Space shuttle

Part 1 Mechanics

MECHANICS

Particle Motion Rotation Oscillation

1. Kinematics: description of motion

1.1 Frame of reference and coordinate system

1.2 Physical quantities

1.3 Ideal model and motion

How does the matter move?

Why does the matter move?

Kinematics Dynamics

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

MECHANICS

Kinematics Dynamics

How does the matter move?

Why does the matter move?

Reference of frameReference of frame

quantities to describe motion quantities to describe motion method to describe motion method to describe motion

linear quantities linear quantities Angular quantitiesAngular quantities calculate method calculate method

Project motion

Project motion

circular motion

circular motion

curve motion

curve motion

Particle motionParticle motion

Chapter1-3 Particle Motion

scalarvectorunit vectormagnitudedirectionlengthcoordinate axisdisplacementdistance

vector additioncomponent vectorscomponents

positivenegativescalar productvector product time intervalinstantcurved lineline-segmentarroworigin point

parallelperpendicular

Key word:

particleframe of referencepositiondisplacementaverage (/instantaneous) velocity average (/instantaneous ) accelerationspeed free fallacceleration due to gravity projectiletrajectoryderivativenormal componenttangential component

Key word:

1.1 Ideal Model

Particle: It is the body that has only the mass, but not its shape and size.

Which one is a particle?

a pingpong the earth

Ideal Models: Simple pendulum, rigid body, point charge, harmonical oscillator…

1. Basic Concepts

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 Eavr

,,,

Represent by: A

AAAAA ˆˆ

M

NA

A

B

C

vectorUnit:A

ACAB

,

A

jAiAA yxˆˆ

22yx AAAA

O x

y

xy AAtg /

ji ˆˆ， :represent unit vectors in direction of +x-axis or +y-axis

xA

yA

yx AAA

1) Components of a vector:

jAAiAA yyxxˆ,ˆ

2) Vector Addition

(1) adding with components;(2) adding by geometrical way.

? BA

B

A

BAC

B

A

BAC

1P2P

3P

4P

...21 PPPP

jBiBBjAiAA yxyxˆˆ,ˆˆ

j)BA(i)BA(BAC yyxx

3) Scalar Product (Dot Product)

cosAB

cosBABAC

B

A

BA

//Suppose:

BABA

Then: 2AAA

BA

//Suppose:

BABA

Then:

BA

A

B

Example:

rdFdWSFW

;

4) Vector Product (Cross Product)

sinABBAC BAC

B

A

BAC

BAorBA

////Suppose:

0BA

Then: 0AA

Example:

vmrLFrM

;

Direction: determined by right-hand rule

c

2. Physics quantities to describe the particle motion2.1 Position Vector , Displacement and Motional Equation2.1 Position Vector , Displacement and Motional Equation

1) position vectors

kzjyixr ˆˆˆ

1coscoscos

/cos;/cos;/cos222

rzryrx

Direction is determined by:

222 zyxrr

Magnitude is determined by:

OPr

BPABOA

o

x

y

z

P(x,y,z)

xy

z

A B

C

r

2) displacement vectors

r

Br

BA

Ar

O x

y Displacement Vector: r

AB rrr

ABr

Caution!

ABr

ABs Displacement is different from distance.

Discussion: A very small displacement during a small time interval

O

y

Ar

Br

BA

x

rd

dsABds

dsrd When time interval approaches to 0:

ABrd A very small displacement:

A very small distance:

Cdr

rdAC

rdrˆrdCBACrd

OCOA Let:

CBACAB

d

rd

drCB

Caution!

drrd

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

Motional equation

)(),(),( tzztyytxx

ktzjtyitxtrr ˆ)(ˆ)(ˆ)()(

x2+y2=62

x

y

Pathequation

Path graph

kzjyixr ˆˆˆ

ty

tx

2sin6

2cos6Example:

2.2 Velocity and Speed2.2 Velocity and Speed

S

r

x

Az

yo

C

B)t(r

)tt(r

1) Average Velocity

t

trttr

t

rvav

)()(

kt

zj

t

yi

t

xvav

ˆˆˆ

t

svav

2) Average Speed

sr

avav vv

r

)()( trttrr

S

r

x

Az

yo

C

B)t(r

)tt(r

3) Instantaneous Velocity

dt

rd

t

rv

t

0

lim

kdt

dzj

dt

dyi

dt

dxv ˆˆˆ

dt

ds

t

slimv

0t

4) Instantaneous Speed

vv

Caution!

dsrd sr

Example: Chose the correct equation

dt

rdv

)1(

dt

rdv

)2(

dt

rdv

)3(

dt

drv )4(

Example: How to determine the direction of V in the curved-line motion?

x

y

Example: How to find Velocity on an x-t graph?

t

x

P

O

Slope of tangent = instantaneous speed

t1

02 vQ=tg2=01 vp= tg1

tangentVp=?

A

BVA=?

VB=?

Q

t2

VQ=?

)( ttv

)(tv v

)(tv

Instantaneous acceleration

t

vlima

0t

kdt

vdj

dt

ydi

dt

xd

kdt

dvj

dt

dvi

dt

dva zyx

ˆˆˆ

ˆˆˆ

2

2

2

2

2

2

A

)( ttv

B

)t(v)tt(vv

t

)t(v)tt(v

t

vaav

Average acceleration

2

2

dt

rd

dt

vd

2.3 Acceleration2.3 Acceleration1) acceleration in Cartesian coordinates1) acceleration in Cartesian coordinates

kdt

zdj

dt

ydi

dt

xdk

dt

dvj

dt

dvi

dt

dv

dt

vda zyx ˆˆˆˆˆˆ

2

2

2

2

2

2

2z

2y

2x

2

2z

z2

2y

y2

2x

x

aaaaa

dt

zd

dt

dva,

dt

yd

dt

dva,

dt

xd

dt

dva

kzjyixr ˆˆˆ

Components of velocity and acceleration

kdt

dzj

dt

dyi

dt

dx

dt

rdv ˆˆˆ

2z

2y

2x

zyx

vvvvv

dt

dzv,

dt

dyv,

dt

dxv

Example : direction of acceleration

Concave side of the path

aa

a t

x

OA

B

C EF

D

Inflection point

G

Example : Chose the correct equation:

dt

dva )1(

dt

vda

)2(

2

2

)3(dt

rda

dt

vda

)4(

Example: The position of a particle is given by

(1) calculate: when t=2s.

(2)when is the velocity perpendicular to acceleration.

jtitr ˆ)21(ˆ2

av

,

jtitr

)21(2

idt

vda

jitdt

rdv

2

22

(1)

0t(2)

Solution:

Example: The motion of a particle is described by the function

What kind of motion does it undergo?

j)1t2(i)t23(r 22

Tow kinds of problems in kinematics

dt

dr

dt

da ktzjtyitxr ˆ)(ˆ)(ˆ)(

derivative

integral

Calculus-based-physics!

asvv

attvs

atvv

2

2

1

20

2

20

0

Example: deduce the following equation if particle move in straight line with a=c, and t=0, v=v0, x=x0 .

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: )5t)(1t(315t18t3dt

dxv)a 2

s/m15v,0t

v=0, the object turns around at t=1,t=5

Because condition for turning around is:

m2m)13()0(x)2(xx)b

m12)1(x)2(x)0(x)1(xS

Example: The position of a particle is given by .a) What kind of motion does it undergo?b) Find the displacement and the distance traveled for

the time interval t=/ to t= 2/.

jtsinRitcosRr

Solution: tsinRy,tcosRx)a 222 Ryx

The particle moves along a circle with constant speed

,jtcosRitsinRdt

rdv

jtsinRitcosRdt

vda 22

r2

Ra,Rvvv 22y

2x

iR2)/(r)/2(rr)b RS

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, the

n 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 xy-pla

ne?e) What is the distance for t=0s to t=1s.

Solution:

j)t219(it2r)a 2

m)j11i4(r,s2t,m)j17i2(r,s1t 21

s/m)j6i2(t

rv,m)j6i2(rrr av12

j4dt

vda,jt4i2

dt

rdv)b

2s/mj4a,s/m)j4i2(v,s1t

Solution:

2/x19y)c 2 This is a parabola

22222 )219(4) ttyxrd

minrr,0dt

drwhen

22 )t4(2dt

dsvjt4i2v)e

m08.6rr,s3t min

vdtds

mttt

tdtvdtst

t

t

t

34.1)])2(12ln()2(1[2

1

)2()2(1

10

22

22

1

2

1

20 at

2

1tvs

Example:The motion of an object falling from rest in aresisting medium is described by the equation dv/dt=A-Bv,Where A and B are constants. In terms of A and B, finda) The initial acceleration.b) The velocity at which the acceleration becomes zero (th

e terminal velocity).c) Show that the velocity at any given t is given by

)e1(B

Av Bt

Solution: 0v,BvAdt

dva)a 0

t

0

v

0dt

BvA

dv)e1(

B

Av Bt

Aa0

B

Av,0BvA

dt

dva)b

dt)BvA(dv)c

h

x

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.

22 hrx,jhixr

idt

dx

dt

rdv

x

hxvi

22

3

22ˆ

x

hvi

dt

da

ox

y

r

2) Tangential and Normal Acceleration2) Tangential and Normal Acceleration

P

R

kdt

zdj

dt

ydi

dt

xdk

dt

dvj

dt

dvi

dt

dv

dt

vda

2

2

2

2

2

2yyx

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?

v

tevv

dt

)ev(d

dt

vda t

tt e

dt

dv

dt

edv

:e

:e

n

t tangential direction

normal direction

ne

te

O

te)t(et

)tt(et

tt e

dt

dv

dt

edva

P

R)t(v

One

)t(et

for a very small time interval

0,0 t

,ee tt nt eded

tn edt

dve

dt

dva

v

dt

ds

ds

d

dt

d

tn

2

edt

dve

va

Q

)( ttv

)tt(et

ta

tn

2

edt

dve

va

tn aa

n

2

n ev

a

tt e

dt

dva

Direction: normal direction

Describe the change rate of the magnitude of velocity with time

2

n

va Magnitude:

Describe the change rate of direction of velocity with time

dt

dvat Magnitude:

Direction: tangential direction

t

n

222

2t

2n

a

atg

)dt

dv()

v(aaa

tn

2

tn edt

dve

vaaa

na

ta

v

a

dtdva ）1

dtvdat

）2

dt

vdat

）3

Example： correct the following formula

1) In straight line motion

2) In free fall motion

3) In projectile motion

4) In uniform circular motion

5)In nonuniform circular motion

tn aaa 0

0 nt agaa

sincos gagaga tn

02

tn aR

vaa

222

tntn aaadt

dva

R

va

Example： what is the character of aaa nt

Example: A particle moves in a circle of radius R. The distance is described by the equation

(b,c are constants, b2>Rc)

a) When an= at?b) When a= c?

2ct2

1bts

Solution:

ctbdt

ds)a

cdt

dat

R

)ctb(

Ran

22

For an = | at | c

R

c

bt

2t

2n aaa)b

c

bt

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

x

y

g

0v

Projectile motion can be considered as a combination of horizontal motion and vertical motion

cosox

gtoy sin 22

yx

ga

xvyv

nata

x

n gga cos

yt gga sin

x

n gga cos

yt gga sin

22 sincos

sin

)gt()(

)gt(g

oo

o

22 sincos

cos

)gt()(

g

oo

o

cos

sincos

o

2

32

o2

o

n

2

g

)gt()(

a

Another solution: ,ox cos ,gtoy sin 22

yx

dt

dat

2o

2o gt)-sin()cos(

sin

)gt(g o

22

tn aag

na

2 2t

2n aga

x

y

g

0v

xvyv

nata

y

xo

A

R

Suppose a particle moves in a circle of radius R. We can use the single quantity as a coordinate, 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.

1) Angular Displacement, Velocity and Acceleration

(2) Angular velocity:

3. Angular quantities to describe the particle motion

(1) Angular Displacement

)()( ttt

Average angular velocitytav

Instantaneous angular velocitydt

d

Caution :

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.

dtd

If rotation is counter clockwise increasing positive

If rotation is clockwise decreasing negative

dtd

(3) Angular Acceleration:

dt

d

Average Angular Acceleration

Instantaneous Angular Acceleration

tav

find the angular velocity and angular acceleration

812

68 2

tdt

d

ttdt

d

Solution:

Example: The angular position

32 242 tt

Rdt

da,R

Ra,R t

22

n

2) Relationship between Linear Quantities and Angular Quantities

r

in circular motion

prove：

General rule:

rvrdt

d

dt

dsrdds

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. =const.atvv 0

200 at

2

1tvxx

)(2 020

2 xxavv

t0 2

00 t2

1t

)(2 020

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

t

T

TtT

t

2

2)2

(

137

5

104101.3

1026.1

t

T

29 /103.2 srad

sT

twhen

t

10

0

0

103.82

/0

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:

,t312dt

d)a 2 t

dt

d 6

)SI(6a,81a,s1t

t6Ra,)t312(R

t2

n

t222

n

0t312,stopwhen)b 2 s2t

16

2,0,18,2 12

tst

4. Relative Velocity

o x

?v,h/km30v,h/km20v

)earth(E),dog(D),car(C:let

CEDECD

4. Relative velocity

yA

zA

xA

OA

yB

zB

xBOB

PAr

P

PBr

BAPBPA rrr

BAv

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:

BAPBPA vvv

BAPBPA aaa

APPA vvv

Caution :

BAPBPA vvv

BAPBPA aaa

(1) They are vector addition.

(2) You can not change the sequence of the subscripts.

APPA vv

vPA :velocity of P relative to A

B B

ECCE vvv

?v

,h/km30v,h/km20v

)earth(E),dog(D),car(C:let

CE

DECD

D D

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

E

S

W

)earth(E),air(A),plane(P:let

PAv

AEv

?vPE

EPPE vvv A A

,ˆ/100 ihkmvAE

?PEv

,ˆ/240 jhkmvPA

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?

ism10vTEˆ/

030?RTv

ERRE vvv T T

)earth(E),rain(R),train(T:letSolution:

?REv

sm10v0v RTRE /, ////

s/m20v

s/m310v

RT

RE