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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.!!
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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!!)
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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?
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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?
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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)
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
11
Instantaneous Velocity
Time
Average Velocity
Instantaneous Velocity
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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!!
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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
Wed., Jan. 16, 2002 PHYS 1443-501, Spring 2002Dr. Jaehoon Yu
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