Wednesday, June 8, 2011
Dr. Jae hoon 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
1

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

• Derive the quadratic equation for Bx 2 -Cx+A=0
 5 points 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
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
3

• 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!!
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
4

• Free fall motion is a motion under the influence of the gravitational pull (gravity) only; Which direction is a freely falling object moving?
Yes, down to the center of the earth!!
– 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/s 2 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/s 2 when +y points upward
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
5

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,
What is the acceleration in this motion?
g=-9.80m/s 2
(a) Find the time the stone reaches at the maximum height.
What is so special about the maximum height?
V=0 t
 20.0
9.80
 2.04
s v
 f
v yi
 a y t
 
20.0 9.80
t

0.00 / Solve for t
(b) Find the maximum height.
y f

1 a y t
2  y i
 v yi t


50.0 20 2.04
1
2
 
70.4( )
2
2
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
6

(c) Find the time the stone reaches back to its original height.
t

2.04 2 4.08
s
(d) Find the velocity of the stone when it reaches its original height.
Velocity
v yf

v yi
 a y t
   
20.0( / )
(e) Find the velocity and position of the stone at t=5.00s.
Position
v yf
y f

v yi
 a y t

20.0
    
29.0( / )
1
 y i
 v yi t

2 a y t
2

50.0

20.0

5.00
( 9.80)

(5.00)
2
 
27.5( m )
2
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
7

• 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,   (r  )
• Vectors become a lot easier to express and compute
+y
How are Cartesian and
Polar coordinates related?
y
1
(x
1
,y
1
) =(r
1

1
) r
1

1
O (0,0)
Wednesday, June 8, 2011 x
1
+x x

1 y
1
 r
1 cos

1 r
1 sin

1
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu r

1 tan

1
 x
1
2
 y
1
2 y
1

 x
1

1
 tan

1

 y
1
 x
1

8

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.
 s
 r
(-3.50,-2.50)m y x r


 
 x

2
 y 2
3.50

2


2.50

18.5

4.30( m )

2
 
180
  s
 tan
 s


2.50

3.50
 
 s



1
 tan
180



5
7 s





5
7
35.5
o
180 o 
35.5
o 
216 o
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
9

Vector quantities have both magnitudes (sizes) and directions
Force, gravitational acceleration, momentum
Normally denoted in BOLD letters, F , or a letter with arrow on top ur
F
Their sizes or magnitudes are denoted with normal letters, ur
F , or absolute values:
F or F
Scalar quantities have magnitudes only Energy, heat, mass, time
Can be completely specified with a value and its unit Normally denoted in normal letters, E
Wednesday, June 8, 2011
Both have units!!!
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
10

• Two vectors are the same if their and the are the same, no matter where they are on a coordinate system!!
sizes directions y
Which ones are the
D same vectors?
F
A=B=E=D
A Why aren’t the others?
B
E
C x C: The same magnitude but opposite direction:
C=-A:A negative vector
F: The same direction
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu

• 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+B
A
B =
B
A
A+B
OR
B
A
A+B
• Subtraction:
– The same as adding a negative vector:A - B = A + (-B)
A
A-B
-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
B 
2
A
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
B=2A
12

A car travels 20.0km due north followed by 35.0km in a direction 60.0
o west of north. Find the magnitude and direction of resultant displacement.
Bsin60 o
N
B
60 o
Bcos60 o r

20
A
Wednesday, June 8, 2011 r






A
A

B cos
2 
B
2

 cos
2

2




B sin sin
2




2

2 AB cos

A
2 
B
2 
2 AB cos


20 .
0
 
35 .
0

2 
2

20 .
0

35 .
0 cos 60
2325

48 .
2 ( km )
E B sin 60
  tan

1
A

B cos 60
 tan

1
35 .
0 sin
20 .
0
60

35 .
0 cos 60
 tan

1
30 .
3
37 .
5

38 .
9

to W wrt N
Do this using components!!
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
13

Coordinate systems are useful in expressing vectors in their components y
(-,+)
(-,-)
A y
(+,-)

A
(+,+)
(A x
,A y
)
A x x
A
 x
A
 y ur
A ur
A cos
 sin
 ur
A

A x
2 
A y
2
}
Components
} Magnitude
A


A cos


2


A sin


2 r

A
2  cos
2
  sin
2

 r

A
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
14

• 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 r r r i , j , k
So a vector A can be expressed as ur
A

A x i r

A y r j ur

A cos
 r i
 ur
A sin
 r j
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
15

Find the resultant vector which is the sum of A=(2.0i+2.0j) and B =(2.0i-4.0j) ur
C ur

A ur
C ur

B




2.0
i
2.0
r
 r

2.0
j
2.0
 i r





2.0
2.0
 r i

 
2  
2.0
2

16

4.0

20

4.5( m ) r

4.0
j
4.0
 r j
 


4.0
i r tan

1 r

2.0
j
C
C x y  tan

1

2.0
4.0
 
27 o
Find the resultant displacement of three consecutive displacements: ur
D d
1
=(15i+30j +12k)cm, d


 ur d
15
1
 d
 ur
2
23


Magnitude d ur
3
13


 r
15 i i ur
D
 r



30
2
=(23i+14j -5.0k)cm, and d
3
30
 r j

14
25
2
12

 k r 

15
59

 r
23 j
2 i r

 r

14 j

12
7.0

5.0
k r

2
5.0

=(-13i+15j)cm

 r k
 


65( cm )
13
25 i r i r r

15 j
 r

59 j

7.0
k r
( cm )
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
16

Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
• Average Velocity:
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration: r
  r r r f
 r i r v
 r v r a



 r r t
 r r f
 r r t f
 t i i lim
0

 r v t
 r r
 t

 r d r dt r r v f
 v i t f
 t i r a
 lim t 0
 r v
 t
 r d v dt

How is each of these quantities defined in 1-D?
d
 r d r
 dt
 dt
  r
2 d r dt 2
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
17

r r o

initial position r r

final position
Displacement:  r r

r r
 r r o
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
18

Average velocity is the displacement divided by the elapsed time.
r v
 r r
 t
 t r r o o


 t r r
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
19

The instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time.
r v

lim
 t

0
 r r
 t
Wednesday, June 8, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
20

r v
 r v
Wednesday, June 8, 2011
r v o r a
 r v
 t
 t r v o o
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu

 r v
 t
21

Quantities 1 Dimension 2 Dimension
Displacement
Average Velocity
Inst. Velocity
Average Acc.
Inst. Acc.
Wednesday, June 8, 2011 v v a x x x a
 x


 x



 t x lim
 t

0
 v
 t lim x
 t

0 x





 v f t x x t x v t t f f xf
 f





 t dv dt x t x x v
PHYS 1443-001, Spring 2011 Dr. i i dt i i dx xi r v r a r v



 r a r r

 r
 r
 t lim
 t

0
 r v t
 lim
 t

0
What is the difference between 1D and 2D quantities?
r r
 f

 t t r r r
 r t r
 v r
 v t f f
 r r i r
 r f i r f
 d r dt
 v i

 t i
 t r i d v dt
22
Jaehoon Yu