1443-501 spring 2002 lecture #6 dr. jaehoon yu 1.motion with resistive forces 2.resistive force in...

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!!

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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.


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


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


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


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…


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


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

http://www.digitaldutch.com/unitconverter/


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


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


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!!


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


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


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


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


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


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?


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.


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


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!!


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


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


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


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


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?


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.

1443-501 spring 2002 lecture #6 dr. jaehoon yu 1.motion with resistive forces 2.resistive force in...

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