1443-501 spring 2002 lecture #6 dr. jaehoon yu 1.motion with resistive forces 2.resistive force in...
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1443-501 Spring 2002Lecture #6
Dr. Jaehoon Yu1. Motion with Resistive Forces2. Resistive Force in Air at High Speed 3. Analytical and Numerical Methods4. Review Examples
1st term exam on Monday Feb. 11, 2002, at 5:30pm, in the classroom!! Will cover chapters 1-6!!Will cover chapters 1-6!!
Physics Clinic Hours Extened!!Physics Clinic Hours Extened!!Mon. – Thu. till 6pmMon. – Thu. till 6pm
Thursday: 6-9pm by appointments Thursday: 6-9pm by appointments ((e-mail to Andre Litvin andrey_litvin@yahoo.com
Mustafa symmetry80@hotmail.com ) )
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
2
Resistive Force Proportional to Speed Since the resistive force is proportional to speed, we can write R=bv
m
Let’s consider that a ball of mass m is falling through a liquid.
R
mgv
vm
bg
dt
dvdt
dvmmabvmgFF
RFF
yx
g
;0This equation also tells you that
0 when , vgvm
bg
dt
dv
The above equation also tells us that as time goes on the speed increases and the acceleration decreases, eventually reaching 0. An object moving in a viscous medium will obtain speed to a certain speed (terminal speed) and then maintain the same speed without any more acceleration.
What does this mean?
What is the terminal speed in above case?
b
mgvv
m
bg
dt
dvt ;0
How do the speed and acceleration depend on time?
vm
bge
m
b
b
mge
m
b
b
mg
dt
dv
tgageem
b
b
mg
dt
dva
tveb
mgv
tt
tm
bt
mbt
11
;0 when ;
;0 when 0 ;1 The time needed to reach 63.2% of the terminal speed is defined as the time constant, m/b.
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
3
Example 6.11A small ball of mass 2.00g is released from rest in a large vessel filled with oil, where it experiences a resistive force proportional to its speed. The ball reaches a terminal speed of 5.00 cm/s. Determine the time constant and the time it takes the ball to reach 90% of its terminal speed.
mR
mgv
sskg
kg
b
m
skgsm
smkg
v
mgb
b
mgv
t
t
33
2
23
1010.5/392.0
1000.2
/392.0/1000.5
/80.91000.2
Determine the time constant .
Determine the time it takes the ball to reach 90% of its terminal speed.
mst
ee
evv
eveb
mgv
tt
t
tt
t
t
t
7.111010.530.230.21.0ln
1.0 ;9.01
19.0
11
3
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
4
Air Drag For objects moving through air at high speed, the resistive force is roughly proportional to the square of the speed.The magnitude of such resistive force
2
2
1AvDR
D: Drag Coefficient (dim.less): Density of Air A: Cross section of the object
Let’s analyze a falling object through the air
m
R
mgv
mgAvDmaF yy 2
2
1
gvm
ADay
2
2
AD
mgvgv
m
ADa ty
2 ;0
22
Force
Acceleration
Terminal Speed
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
5
Example 6.4A pitcher hurls a 0.145 kg baseball past a batter at 40.2m/s (=90 mi/h). Determine the drag coefficients of the air, density of air is 1.29kg/m3, the radius of the ball is 3.70cm, and the terminal velocity of the ball in the air is 43.0 m/s.
Find the resistive force acting on the ball at this speed.
277.0
0.431030.429.1
80.9145.022
1030.40370.0
232
2322
tAv
mgD
mrA
Using the formula for terminal speed
NAvDR 24.12.401030.429.1277.02
1
2
1 232
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
6
Analytical Method vs Numerical Method
Analytical Method:
The method of solving problems using cause and effect relationship and mathematical expressions
When solving for a motion of an object we follow the procedure:1. Find all the forces involved in the motion2. Compute the net force3. Compute the acceleration4. Integrate the acceleration in time to obtain velocity5. Integrate the velocity in time to obtain position
But not all problems are analytically solvable due to complex conditions applied to the given motion, e.g. position dependent acceleration, etc.
Numerical Method:
The method of solving problems using approximation and computational tools, such as computer programs, stepping through infinitesimal intervals Integration by numbers…
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
7
The Euler MethodSimplest Numerical Method for solving differential equations, by approximating the derivatives as ratios of finite differences.Speed and the magnitude of Acceleration can be approximated in a small increment of time, t
t
tvttv
t
vta
ttatvttv Thus the Speed at time t+t is
In the same manner, the position of an object and speed are approximated in a small time increment, t
t
txttx
t
xtv
And the position at time t+t is
02
1
0
2
2
ttvtxttx
ttattvtxttx
t
The (t)2 is ignored in Euler method because it is closed to 0.
We then use the approximated expressions to compute position at infinitesimal time interval, t, with computer to find the position of an object at any given time
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
8
Unit Conversion: Example 1.4• US and UK still use British Engineering units: foot, lbs, and
seconds– 1.0 in= 2.54 cm, 1ft=0.3048m=30.48cm– 1m=39.37in=3.281ft~1yd, 1mi=1609m=1.609km– 1lb=0.4535kg=453.5g, 1oz=28.35g=0.02835kg– Online unit converter: http://www.digitaldutch.com/unitconverter/
• Example 1.4: Determine density in basic SI units (m,kg )
M=856g
L=5.35cmD=
5.35
cm
H=
5.3
5cm
3336
363
33
33
/1059.51013.153
856.0
856.0/1000
856856
1013.153)/100(
13.15313.153
)35.5()35.5()35.5()35.5(
mkgm
kg
V
M
kgkgg
ggM
mmcm
cmcm
cmcmcmcmHDLV
V
M
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
9
Example 2.1• Find the displacement,
average velocity, and average speed.
)(833053 mxxx if • Displacement:
• Average Velocity:
)/(7.150
83sm
t
x
tt
xxv
i
ix
f
f
• Average Speed: Time
xi=+30mti=0sec xf=-53m
tf=50sec
xm=+52m
)/(5.250
127
50
535222Interval Time Total
Traveled Distance Total
sm
v
Fig02-01.ip
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
10
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
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
11
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!!
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
12
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
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
13
Example 3.1 Cartesian Coordinate of a point in the xy plane are (x,y)= (-3.50,-2.50)m. Find the polar coordinates of this point.
y
x
(-3.50,-2.50)m
rs
)(30.45.18
50.250.322
21
21
m
yxr
2165.35180180
5.357
5tan
7
5
50.3
50.2tan
180
1
s
s
s
s
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
14
2-dim Motion Under Constant Acceleration
• Position vectors in xy plane: jyixr iii jyixr fff
• Velocity vectors in xy plane: jvivv yixii jvivv yfxff
tavjtavitavv
tavvtavv
iyyixxif
yyiyfxxixf
,
• How are the position vectors written in acceleration vectors?
2
22
22
2
1
2
1
2
12
1,
2
1
tatvr
jtatvyitatvxr
tatvyytatvxx
i
yyiixxiif
yyiifxxiif
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
15
Example 4.1A particle starts at origin when t=0 with an initial velocity vv=(20ii-15jj)m/s. The particle moves in the xy plane with ax=4.0m/s2. Determine the components of velocity vector at any time, t.
smjittv
smtavv
smttavv
yyiyf
xxixf
/150.420)(
/15
/0.420
Compute the velocity and speed of the particle at t=5.0 s.
smjismjiv /1540/150.50.420 sm
vvvspeed yx
/431540 22
22
218
3tan
40
15tan 11
Determine the x and y components of the particle at t=5.0 s.
mjijyixr
mtvymtatvx
fff
yifxxif
75150
75515 ,150542
1520
2
1 22
Ex04-01.ip
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
16
Example 4.5Ex04-05a.ip
• A stone was thrown upward from the top of a building at an angle of 30o to horizontal with initial speed of 20.0m/s. If the height of the building is 45.0m, how long is it before the stone hits the ground?
st
stst
t
tttgt
gttvy
smvv
smvv
yif
iiyi
ixi
22.4
22.4or 18.280.92
)90(80.94200.20
00.900.2080.90.900.20
2
10.45
/0.1030sin0.20sin
/3.1730cos0.20cos
2
22
2
• What is the speed of the stone just before it hits the ground?
smvvv
smgtvgtvv
smvvv
yfxf
iiyiyf
ixixf
/9.354.313.17
/4.3122.480.90.10sin
/3.1730cos0.20cos
2222
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
17
Uniform Circular Motion• A motion with a constant speed on a circular path.
– The velocity of the object changes, because the direction changes
– Therefore, there is an acceleration
vi
vj
vr
vivj
rr2
r1
The acceleration pulls the object inward: Centripetal Acceleration
t
v
tt
vva
if
if
Average Acceleration r
r
t
va
r
rvv
r
r
v
v
, ,
Angle is
r
v
r
vv
r
v
t
raa
ttr
2
00limlim
Instantaneous Acceleration
Is this correct in dimension?
What story is this expression telling you?
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
18
Relative Velocity and AccelerationThe velocity and acceleration in two different frames of references can be denoted:
O
Frame S
r’
O’
Frame S’v0
v0t
r0
0
0
'
'
'
vvv
vdt
rd
dt
rd
tvrr
Galilean transformation equation
constant is when ,'
'
'
0
0
0
vaa
dt
vd
dt
vd
dt
vd
tvrr
What does this tell you?The accelerations measured in two frames are the same when the frames move at a constant velocity with respect to each other!!!
The earth’s gravitational acceleration is the same in a frame moving at a constant velocity wrt the earth.
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
19
Example 4.9A boat heading due north with a speed 10.0km/h is crossing the river whose stream has a uniform speed of 5.00km/h due east. Determine the velocity of the boat seen by the observer on the bank.
N
EvvBRBR
vvRR
6.260.10
00.5tantan
0.1000.5
00.5 and 0.10
/2.1100.50.10
11
2222
BBx
BBy
BB
BR
RBRBB
BRBB
v
v
jiv
ivjv
hkmvvv
vvv
R
R
How long would it take for the boat to cross the river if the width is 3.0km? min1830.0
6.26cos2.11
0.3
cos
0.3
cos
hrsv
t
kmtv
BB
BB
vBB
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
20
ForceWe’ve been learning kinematics; describing motion without understanding what the cause of the motion was. Now we are going to learn dynamics!!
Can someone tell me what FORCE is?
FORCEs are what cause an object to move
FORCEs are what cause any change in the velocity of an object!!
The above statement is not entirely correct. Why?Because when an object is moving with a constant velocity no force is exerted on the object!!!
What does this statement mean?
When there is force, there is change of velocity. Forces cause acceleration.
What happens there are several forces being exerted on an object?
Forces are vector quantities, so vector sum of all forces, the NET FORCE, determines the motion of the object.F1 F2 NET FORCE,
F= F1+F2
When net force on an objectis 0, the has constant velocity and is at its equilibrium!!
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
21
Newton’s LawsIn the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity.
The acceleration of an object is directly proportional to the net force exerted on it and inversely proportional to the object’s mass.
amFi
i
If two objects interact, the force, F12, exerted on object 1 by object 2 is equal magnitude to and opposite direction to the force, F21, exerted on object 1 by object 2.
1221 FF
1st Law: Law of Inertia
2nd Law: Law of Forces
3rd Law: Law of Action and Reaction
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
22
F
F
Example 5.1Determine the magnitude and direction of acceleration of the puck whose mass is 0.30kg and is being pulled by two forces, F1 and F2, as shown in the picture, whose magnitudes of the forces are 8.0 N and 5.0 N, respectively.
NFF
NFF
y
x
9.660sin0.8sin
0.460cos0.8cos
11
11
NFF
NFF
y
x
7.120sin0.5sin
7.420cos0.5cos
222
222
yyyy
xxxx
maNFFF
maNFFF
2.57.19.6
7.87.40.4
21
21
Components of F1
Components of F2
Components of total force F
2
11222
22
/1729
3029
17tantan ,/341729
/173.0
2.5 ,/29
3.0
7.8
smjijaiaa
a
asma
smm
Fasm
m
Fa
yx
x
y
yy
xx
Magnitude and direction of acceleration a
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
23
Example 5.4A traffic light weighing 125 N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37.0o and 53.0o with the horizontal. Find the tension in the three cables.
NTTNT
NTT
TTT
TT
mgTT
TFTFTTTFi
iiyy
i
iixx
4.75754.0 ;100
12525.137sin754.053sin
754.037cos
53cos
053cos37cos
053sin37sin
0 ;0 ;
212
22
221
21
21
3
1
3
1321
Free-bodyDiagram
T1 T2
T3
53o37o
53o37o
x
y
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
24
Example 5.12Suppose a block is placed on a rough surface inclined relative to the horizontal. The inclination angle is increased till the block starts to move. Show that by measuring this critical angle, c, one can determine coefficient of static friction, s.
cc
ccs
csssgxx
cgygyyy
sg
Mg
Mg
n
Mg
MgnffFF
MgFnFnMaF
fnFF
tancos
sinsin
sin ;0
cos ;0
Free-bodyDiagram
x
yM a
Fg
nn
F=Ma
F= -Mg
fs=kn
Ex05-12.ip
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
25
Example 6.8A ball of mass m is attached to the end of a cord of length R. The ball is moving in a vertical circle. Determine the tension of the cord at any instant when the speed of the ball is v and the cord makes an angle with vertical.
Tm
What are the forces involved in this motion?
cos
cos
sin
sin
2
2
gR
vmT
R
vmmamgTF
ga
mgmaF
rr
t
tt
The gravitational force Fg
and the radial force, T, providing tension.
R Fg=mg
At what angles the tension becomes maximum and minimum. What are the tension?
Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #6
26
Non-Inertial Frame
Example 6.9A ball of mass m is is hung by a cord to the ceiling of a boxcar that is moving with an acceleration a. What do the inertial observer at rest and the non-inertial observer traveling inside the car conclude? How do they differ?
m
This is how the ball looks like no matter which frame you are in.
tan ;cos
0cos
sin
gamg
T
mgTF
TmamaF
TFF
c
y
cxx
g
Inertial Frame
How do the free-body diagrams look for two frames?
Fg=mgm
T
Fg=mgm
T
Ffic
a
How do the motions interpreted in these two frames? Any differences?
For an inertial frame observer, the forces being exerted on the ball are only T and Fg. The acceleration of the ball is the same as that of the box car and is provided by the x component of the tension force.
tan ;cos
0cos
sin ;0sin
gamg
T
mgTF
TmaFFTF
FTFF
fic
y
ficficficx
ficg
In the non-inertial frame observer, the forces being exerted on the ball are T, Fg, and Ffic.
For some reason the ball is under a force, Ffic, that provides acceleration to the ball.While the mathematical expression of the acceleration of the ball is identical to that of inertial frame observer’s, the cause of the force is dramatically different.
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