Thursday, June 9, 2011
Dr. Jae hoon Yu
• Motion in Two Dimensions:
• Motion under a constant acceleration
• Projectile Motion
• Maximum range and Maximum height
• Newton’s Laws of Motion
• Force; Mass; Newton’s 1 st – 3 rd Laws
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
1

• Homework
– 100% of you have registered and submitted!!
– Excellent job!
– 6/27 already started working on HW#2, Great!!
• Quiz 1 results
– Class average: 7.5/12
• Equivalent to 62.5/100
– Top score: 11.2/12
• Quiz #2 coming Tuesday, June 14
– Covers: CH 1.1 – what we finish on Monday, June 13
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
2

Reminder: 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 21 of this lecture note!!
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
3

This is a motion that could be viewed as two motions combined into one. (superposition…)
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
4

v x v x
2
 v xo
 v
2 xo
Thursday, June 9, 2011
 a x t

2 a x x x

1
2
 v xo
 v x x
 v xo t

1
2
 a x t
2 t
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
5

v y
 v yo y

1
2
 a y t
 v
 t yo
 v y v
2 y
 v
2 yo

2 a y y y
 v yo t

1
2 a y t
2
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
6

Imagine you add the two 1 dimensional motions on the left.
It would make up a one 2 dimensional motion on the right.
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
7

v x
 v xo
 a x t x

1
2
 v
 t xo
 v x v x
2
x-component y-component
 v
2 xo

2 a x x v y
 v yo y

1
2
 a y t
 v
 t yo
 v y v
2 y
 v
2 yo

2 a y y
 x
 v xo t

1
2
Thursday, June 9, 2011 a x t
2
 y
 v yo t

1
2
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu a y t
2
8

Ex. A Moving Spacecraft
In the x direction, the spacecraft in zero-gravity zone has an initial velocity component of +22 m/s and an acceleration of +24 m/s 2 . In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s 2 . Find (a) x and v x
, (b) y and v y
, and (c) the final velocity of the spacecraft at time 7.0 s.
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
9

1. Visualize the problem  Draw a picture!
2. Decide which directions are to be called positive (+) and negative (-).
3. Write down the values that are given for any of the five kinematic variables associated with each direction.
4. Verify that the information contains values for at least three of the kinematic variables. Do this for x and y
separately. Select the appropriate equation.
5. When the motion is divided into segments in time, remember that the final velocity of one time segment is the initial velocity for the next.
6. Keep in mind that there may be two possible answers to a kinematics problem.
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
10

In the x direction, the spacecraft in a zero gravity zone has an initial velocity component of +22 m/s and an acceleration of +24 m/s 2 . In the y direction, the analogous quantities are +14 m/s and an acceleration of +12 m/s 2 . Find (a) x and v x
, (b) y and v y
, and (c) the final velocity of the spacecraft at time 7.0 s.
?
x a x
+24.0 m/s 2 v x
?
v ox
+22.0 m/s t
7.0 s
?
y
Thursday, June 9, 2011 a y
+12.0 m/s 2 v y
?
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu v oy
+14.0 m/s t
7.0 s
11

?
x
First, the motion in x-direciton… a x
+24.0 m/s 2 v x
?
v ox
+22 m/s t
7.0 s
 x
 v ox t

1
2 a x t
2


22 m s

7.0 s

1
2

24 m s
2

7.0 s
2  
740 m v
 x 
 v ox
 a x
22m s
 t


24m s
2

7.0 s
 
190m s
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
12

?
y
a y
+12.0 m/s 2 v y
?
v oy
+14 m/s t
7.0 s
 y
 v oy t

14 m s

1
2 a y
 
7.0 s t
2
 

1
2

12 m s
2

7.0 s
2  
390 m v
 y

 v oy
 a
14m s y t



12m s
2

7.0 s
 
98m s
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
13

v v y

98 m s v x


190 m s v




190 m s tan

1

98 190
Yes, you are right! Using the components and unit vectors!!
Thursday, June 9, 2011 v r


2
 v

 x r i

98m s
27
 o v y
A vector can be fully described when j r
PHYS 1443-001, Spring 2011 Dr.


2
190

210 m s the magnitude and the direction are given. Any other way to describe it?
 i r

98 r j
 m s
14
Jaehoon Yu

If we visualize the motion…
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
15

2-dim Motion Under Constant Acceleration
• Position vectors in x-y plane: r r i
 x i i r
 r y j i r r f
 r r x i f
 y j f
• Velocity vectors in x-y plane: r v i
 v i xi r
 r v j yi r v f
 v i xf r
 r v j yf
Velocity vectors in terms of the acceleration vector
X-comp v r f


 v r v
 xf v xi i
 a x

Thursday, June 9, 2011 r at t
 r i v xi
 a t x

 v yi
 a y t
 r j

Y-comp
 v xi r i
 v
PHYS 1443-001, Spring 2011 Dr. v
 yf yi j r 

 a x v yi
 a t y r i
 a y j r  t
16

Jaehoon Yu

2-dim Motion Under Constant Acceleration
• How are the 2D position vectors written in acceleration vectors?
Position vector components x
 f i
 xi

1 x v t a t
2 x
2 y
 f i
 yi

1 y v t a t
2 y
2
Putting them together in a vector form
Regrouping the above
Thursday, June 9, 2011 r r f
 x i f r
 r y j f


 x v t a t i i
 xi

1
2 x
2 r


 r i r x i r i
 y i
 r v t i j r



1
2
 v i r at xi
2 r


 i
 yi

1
2 r y v t a t j
2 y v j t yi r 

1
2
 a i x r
 y r 
2
2D problems can be interpreted as two 1D problems in x and y
PHYS 1443-001, Spring 2011 Dr. 17
Jaehoon Yu

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 a x
=4.0m/s components of the velocity vector at any time t .
2 . Determine the v xf
 v xi
 a t x

20

4.0

/
 v yf
 v yi
 a t y
 
15

0 t
 
15

/

Velocity vector r  
 v x
  r
 v y
  r
 
20

4.0
 r
 r
15 ( / )
Compute the velocity and the speed of the particle at t=5.0 s .
v r t

5
 v x , t

5 r speed
 v i r
 v y , t

5 j r


20

4.0

5.0

  x
2 
  y
2

 i r

15 j r


40 r i

15 r j
 m / s
  
15

2  m s
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
18

Angle of the
Velocity vector
  tan

1 v v
 tan

1

 
15
 
40 
  tan

1

 
3
 
8

  
21 o
Determine the x and y components of the particle at t=5.0 s .
x
 f y
 f v t xi
 v t
 yi
1 a t x
2
2
15 5

75 ( )
1
2
4 5
2 
150( )
Can you write down the position vector at t=5.0s?
r r f
 x f i r
 y f r j

150 i r r

75 j
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
19

• A 2-dim motion of an object under the gravitational acceleration with the following assumptions
– Free fall acceleration, g , is constant over the range of the motion
• r g
 
9.8
j r  m s 2

– Air resistance and other effects are negligible
• A motion under constant acceleration!!!!  Superposition of two motions
– Horizontal motion with constant velocity ( no acceleration )
– Vertical motion under constant acceleration ( g )
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
20

Show that a projectile motion is a parabola!!!
x-component v
 xi v i cos

 y-component v
 yi v i sin
 i a x
=0 a
 a i x
 a j y
  gj x f
 v xi t
 v i cos
 i t t
 v i x f cos
 i
In a projectile motion, the only acceleration is gravitational one whose direction is always toward the center of the earth (downward).
y f
 v t yi

1
2
  t
2
 v i sin
 i

1 t gt
2
2
Plug t into the above y
 f y f
Thursday, June 9, 2011 v i sin
 i

 v i x f cos
 i



1

2 g

 v i x f cos
 i


2
 x f tan
 i



2 v
2 g cos
2
 i i
PHYS 1443-001, Spring 2011 Dr.


Jaehoon Yu x f
2
What kind of parabola is this?
21

Thursday, June 9, 2011
The only acceleration in this
PHYS 1443-001, Spring 2011 Dr. motion. It is a constant!!
22

A ball is thrown with an initial velocity v=(20i+40j)m/s . Estimate the time of flight and the distance the ball is from the original position when landed.
Which component determines the flight time and the distance?
Flight time is determined by the y component, because the ball stops moving when it is on the ground after the flight.
t y
 f

80
40 t

1


2
 gt

 
0
2 
0 m
So the possible solutions are…
Distance is determined by the x component in 2-dim, because the ball is at y=0 position when it completed it’s flight.
 t

0 or t
 
8sec x
 f
 v t
 xi
80

8 sec g
Why isn’t 0 the solution?
20 8 160
 
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
23

• Based on what we have learned, one can analyze a projectile motion in more detail
– Maximum height an object can reach
– Maximum range
What happens at the maximum height?
At the maximum height the object’s vertical motion stops to turn around!!
v i h v
 yf
 v i v yi
 a t y sin


 gt
A

0

Solve for t
A
 t
A
 v i sin

 g
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu

Since no acceleration is in x direction, it still flies even if v y
=0.
R
 v t xi
 v xi
 
A

2 v i cos
 i

 v i sin
 i g


Range
R


 v i
2 sin g
2
 i

 y
 f h
 v t yi

1
2
 
2  v i sin
 i

 v i sin
 i g



1
2 g

 v i sin
 i g


2
Height y f
 h

 v i
2 sin
2
 i
2 g


Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu

• What are the conditions that give maximum height and range of a projectile motion?
h


 v i
2 sin
2
 i
2 g


This formula tells us that the maximum height can be achieved when θ i
=90 o !!!
R



 v i
2 sin g
2
 i



Thursday, June 9, 2011
This formula tells us that the maximum range can be achieved when 2θ i i.e., θ i
=45 o !!!
=90 o ,
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu

• A stone was thrown upward from the top of a cliff at an angle of 37 o to horizontal with initial speed of 65.0m/s. If the height of the cliff is
125.0m, how long is it before the stone hits the ground? v xi
 v i cos


  
51.9 / v yi
 v i sin
 i
  
39.1 / y
 f

125.0
 yi

1 v t gt
2
2 gt
2 
78.2
t

250

Becomes
9.80
t
2 
78.2
t

250

0 t

78 .
2



78 .
2

2 
4

9 .
80

(

250 )
2

9 .
80 t t
 
2 .
43 s or t

10 .
4 s
Since negative time represents the
PHYS 1443-001, Spring 2011 Dr. were thrown from the ground.

• What is the speed of the stone just before it hits the ground? v xf
 v xi
 v i cos



65.0
 cos 37

51.9
/ v
 yf v yi
 gt
 v i sin
   i gt
   
62.8 / v
 v xf
2  v yf
2 
51.9
2

62.8

2 
81.5 /
• What are the maximum height and the maximum range of the stone?
Thursday, June 9, 2011 PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu