  2 î  2 ĵ  2 k̂
dt
dt
dt
dt
dt
dt
dt
Can we represent the acceleration of a particle
moving in a curved path in terms of components parallel
and perpendicular to the velocity at each point?
ê t
ê n

v
: tangential direction
: normal direction

v  vêt

dê t
dv
d ( vê t )
 dv
v

ê t
a

dt
dt
dt
dt
êt
P
ên O
R
dêt dv

av

êt
dt
dt

v( t )
for a very small time interval
t  0,   0
êt  êt , dêt  dên
d
dv

av
ê n 
êt
dt
dt
d
d ds
v



dt
ds dt

2
dv
 v
a
ên 
êt

dt

v ( t  t )
êt ( t  t )
R
êt ( t ) Q
P
 
ên O
êt ( t )
êt

êt ( t  t )
dv
 v


a
ên 
êt  an  at

dt

at
2
dv

at 
ê t
dt
2
v

an 
ên

Magnitude: a n 
v2

dv
Magnitude:a t 
dt
Direction: normal direction
Direction: tangential direction
Describe the change rate
of direction of velocity
with time
Describe the change rate of
the magnitude of velocity
with time

at
2
v
dv
 

a  an  at 
ên 
êt

dt


an
2
v 2
dv 2

2
2
a  an  at  (
) (
)

dt
an
tg 
at
Example：correct the following formula
1）a  dv
dt

2）at  dv
dt

v

dv
3）at 
dt

a
Example：what is the character of
at

a
an
1) In straight line motion
an  0 a  at
2) In free fall motion
a  at   g an  0
3) In projectile motion
a  g an  g cos
4) In uniform circular motion
at  g sin
2
v
a  an 
R
at  0
5)In nonuniform circular motion
v2
an 
R
dv
at 
dt
a  an  at2
2
Example: A particle moves in a circle of radius R. The
distance is described by the equation
1 2 (b,c are constants, b2>Rc)
s  bt 
2
ct
a) When an= at?
b) When a= c?
Solution: a )  ds  b  ct
dt
d
at 
 c
dt
 2 ( b  ct )2
an 

R
R
b
R
For an = | at |
t 
c
c
b ) a  an2  at2
b
t 
c
Example8: Find an , at and  of projectile motion at any time.
Suppose t=0, v=v0 , and makes an angle  with +x.
Solution: Set up x,y coordinate axis
Projectile motion can be considered as a combination
of horizontal motion and vertical motion
 x   o cos 
 y   osin  gt
y

at
v0

vx
 an
g
vy
  x y
2
2
a  g
x
 an  gcos  g

x
y
 at   gsin   g

g o cos
x 
a n  gcos  g
(  o cos )2  (  o sin  gt )2

y
 g(  o sin  gt )

at   gsin   g

(  o cos )2  (  o sin  gt )2



(  cos )

2
o
an
 (  o sin  gt )
g o cos
2

3
2 2
Another solution:
y
 x  ocos , y   osin  gt ,    x 2   y 2

at
v0

vx
 an
g
vy
d
 g(  o sin  gt )
at 

dt
( o cos ) 2  ( o sin - gt)2
 g  an  at
2
2
 an  g  at
2
x
2

2
an
3. Angular quantities to describe the particle motion
1) Angular Displacement, Velocity and Acceleration
Suppose a particle moves in a circle of radius
R. We can use the single quantity  as a coordinate,
 is called angular coordinate, and usually measured
in radians.
y
(1) Angular Displacement
   ( t  t )   ( t )
(2) Angular velocity:

Average angular velocity  av 
t
d
Instantaneous angular velocity  
dt
R
o
A

x
Caution :
  d
dt
Angular velocity is a vector!
Right Hand Rule: the vector is
represented by an arrow drawn so
that if curled fingers of the right
hand point in the direction of the
rotation, the direction of the vector
coincides with the direction of the
extended thumb.
If rotation is counter clockwise
  increasing  positive
If rotation is clockwise  
decreasing  negative
  d
dt
(3) Angular Acceleration:
Average Angular Acceleration
 av


t
Instantaneous Angular Acceleration
d

dt
Example: The angular position
  2  4t 2  2t 3
find the angular velocity and angular acceleration
Solution:
d

 8t  6t 2
dt
d

 12t  8
dt
2) Relationship between Linear Quantities and
Angular Quantities
in circular motion
2
d
  R , a n 
 R , at 
 R
R
dt
2
prove：
ds  rd
ds d

r
dt dt
General rule:
v  r
 
  r


Compare circular motion with straight-line motion
Straight-line motion
Circular motion
x

v = dx/dt
 = d/dt
a = dv/dt
 = d/dt
a=const.
v  v 0  at
 =const.
   0  t
v 2  v02  2a( x  x0 )
1 2
   0   0 t  t
2
2
2
   0  2  (   0 )
1 2
x  x0  v0 t  at
2
Example: An early method of measuring speed of light
makes use of a rotating slotted wheel. A beam of light pass
through a slot at the outside edge of the wheel, travel to a
distant mirror, and return to the wheel just in time to pass
through the next slot in the wheel. r=5cm and there are
500slots at wheel’s edge, L=500m, c=3*108m/s. What is the
constant angular speed of the wheel and linear speed of a
point on the edge of wheel.
Example: A pulsar is a rapidly rotating neutron star that
emits a radio beam like a lighthouse emit a light beam. We
receive a radio pulse for each rotation of the star. The period
T of rotation is found by measuring the time between pulses.
The pulsar in the crab nebula has a period of rotation of
T=0.033s that is increasing at the rate of 1.26*10-5s/y.
1) What is the average pulsar’s angular acceleration?
2) If its acceleration is constant, how many years from now
will the pulsar stop rotating
2
(
)

2 T
T
 

 2
t
t
T t
T 1.26  105
13


4

10
t
3.1  107
  2.3  109 rad / s 2
   0  t
2
when   0 t   0 /   
 8.3  1010 s
T
Example: A particle move around a circle of radius R=1.0m.
The angular coordinate vary with time according to
 =2 + 12 t - t3 (SI).
a) Find the normal acceleration and tangential acceleration
in rim of wheal at time t=1s.
b) When will the particle stop? How many circles does the
particle turn?
Solution: a )  d  12  3t 2 ,
dt
d

 6t
dt
 n  R  ( 12  3t ) , at  R  6t
2
2
2
t  1s , an  81 2 , at  6 ( SI )
b ) when
stop ,  12  3t  0  t
t  2 s, 2  18 , t  0, 1  2
   16
2
 2s
4. Relative Velocity
let : C ( car ), D( dog ), E ( earth )
vCD  20 km / h , v DE  30 km / h , vCE  ?
o
x
4. Relative velocity
Suppose here are two reference frame A and B.
xA//xB, yA//yB , zA//zB . B moves with constant velocity
along x axis relative to A.
for point P:

v BA
yB



rPA  rPB  rBA



v PA  v PB  v BA



a PA  a PB  aBA
yA

rPA
P

rPB
xB
OB
OA
zB
zA
xA



v PA  v PB  v BA



a PA  a PB  aBA
Caution :
vPA :velocity of P relative to A
(1) They are vector addition.
(2) You can not change the sequence of the subscripts.


v PA  v AP



v PA  v PB  vB A
let : C ( car ), D( dog ), E ( earth )
vCD  20 km / h , v DE  30 km / h ,
vCE  ?



vCE  vCD  vD E
Example: The compass of an airplane indicates that it is
headed due north, and its airspeed indicator shows
that it is moving through the air at 240km/h. If there is
a wind of 100km/h from west to east , what is the
velocity of the airplane relative to the earth?
Solution:
N
let : P ( plane ), A( air ), E ( earth )



v PE  v P A  vA

v PA
E

v AE  100km / hiˆ ,

v PA  240km / hˆj ,

v PE  ?
W

v AE

v PE  ?
E
S
Example: When a train’s speed is 10m.s-1 eastward,
raindrops that are falling vertically with respect to the
earth make traces that are inclined 30o to the vertical on
the windows of the train.
a) What is the horizontal component of a drop’s velocity
with respect to the earth? With respect to train?
b) What is the magnitude of the velocity of the raindrop
with respect to the earth? With respect to train?
Solution:
let : T ( train ), R( rain ), E ( earth )



v RE  v R T  vT E
v RE //  0, v RT //  10 m / s
v RE  10 3m / s
v RT  20 m / s

vTE  10 m / siˆ

v RT  ?
  30 0

v RE  ?