wed., jan. 16, 2002phys 1443-501, spring 2002 dr. jaehoon yu 1 phys 1443 – section 501 lecture #2...

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.!!

http://www.economist.com/surveys/displaystory.cfm?story_id=922278


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

3

4rVsphere

3

3

4rVM spheresphere r1

r23

13

4rVM innerinner

323

4rVM outerouter

)(3

4 31

32 rr

MMM inneroutershell

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

22

;02 ,1

222

mn

mnm

tltlttt

ll nmmnmn

m


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

cm

g

V

Μ

• 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

m

A

VThickness 4

2

33

1051.10.25

1078.3

• 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 motion

t

x

tt

xxv

i

ix

f

f

Average speed is defined as:

Interval Time Total

Traveled Distance Totalv

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: )/(020

0sm

t

x

tt

xxv

i

ix

f

f

Average Speed: )/(320

60

Interval Time Total

Traveled Distance Totalsmv


8

Example 2.1• Find the displacement,

average velocity, and average speed.

Time

xi=+30mti=0sec xf=-53m

tf=50sec

)(833053 mxxx if • Displacement:

• Average Velocity:

)/(7.150

83sm

t

x

tt

xxv

i

ix

f

f

• Average Speed:

xm=+52m

)/(5.250

127

50

535222Interval Time Total

Traveled Distance Total

sm

v

Fig02-01.ip


9

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:

dt

dx

t

xvx

lim0Δt

dt

dx

t

xvx

lim0Δt

*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: )/(

1

21,0 smt

xv tx

)/(42

83,1 smt

xv tx

• Compute the instantaneous velocity at t=2.5s:

tttdt

d

dt

dx

t

xtvx 4424 2

lim0Δt

)/(6)5.2(445.2 smtvx

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


13

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

t

v

tt

vva

x

f

xf

i

xix

t

x

tt

xxv

i

ix

f

f

analogs to

2

2

dt

xd

dt

dx

dt

d

dt

dv

t

va

xxx

lim0Δt dt

dx

t

xvx

lim0Δt

analogs to


14

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

02

4020

)/(202540)0.2(

)/(40)0(

2

2

smt

v

tt

vva

smtv

smtv

x

if

xixfx

fxf

ixi

ttdt

d

dt

dvta

xx 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)

t

vv

tt

vva

xi

i

xix

xf

f

xf

If tf=t and ti=0 tavv xxixf

For constant acceleration, simple numeric average

t

xx

tt

xxv

i

i

ix

f

f

f

If tf=t and ti=0

2

2

1tatvxtvxx xxiixif

Resulting Equation of Motion becomes

tavtavvv

v xxixxixfxi

x

2

1

2

2

2

tvxx xif


17

Kinetic Equation of Motion in a Straight Line Under Constant Acceleration tavtv xxixf

2

2

1tatvxx xxiif

tvvtvxx xixfxif 2

1

2

1

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 00.3

2

1

2

1TaTx Police

f )1(0.451 TTvx Carf

possible)(not 00.1or (possible) 0.31

;00.300.3000.1

);1(0.4500.32

1;

2

2

TT

TT

TTxx Policef

Policef

a

acbbx

cbxax

2

4

are 0for Solutions

2

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(2

104.2200.50

2

1

2

2

m

tatvyy yyiif

2

st 08.4204.2 3 Other ways?

)/(0.2008.4)80.9(0.20 smtavv yyiyf 4

)/(0.29

00.5)80.9(0.20

sm

tavv yyiyf

)(5.27)00.5()80.9(2

100.50.200.50

2

1

2

2

m

tatvyy yyiif

5-Position5-Velocity

wed., jan. 16, 2002phys 1443-501, spring 2002 dr. jaehoon yu 1 phys 1443 – section 501 lecture #2...

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