PHYS 1443 – Section 001 Lecture #3

PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu

1Wednesday, June 8, 2011

PHYS 1443 – Section 001Lecture #3

Wednesday, June 8, 2011Dr. Jaehoon Yu

• One Dimensional Motion: Free Fall• Coordinate System • Vectors and Scalars and their operations• Motion in Two Dimensions: Motion under constant

acceleration; Projectile Motion; Maximum ranges and heights

• Newton’s Laws of Motion

Today’s homework is homework #2, due 10pm, Monday, June 13!!

Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr. Jaehoon Yu

2

Announcements• Homework registration and submissions

– 27/28 registered but only 9 completed the submission– The due for homework #1, the freebee, is 10pm tonight!

• Quiz #2 coming Tuesday, June 14– Covers: CH 1.1 – what we finish on Monday, June 13

• Reading assignment– CH3.9


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Reminder: Special Problems for Extra Credit• Derive the quadratic equation for Bx2-Cx+A=0

5 points• Derive the kinematic equation

from first principles and the known kinematic equations 10 points

• You must show your work in detail to obtain full credit

• Due at the start of the class, Thursday, June 9

ifxf xxavv xxi 222

Wednesday, June 8, 2011 4

Special Project for Extra Credit• Show that the trajectory of a projectile motion

is a parabola!!–20 points–Due: next Tuesday, June 14–You MUST show full details of your OWN

computations to obtain any credit• Much beyond what was covered in page 40 of

this lecture note!!



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Free Fall• Free fall motion is a motion under the influence of the

gravitational pull (gravity) only; Which direction is a freely falling object moving?– A motion under constant acceleration – All kinematic formula we learned can be used to solve for falling

motions. • Gravitational acceleration is inversely proportional to the

distance between the object and the center of the earth• The magnitude of the gravitational acceleration is g=9.80m/s2

on the surface of the earth, most of the time.• The direction of gravitational acceleration is ALWAYS 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 when +y points upward

Yes, down to the center of the earth!!


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Example for Using 1D Kinematic Equations on a Falling object (similar to Ex. 2.16)

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

fv

fy

(a) Find the time the stone reaches at the maximum height.

What is the acceleration in this motion?

What is so special about the maximum height? V=0

t

(b) Find the maximum height.

vyi ayt 20.0 9.80t 0.00 /m s Solve for t20.09.80

2.04s

2150.0 20 2.04 ( 9.80) (2.04)2

50.0 20.4 70.4( )m yi vyit 1

2ayt2


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Example of a Falling Object cnt’d

t

vyf

yf Position

vyf Velocity

(c) Find the time the stone reaches back to its original height.

(d) Find the velocity of the stone when it reaches its original height.

(e) Find the velocity and position of the stone at t=5.00s.

2.04 2 4.08s

vyi ayt20.0 ( 9.80) 4.08 20.0( / )m s

vyi ayt20.0 ( 9.80) 5.00 29.0( / )m s

2150.0 20.0 5.00 ( 9.80) (5.00)2

yi vyit 1

2ayt2

27.5( )m


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2D Coordinate Systems• They make it easy and consistent to express locations or positions• Two commonly used systems, depending on convenience, are

– Cartesian (Rectangular) Coordinate System• Coordinates are expressed in (x,y)

– Polar Coordinate System • Coordinates are expressed in distance from the origin ® and the angle measured

from the x-axis, θ (r,θ)• Vectors become a lot easier to express and compute

O (0,0)

(x1,y1)r1

θ1

1x

How are Cartesian and Polar coordinates related?

1r

y1

x1

+x

+y

1y tanθ1

x1

2 y12

1y r1

r1

= (r1,θ1)

1x

θ1 tan−1 y1

x1

⎛⎝⎜

⎞⎠⎟

cosθ1

sinθ1


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Example Cartesian Coordinate of a point in the xy plane are (x,y)= (-3.50,-2.50)m. Find the equivalent polar coordinates of this point.

y

x

(-3.50,-2.50)mr

θ

θs

r

θ 180 +θ s

tanθs

sθ

∴θ

−3.50

2

−2.50 2

18.5 4.30m

x 2 y2

−2.50−3.50

57

tan−1 5

7⎛⎝⎜

⎞⎠⎟ 35.5o

180 θs 180o 35.5o 216o


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Vector and ScalarVector quantities have both magnitudes (sizes) and directions

Scalar quantities have magnitudes onlyCan be completely specified with a value and its unit

Force, gravitational acceleration, momentumNormally denoted in BOLD letters, F, or a letter with arrow on top F

urTheir sizes or magnitudes are denoted with normal letters, F, or absolute values:

Fur

or F

Energy, heat, mass, timeNormally denoted in normal letters, E

Both have units!!!


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Properties of Vectors• Two vectors are the same if their and the

are the same, no matter where they are on a coordinate system!!

x

y

AB

E

D

C

F

Which ones are the same vectors?

A=B=E=D

Why aren’t the others?

C: The same magnitude but opposite direction: C=-A:A negative vector

F: The same direction but different magnitude

sizes directions


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Vector Operations• Addition:

– Triangular Method: One can add vectors by connecting the head of one vector to the tail of the other (head-to-tail)

– Parallelogram method: Connect the tails of the two vectors and extend– Addition is commutative: Changing order of operation does not affect the results

A+B=B+A, A+B+C+D+E=E+C+A+B+D

A

BA

B=A

BA+B

• Subtraction: – The same as adding a negative vector:A - B = A + (-B)

A-B Since subtraction is the equivalent to adding a

negative vector, subtraction is also commutative!!!

• Multiplication by a scalar is increasing the magnitude A, B=2A

A B=2A

AB 2

A+BA+B

A-B

OR


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Example for Vector AdditionA car travels 20.0km due north followed by 35.0km in a direction 60.0o west of north. Find the magnitude and direction of resultant displacement.

N

E

60o

θr

r

20A

B

60cos

60sintan 1

BA

B

θ

Do this using components!!

θθθ cos2sincos 2222 ABBA

θcos222 ABBA

60cos0.350.2020.350.20 22

)(2.482325 km

60cos0.350.2060sin0.35tan 1

N W wrt to9.385.373.30tan 1

Bcos60o

Bsin60o A Bcosθ

2 Bsinθ

2


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Components and Unit VectorsCoordinate systems are useful in expressing vectors in their components

Aur

Ax2 Ay

2

(Ax,Ay)A

θ

Ay

Ax

x

y

xA

Aur

}Components

(+,+)

(-,+)

(-,-) (+,-)

yA

} Magnitude

Aur

cosθ 2

Aursinθ

2

A

ur2

cos2θ sin2θ Aur

Aur

cosθ

Aur

sinθ


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Unit Vectors• Unit vectors are the ones that tells us the

directions of the components• Dimensionless • Magnitudes these vectors are exactly 1• Unit vectors are usually expressed in i, j, k or

ir, jr, kr

AurSo a vector A can be

expressed as yA

A

urcosθ

Aur

sinθ xA ir

ir

jr

jr


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Examples of Vector OperationsFind the resultant vector which is the sum of A=(2.0i+2.0j) and B =(2.0i-4.0j)

Cur

Find the resultant displacement of three consecutive displacements: d1=(15i+30j +12k)cm, d2=(23i+14j -5.0k)cm, and d3=(-13i+15j)cm

Dur

Cur

Dur

θ

2.0i

r 2.0 j

r 2.0i

r−4.0 j

r

2.0 2.0 ir

tan−1

Cy

Cx

tan−1 −2.0

4.0−27o

15 23−13 ir

25ir d

ur1 d

ur2 d

ur3

15i

r 30 j

r12k

r 23i

r14 j

r−5.0k

r −13i

r15j

r

25

2 59

2 7.0

2 65(cm)

4.0

2 −2.0

2

16 4.0 20 4.5m

2.0 −4.0 jr

4.0ir

−2.0 jrm

Aur Bur

30 14 15 jr

12−5.0 kr

59 jr 7.0k

rcm

Magnitude

Wednesday, June 8, 2011

Displacement, Velocity, and Acceleration in 2-dim

• Displacement: r r

• Average Velocity: v r

• Instantaneous Velocity:

v r

• Average Acceleration

a r

• Instantaneous Acceleration: a

r

How is each of these quantities defined in 1-D?

f ir rr r

rt

rf i

f i

r rt t

r r

0lim

t

rt

rdrdt

r

vt

rf i

f i

v vt t

r r

0lim

t

vt

rdvdt

rd drdt dt

r2

2

d rdt

r


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initial positiono rr

final positionrr

Displacement:

2D Displacement

rr rr−r

ro


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rv

Average velocity is the displacement divided by the elapsed time.

rr−r

ro

t −to

rr

t

2D Average Velocity


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The instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time.

rv

limt→ 0

rr

t


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ra

rvo

rv

rv

2D Average Acceleration

rv−r

vo

t −to rv

tPHYS 1443-001, Spring 2011 Dr.

Jaehoon Yu21


Kinematic Quantities in 1D and 2DQuantities 1 Dimension 2 Dimension

Displacement

Average Velocity

Inst. Velocity

Average Acc.

Inst. Acc.

x rr

vx xt

xf −xitf −ti

vr≡rr

trr

f −rri

tf −ti

vr≡lim

t→ 0

rr

tdrr

dt vx ≡lim

t→ 0

xt

dxdt

ar≡vr

tvr

f −vri

tf −tiax ≡

vxt

vxf −vxitf −ti

ar≡lim

t→ 0

vr

tdvr

dt ax ≡lim

t→ 0

vxt

dvxdt

What is the difference between 1D and 2D quantities?

x f −xi r

rf −r

ri


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PHYS 1443 – Section 001 Lecture #3