PHYS 1443 – Section 501 Lecture #2yu/teaching/spring02-1443-501/lectures/phys144… · Some...

Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu

1

PHYS 1443 – Section 501Lecture #2

Monday, Jan. 16, 2002Dr. Jaehoon Yu

1. Some Chapter 1 problems2. Some fundamentals3. Displacement, Velocity, and Speed4. Acceleration5. Kinetic Equation of Motion

Reading of the dayhttp://www.economist.com/surveys/displaystory.cfm?story_id=922278

Use the CD-Rom in your book for demonstration.!!


2

Problems 1.4 and 1.13• The mass of a material with density, ? , required to make a

hollow spherical shell with inner radius, r1, and outer radius, r2?

• Prove that displacement of a particle moving under uniform acceleration is, s=kamtn, is dimensionally correct if k is a dimensionless constant, m=1, and n=2.

3

34

rVsphere

??

3

34

rVM spheresphere ??

? ??r1

r23

134

rVM innerinner ??

? ??

323

4rVM outerouter ?

?? ??

)(3

4 31

32 rr

MMM inneroutershell

??

??

??

Displacement: Dimension of LengthAcceleration a:Dimension of L/T2 ? ? ? ? ? ? ? ? ? ? ? ?

22;02 ,1

222

???????

?????

???? ???

mnmnm

tltltttl

l nmmnmnm


3

Problems 1.25 & 1.31• Find the density, ? , of lead, in SI unit, whose mass is 23.94g and

volume, V, is 2.10cm3.

3333 /104.11

1001

10001

4.1110.2

94.23 Density; mkg

m

kg

cmg

V?

??

???

???

?????

• Find the thickness of the layer covered by a gallon (V=3.78x10-3

m3) of paint spread on an area of on the wall 25.0m2.

mm

mAV

Thickness 42

33

1051.10.251078.3 ?

?

???

??

• Thickness is in the dimension of Length.• A gallon (V=3.78x10-3 m3) of paint is covering 25.0m2.

A

}Thickness (OK, it is very skewed view!!)


4

Some Fundamentals• Kinematics: Description of Motion without understanding the

cause of the motion• Dynamics: Description of motion accompanied with

understanding the cause of the motion• Vector and Scalar quantities:

– Scalar: Physical quantities that require magnitude but no direction • Speed, length, mass, etc

– Vector: Physical quantities that require both magnitude and direction• Velocity, Acceleration• It does not make sense to say “I ran at a velocity of 10miles/hour.”

• Objects can be treated as point-like if their sizes are smaller than the scale in the problem– Earth can be treated as a point like object (or a particle)in celestial

problems– Any other examples?


5

Some More Fundamentals• Motions:Can be described as long as the position is

known at any time (or position is expressed as a function of time)– Translation: Linear motion– Rotation: Circular or elliptical motion– Vibration: Oscillation

• Dimensions– 0 dimension: A point– 1 dimension: Linear drag of a point, resulting in a line ?

Motion in one-dimension is a motion on a line– 2 dimension: Linear drag of a line resulting in a surface– 3 dimension: Perpendicular Linear drag of a surface, resulting

in a stereo object


6

Velocity and SpeedOne dimensional displacement is defined as:

Displacement is the difference between initial and final potions of motion and is a vector quantity

ixxx f ???

Average velocity is defined as:

Displacement per unit time in the period throughout the motiontx

ttxx

vi

ix

f

f

??

???

?

Average speed is defined as:

Interval Time TotalTraveled Distance Total

?v

Can someone tell me what the difference between speed and velocity is?


7

The difference between Speed and Velocity

• Let’s take a simple one dimensional translation that has many steps:

Let’s call this line as X-axis

Let’s have a couple of motions in total time interval of 20seconds

+10m +15m

-5m -15m-10m

+5m

Total Displacement: 0?????? iif xxxxx i

Total Distance: )(605101551510 mD ???????

Average Velocity: )/(0200 sm

tx

ttxxvi

ix

f

f ?????

???

Average Speed: )/(32060

Interval Time TotalTraveled Distance Total

smv ???


8

Example 2.1• Find the displacement,

average velocity, and average speed.

Time

xi=+30mti=0sec

xf=-53mtf=50sec

)(833053 mxxx if ????????• Displacement:

• Average Velocity:

)/(7.15083 sm

tx

ttxxvi

ix

f

f ???????

???

• Average Speed:

xm=+52m

)/(5.250

12750

535222Interval Time TotalTraveled Distance Total

sm

v

????

??

Fig02-01.ip


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Instantaneous Velocity and Speed• Here is where calculus comes in to help

understanding the concept of “instantaneous quantities”

• Instantaneous velocity is defined as:– What does this mean?

• Displacement in an infinitesimal time interval• Mathematically: Slope of the position variation as a

function of time

• Instantaneous speed is the size (magnitude) of the velocity vector:

dtdx

tx

vx ????

?

lim0?

t

dtdx

tx

vx ???

??

lim0?t

*Magnitude of Vectors are Expressed in absolute values


10

Position vs Time Plot

time

Position

x=0t1 t2 t3t=0

x11 2 3

1. Running at a constant velocity (go from x=0 to x=x1 in t1, Displacement is + x1 in t1 time interval)

2. Velocity is 0 (go from x1 to x1 no matter how much time changes)3. Running at a constant velocity but in the reverse direction as 1. (go

from x1 to x=0 in t3-t2 time interval, Displacement is - x1 in t3-t2 time interval)


11

Instantaneous Velocity

Time

Average Velocity

Instantaneous Velocity


12

Example 2.2• Particle is moving along x-axis following the expression:• Determine the displacement in the time intervals t=0 to t=1s and

t=1 to t=3s:

224 ttx ???

)(202

2)1(2)1(4,0

011,0

210

mxxx

xx

ttt

tt

????????

????????

???

??

)(826

6)3(2)3(4,2

133,1

231

mxxx

xx

ttt

tt

??????????????

???

??

For interval t=0 to t=1s

For interval t=1 to t=3s

• Compute the average velocity in the time intervals t=0 to t=1s and t=1 to t=3s: )/(

121,0 sm

tx

v tx

??

??

? ? )/(4283,1 sm

tx

v tx ???

??

? ?

• Compute the instantaneous velocity at t=2.5s:

? ? ? ? tttdtd

dtdx

tx

tvx 4424 2 ??????????

?

lim0?

t

? ? )/(6)5.2(445.2 smtvx ???????

Instantaneous velocity at any time t Instantaneous velocity at t=2.5s


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Acceleration• Change of velocity in time• Average acceleration:

• Instantaneous Acceleration

• In calculus terms: A slope (derivative) of velocity with respect to time or change of slopes of position as a function of time

tv

ttvv

ax

f

xf

i

xix

??

???

?tx

ttxx

vi

ix

f

f

??

???

?analogs to

2

2

dtxd

dtdx

dtd

dtdv

tv

axx

x ????

?????

??

??

lim0?

t dtdx

tx

vx ????

?

lim0?

tanalogs to


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Example 2.4• Velocity, vx, is express in: • Find average acceleration in time interval, t=0 to t=2.0s

? ? smttvx /540)( 2??

•Find instantaneous acceleration at any time t and t=2.0s

? ?)/(10

024020

)/(202540)0.2(

)/(40)0(

2

2

smt

vtt

vva

smtv

smtv

x

if

xixfx

fxf

ixi

???

??

??

?

??

?????

??

? ? ? ? ttdtd

dtdv

tax

x 10540 2 ?????

)/(20

)0.2(10)0.2(

2sm

tax

??

????

Instantaneous Acceleration at any time

Instantaneous Acceleration at any time t=2.0s


15

Meanings of Acceleration• When an object is moving in a constant velocity (v=v0),

there is no acceleration (a=0)– Is there any acceleration when an object is not moving?

• When an object is moving faster as time goes on, (v=v(t) ), acceleration is positive (a>0)

• When an object is moving slower as time goes on, (v=v(t) ), acceleration is negative (a<0)

• In all cases, velocity is positive, unless the direction of the movement changes.

• Is there acceleration if an object moves in a constant speed but changes direction? The answer is YES!!

Fig02-09.ip


16

One Dimensional Motion• Let’s start with simplest case: acceleration is constant (a=a0)

• Using definitions of average acceleration and velocity, we can draw equation of motion (description of motion, position wrt time)

tvv

ttvv

axi

i

xix

xf

f

xf ??

??

? If tf=t and ti=0 tavv xxixf ??

For constant acceleration, simple numeric average

txx

ttxx

vi

i

ix

f

f

f ??

??

? If tf=t and ti=0

2

21

tatvxtvxx xxiixif ??? ??Resulting Equation of Motion becomes

tavtavvv

v xxixxixfxi

x

21

22

2?

?????

tvxx xif ??


17

Kinetic Equation of Motion in a Straight Line Under Constant Acceleration

? ? tavtv xxixf ??

2

21

tatvxx xxiif ???

? ?tvvtvxx xixfxif ????21

21

? ?ifxf xxavv xxi ??? 222

Velocity as a function of time

Displacement as a function of velocity and time

Displacement as a function of time, velocity, and acceleration

Velocity as a function of Displacement and acceleration

You may use different forms of Kinetic equations, depending on the information given to you for specific physical problems!!


18

Example 2.8• Problem: Car traveling at constant speed of 45.0m/s (~162km/hr

or ~100miles/hr), police starts chasing the car at the constant acceleration of 3.00m/s2, one second after the car passes him. How long does it take for police to catch the violator?

• Let’s call the time interval for police to catch; T• Set up an equation:Police catches the violator when his final

position is the same as the violator’s.

22 00.321

21

TaTx Policef ??? ? ? )1(0.451 ???? TTvx Car

f

possible)(not 00.1or (possible) 0.31;00.300.3000.1

);1(0.4500.321

;

2

2

??????

????

TTTT

TTxx Policef

Policef

aacbb

x

cbxax

24

are 0for Solutions2

2

????

???

Ex02-08.ip


19

Free Fall• Free fall is a motion under the influence of gravitational

pull (gravity) only; Which direction is a freely falling object moving?

• Gravitational acceleration is inversely proportional to the distance between the object and the center of the earth

• The gravitational acceleration is g=9.80m/s2 on the surface of the earth, most the time

• The direction of gravity is toward the center of the earth, which we normally call (-y); where up and down direction are indicated as the variable “y”

• Thus the correct denotation of gravitational acceleration on the surface of the earth is g=-9.80m/s2

Falling body with variable initial velocity.IP


20

Example 2.12Stone was thrown straight upward at t=0 with +20.0m/s initial velocity on the roof

of a 50.0m high building,1. Find the time the stone reaches at maximum height (v=0)2. Find the maximum height3. Find the time the stone reaches its original height4. Find the velocity of the stone when it reaches its original height5. Find the velocity and position of the stone at t=5.00s

st

ttavv yyif

04.280.9

0.20

00.080.90.20

??

????? ?

g=-9.80m/s2

1

)(4.704.200.50

)04.2()80.9(21

04.2200.50

21

2

2

m

tatvyy yyiif

???

???????

???2

st 08.4204.2 ???3 Other ways?

)/(0.2008.4)80.9(0.20 smtavv yyiyf ????????4

)/(0.2900.5)80.9(0.20

sm

tavv yyiyf

??????

??

)(5.27)00.5()80.9(21

00.50.200.50

21

2

2

m

tatvyy yyiif

?????????

??? 5-Position5-Velocity

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