1443-501 Spring 2002
Lecture #6
1.
2.
3.
4.
Dr. Jaehoon Yu
Motion with Resistive Forces
Resistive Force in Air at High Speed
Analytical and Numerical Methods
Review Examples
Physics Clinic Hours Extened!!
Mon. – Thu. till 6pm
Thursday: 6-9pm by appointments
(e-mail to Andre Litvin andrey_litvin@yahoo.com
Mustafa symmetry80@hotmail.com )
1st term exam on Monday Feb. 11, 2002, at 5:30pm, in the classroom!!
Will cover chapters 1-6!!
Resistive Force Proportional to Speed
Since the resistive force is proportional to speed, we can write R=bv
R
v m
mg
Let’s consider that a ball of mass m is falling through a liquid.
F  F
g
R
dv
 Fx  0;  Fy  mg  bv  ma  m dt
dv
b
g v
dt
m
This equation also tells you that
dv
b
 g  v  g , when v  0
dt
m
The above equation also tells us that as time goes on the speed
increases and the acceleration decreases, eventually reaching 0.
What does this mean?
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 is the
terminal speed
in above case?
dv
b
mg
 g  v  0; vt 
dt
m
b
Feb. 6, 2002
How do the speed
and acceleration
depend on time?
mg 
bt
1  e m ; v  0 when t  0;

b 
dv mg b bt m
t
a

e
 ge  ; a  g when t  0;
dt
b m
dv mg b  t 
mg b 
b
t

e

1  1  e    g  v

dt
b m
b m
m
v
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
The time needed to
reach 63.2% of the
terminal speed is
defined as the time
constant, m/b.
2
Example 6.11
A 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.
mg
b
mg 2.00 10 3 kg  9.80m / s 2
b 

 0.392kg / s
vt
5.00 10  2 m / s
vt 
R
v
m
Determine the
time constant .
mg
Determine the time it takes
the ball to reach 90% of its
terminal speed.
Feb. 6, 2002
m 2.00 10 3 kg
 
 5.10 10 3 s
b
0.392kg / s
mg 
t 
t 


v
1  e
  vt 1  e  



b 
t
0.9vt  vt 1  e  


1  e  t    0.9; e  t   0.1


t    ln 0.1  2.30  2.30  5.10  10 3  11.7ms 
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
3
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
R
v
m
1
2
R  DAv
2
Let’s analyze a falling object through the air
Force
mg
Acceleration
Terminal Speed
Feb. 6, 2002
D: Drag Coefficient (dim.less)
: Density of Air
A: Cross section of the object
1
2
F

ma

D

Av
 mg
 y
y
2
 DA  2
ay  
v  g
 2m 
 DA  2
ay  
v  g  0; vt 
 2m 
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
2mg
DA
4
Example 6.4
A 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.
Using the
formula for
terminal speed
 
A  r 2   0.0370  4.30 10 3 m 2
2mg
2  0.145  9.80
D

 0.277
2
2
3
Avt 1.29  4.30 10  43.0
2
Find the resistive force acting on the ball at this speed.
1
1
2
2
3
R  DAv   0.277 1.29  4.30 10  40.2  1.24 N
2
2
Feb. 6, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
5
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.
2.
3.
4.
5.
Find all the forces involved in the motion
Compute the net force
Compute the acceleration
Integrate the acceleration in time to obtain velocity
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:
Feb. 6, 2002
The method of solving problems using approximation and
computational tools, such as computer programs, stepping
through infinitesimal intervals  Integration by numbers…
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
6
The Euler Method
Simplest 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, Dt
Thus the Speed at time t+Dt is
In the same manner, the position of
an object and speed are approximated
in a small time increment, Dt
And the position at time t+Dt is
at  
Dv vt  Dt   vt 

Dt
Dt
vt  Dt   vt   at Dt
Dx xt  Dt   xt 
vt  

Dt
Dt
1
2
xt  Dt   xt   vt Dt  a t Dt 
2

 xt  Dt   xt   vt Dt  0
Dt 2  0
We then use the approximated expressions to
compute position at infinitesimal time
interval, Dt, with computer to find the
position of an object at any given time 1443-501 Spring 2002
Feb. 6, 2002
Dr. J. Yu, Lecture #6
The (Dt)2 is ignored in Euler
method because it is closed to 0.
7
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
V
V  L  D  H  (5.35cm)  (5.35cm)  (5.35cm)  (5.35) 3 cm 3
H=5.35cm

153.13cm 3
 153.13cm 
 153.13  10 6 m 3
3
(100cm / m)
856 g
M  856 g 
 0.856kg
1000 g / kg
M
0.856kg


 5.59  103 kg / m 3
6
3
V
153.13  10 m
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
3
M=856g
L=5.35cm
Feb. 6, 2002
8
Example 2.1
• Find the displacement,
average velocity, and
average speed.
• Displacement:
xm=+52m
Dx  xf  xi  53  30  83(m)
• Average Velocity:
xi=+30m
ti=0sec
xf=-53m
tf=50sec
Time
vx 
xf  xi Dx  83


 1.7(m / s)
tf  ti Dt
50
• Average Speed:
Total Distance Traveled

Total Time Interval
22  52  53 127

 2.5(m / s )
50
50
v
Feb. 6, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
9
Example 2.2
• Particle is moving along x-axis following the expression: x  4t  2t 2
• Determine the displacement in the time intervals t=0 to t=1s and
t=1 to t=3s:
For interval xt 0  0, xt 1  4  (1)  2  (1) 2  2
t=0 to t=1s
Dxt 0,1  xt 1  xt 0  2  0  2(m)
2
For interval xt 1  2, xt 3  4  (3)  2  (3)  6
t=1 to t=3s Dxt 1,3  xt 3  xt 1  6  2  8(m)
• Compute the average velocity in the time intervals t=0 to t=1s
and t=1 to t=3s: v  Dx   2 (m / s) v  Dx  8  4(m / s)
x
t 0,1
Dt
1
x
t 1,3
Dt
2
• Compute the instantaneous velocity at t=2.5s:
Instantaneous velocity at any time t
Dx dx d
vxt   lim

  4t  2t 2   4  4t
dt dt
Δt 0 Dt
Feb. 6, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
Instantaneous velocity at t=2.5s
vxt  2.5  4  4  (2.5)  6(m / s)
10
Kinetic Equation of Motion in a Straight
Line Under Constant Acceleration
Velocity as a function of time
v t   vxi  axt
xf
1
1
xf  xi  vxt  vxf  vxi t Displacement as a function
of velocity and time
2
2
1 2
xf  xi  vxit  axt
2
vxf  vxi  2axxf  xi 
2
2
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 2002
Dr. J. Yu, Lecture #6
11
Example 2.12
Stone was thrown straight upward at t=0 with +20.0m/s initial velocity on the roof
of a 50.0m high building,
g=-9.80m/s2
1.
Find the time the stone reaches at maximum height (v=0)
2.
3.
4.
5.
Find the maximum height
Find the time the stone reaches its original height
Find the velocity of the stone when it reaches its original height
Find the velocity and position of the stone at t=5.00s
1 v f  vyi  ayt  20.0  9.80t  0.00
20.0
t
 2.04s
9.80
3
2 yf  yi  vyit  1 ayt 2
2
t  2.04  2  4.08s Other ways?
1
 50.0  20  2.04   (9.80)  (2.04) 2
2
 50.0  20.4  70.4(m)
4 vyf  vyi  ayt  20.0  (9.80)  4.08  20.0(m / s )
5Velocity
vyf  vyi  ayt
 20.0  (9.80)  5.00
 29.0(m / s )
Feb. 6, 2002
yf  yi  vyit 
1
ayt 2
2
 50.0  20.0  5.00 
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
5-Position
1
 ( 9.80)  (5.00) 2  27.5( m)
2
12
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.
r
y
x
2
1
 y1 2  
 3.50   2.50 
2
q
qs
(-3.50,-2.50)m
Feb. 6, 2002
 18.5  4.30(m)
x
r
2
q  180  q s
 2.50 5

 3.50 7
5
qs  tan 1    35.5
7
q  180  q s  180  35.5  216
tan qs 
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
13
2-dim Motion Under Constant Acceleration
• Position vectors in xy plane:
ri  xi i  yi j
rf  x f i  y f j
• Velocity vectors in xy plane:
vi  v xi i  v yi j
v f  v xf i  v yf j
v xf  v xi  a x t , v yf  v yi  a y t
v f  v xi  a x t i  v yi  a y t  j  vi  at
• How are the position vectors written in acceleration vectors?
1
1
a x t 2 , y f  yi  v yi t  a y t 2
2
2
1
1

 

r f   xi  v xi t  a x t 2 i   yi  v yi t  a y t 2  j
2
2

 

1
 ri  vt  at 2
2
x f  xi  v xi t 
Feb. 6, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
14
Example 4.1
A particle starts at origin when t=0 with an initial velocity v=(20i-15j)m/s.
The particle moves in the xy plane with ax=4.0m/s2. Determine the
components of velocity vector at any time, t.
v xf  v xi  a x t  20  4.0t m / s 
v yf  v yi  a y t  15m / s 
v (t )  20  4.0t i  15 j m / s
Compute the velocity and speed of the particle at t=5.0 s.




v  20  4.0  5.0i 15 j m / s  40i 15 j m / s
  15 
1   3 

  tan 
  21
 40 
 8 
q  tan 1 
speed  v 

vx 2  v y 2
402   152
 43m / s
Determine the x and y components of the particle at t=5.0 s.
x f  v xi t 
1
1
a x t 2  20  5   4  52  150m, y f  v yi t  15  5  75m
2
2


rf  x f i  y f j  150i  75 j m
Feb. 6, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
15
Example 4.5
•
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?
v xi  vi cos q  20.0  cos 30   17.3m / s
v yi  vi sin q i  20.0  sin 30   10.0m / s
1
gt 2
2
 20.0t  90.0  9.80t 2  20.0t  90.0  0
y f  45.0  v yi t 
gt 2
 20 2
 4  9.80  ( 90)
2  9.80
t  2.18 s or t  4.22 s
 t  4.22 s
t 
•
20.0 
What is the speed of the stone just before it hits the ground?
v xf  v xi  vi cos q  20.0  cos 30  17.3m / s
v yf  v yi  gt  vi sin q i  gt  10.0  9.80  4.22  31.4m / s
v 
v xf
Feb. 6, 2002
2
 v yf
2

17.32   31.4 
2
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
 35.9m / s
16
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
r
q
r1
r2 Dq Dr
vi
vj
vj
Dv
vi
Angle
is Dq
The acceleration pulls the object inward: Centripetal Acceleration
Average
Acceleration
Instantaneous
Acceleration
a
vf  vi
Dv

t f  ti
Dt
q 
Dv
v

Dr
v Dr
, Dv  v
, a
r
r
Dt r
Dr v
v
v2
ar  lim a  lim
 v 
Dt  0
Dt  0 Dt r
r
r
Dr
Is this correct in
dimension?
What story is this expression telling you?
Feb. 6, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
17
Relative Velocity and Acceleration
The velocity and acceleration in two different frames of
references can be denoted:
Frame S
v0
Frame S’
r’
r
O
v0t
O’
r '  r  v0 t
d v'
d v d v0


dt
dt
dt
a '  a, when v0 is constant
Feb. 6, 2002
Galilean
transformation
equation
r '  r  v0 t
d r' d r

 v0
dt
dt
v'  v  v0
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.
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
18
Example 4.9
A 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.
vBB  vBR  vR
N
vBB 
vR
vBR
q
vBR
 vR
2

10.02  5.002

 11.2km / h

 vBR  10.0 j and vR  5.00 i


vBB  5.00 i  10.0 j
vBB
E
 vBBy 
5.00 

  tan 1 
  26.6
 10.0 
 vBBx 
q  tan 1 
How long would it take for
the boat to cross the river if
the width is 3.0km?
Feb. 6, 2002
2
vBB cos q  t  km
t
3.0
3.0

 0.30hrs  18 min

vBB cos q 11.2  cos 26.6
1443-501 Spring 2002
Dr. J. Yu, Lecture #6


19
Force
We’ve been learning kinematics; describing motion without understanding
what the cause of the motion was. Now we are going to learn dynamics!!
FORCEs are what cause an object to move
Can someone tell me
The above statement is not entirely correct. Why?
what FORCE is?
Because when an object is moving with a constant velocity
no force is exerted on the object!!!
FORCEs are what cause any change in the velocity of an object!!
What does this statement mean?
When there is force, there is change of velocity.
Forces cause acceleration.
Forces are vector quantities, so vector sum of all
What happens there are several
forces being exerted on an object? forces, the NET FORCE, determines the motion of
the object.
F1
Feb. 6, 2002
F2
NET FORCE,
F= F1+F2
When net force on an objectis 0, the has
constant velocity and is at its equilibrium!!
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
20
Newton’s Laws
1st Law:
Law of Inertia
2nd Law:
Law of Forces
 F  ma
i
i
3rd Law:
Law of Action
and Reaction
F21  F12
Feb. 6, 2002
In 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.
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.
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
21
Example 5.1
Determine 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.

Components F1x  F1 cos q1  8.0  cos60   4.0 N
of F1
F1 y  F1 sin q1  8.0  sin 60   6.9 N
F1
q16o
q22o
F2
Components F2 x  F2 cos q 2  5.0  cos 20   4.7 N
of F2
F2 y  F2 sin q 2  5.0  sin  20   1.7 N
Components of
Fx  F1x  F2 x  4.0  4.7  8.7 N  max
total force F
Magnitude and
direction of
acceleration a
Feb. 6, 2002
ax 
a 
Fy  F1 y  F2 y  6.9  1.7  5.2 N  ma y
Fx
8.7

 29m / s 2 ,
m
0.3
29 2
 17 
2
Fy
5.2
 17 m / s 2
m
0.3
 ay 
1  17 

 34m / s 2 , q  tan 1 
  30
a 
  tan 
 29 
 x 
ay 







a  a x i  a y j   29 i  17 j m / s 2


1443-501 Spring 2002
Dr. J. Yu, Lecture #6
22
Example 5.4
A 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.
37o
y
53o
T1
T2
37o
Free-body
Diagram
53o
T3
F  T1  T2  T3 ;
T1 sin 37
x
Fx 
i 3
 Tix  0;
i 1
Fy 
  T sin 53   mg  0
 T cos 37   T cos 53   0
cos53 
T 
T  0.754T
cos 37 
T sin 53   0.754  sin 37   1.25T

i 3
T
i 1
iy
0

2


1
2

1

2

2

2
2
 125 N
T2  100 N ; T1  0.754T2  75.4 N
Feb. 6, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
23
Example 5.12
Suppose 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, qc, one can determine coefficient of static friction, ms.
y
n
n
q
fs=mkn
q
Fg
Free-body
Diagram
F=Ma
x
F= -Mg
F  Fg  n  f s
Fy  Ma y  n  Fgy  0; n   Fgy  Mg cos q c
Fx  Fgx  f s  0; f s  m s n  Mg sin q c
ms 
Feb. 6, 2002
Mg sin q c Mg sin q c

 tan q c
n
Mg cos q c
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
24
Example 6.8
A 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 q with vertical.
What are the forces involved in this motion?
q
T
R
m
Fg=mg
The gravitational force Fg and the
radial force, T, providing tension.
F
t
 mat  mg sin q
at  g sin q
v2
 Fr  T  mg cosq  mar  m R
 v2


T  m

g
cos
q
 R



At what angles the tension becomes maximum and minimum. What are the tension?
Feb. 6, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
25
Example 6.9
A 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
Inertial
Frame
m
Fg=mg
Non-Inertial
Frame
T q
Ffic m
Fg=mg
Feb. 6, 2002
a
q
T q
This is how the ball looks like no matter which frame you are in.
How do the free-body diagrams look for two frames?
How do the motions interpreted in these two frames? Any differences?
F
F
F
T 
 Fg  T
x
 max  mac  T sin q
y
 T cos q  mg  0
mg
; ac  g tan q
cos q
 F  F T  F
 F  T sin q  F  0; F
 F  T cosq  mg  0
g
x
fic
fic
fic
 ma fic  T sin q
y
T
mg
; a fic  g tan q
cosq
1443-501 Spring 2002
Dr. J. Yu, Lecture #6
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.
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
26
force is dramatically different.