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Class 9 – Review of motion in 2D and 3D
Chapter 4 - Monday September 13th
•Exam instructions
•Brief review
•Example problems
Reading: pages 58 thru 75 (chapter 4) in HRW
Read and understand the sample problems
•Exam tonight: 8:20-10:20pm
WebAssign is NO LONGER FREE!! You must register
ˆ ˆ ˆi j kr x y z  
r
2 1
ˆ ˆ ˆi j kr r r x y z        
r r r
ˆ ˆ ˆi j k
dr dx dy dz
v
dt dt dt dt
   
r
r
ˆ ˆ ˆi j kyx z
dvdv dv dv
a
dt dt dt dt
   
r
r
Two-dimensional kinematics
Acceleration:
Position:
Displacement:
Velocity:
 
 
   
0 0 0
0 0
21
0 0 0 2
0 0
22
0 0 0
cos 4 21
cos
sin 4 22
sin 4 23
sin 2 4 24
x
y
y
x x v t
v v
y y v t gt
v v gt
v v g y y





  

   
  
   
Projectile motion
Uniform circular motion
•Although v does not change, the
direction of the motion does, i.e.
the velocity (a vector) changes.
•Thus, there is an acceleration
associated with the motion.
•We call this a centripetal
acceleration.
2
2
Acceleration: Period:
1 1
Frequency : ; 2
2
v r
a T
r v
v v
f f
T r r

 

 
   
•Since v does not change, the acceleration must be
perpendicular to the velocity.
cos sin
ˆ ˆi j
p p
p p
x r y r
r x y
  
 
r
Analyzing the motion
 2 2
t
f t t
T
     
Uniform motion:
Note: 180 2 radianso
  
cos sinp px r t y r t  
sin cos
sin cos
ˆ ˆi j
x y
x y
x y
v r t v r t
v v t v v t
v v v
   
 
  
  
 
r
Analyzing the motion
cos sinp px r t y r t  
p p
x y
dx dy
v v
dt dt
 
[remember, ]
v
v r
r
  
Acceleration
ˆ ˆi j
ˆ ˆ= i j
yx
x y
dvdvdv
a
dt dt dt
a a
  
     
   

r
r
   
2 2
2 22 2
cos sinx y
v v
a a a v
r r
       
sin cos
cos sin
x y
x y
v v t v v t
a v t a v t
 
   
  
   
recall : cos sinp px r t y r t    
2 2
cos sinx y
v v
a t a t
r r
    
v
r
  
Relative motion
PA PB BAr r r 
r r r
Position:
PA PB BA
PA PB BA PB R
dr dr dr
v v v v v
dt dt dt
      
r r r
r r r r r
Velocity:
PA PB R
PA PB
dv dv dv
a a
dt dt dt
   
r r r
r r
Acceleration (if frame B is “inertial”):

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Class9

  • 1. Class 9 – Review of motion in 2D and 3D Chapter 4 - Monday September 13th •Exam instructions •Brief review •Example problems Reading: pages 58 thru 75 (chapter 4) in HRW Read and understand the sample problems •Exam tonight: 8:20-10:20pm WebAssign is NO LONGER FREE!! You must register
  • 2. ˆ ˆ ˆi j kr x y z   r 2 1 ˆ ˆ ˆi j kr r r x y z         r r r ˆ ˆ ˆi j k dr dx dy dz v dt dt dt dt     r r ˆ ˆ ˆi j kyx z dvdv dv dv a dt dt dt dt     r r Two-dimensional kinematics Acceleration: Position: Displacement: Velocity:
  • 3.         0 0 0 0 0 21 0 0 0 2 0 0 22 0 0 0 cos 4 21 cos sin 4 22 sin 4 23 sin 2 4 24 x y y x x v t v v y y v t gt v v gt v v g y y                     Projectile motion
  • 4. Uniform circular motion •Although v does not change, the direction of the motion does, i.e. the velocity (a vector) changes. •Thus, there is an acceleration associated with the motion. •We call this a centripetal acceleration. 2 2 Acceleration: Period: 1 1 Frequency : ; 2 2 v r a T r v v v f f T r r           •Since v does not change, the acceleration must be perpendicular to the velocity.
  • 5. cos sin ˆ ˆi j p p p p x r y r r x y      r Analyzing the motion  2 2 t f t t T       Uniform motion: Note: 180 2 radianso    cos sinp px r t y r t  
  • 6. sin cos sin cos ˆ ˆi j x y x y x y v r t v r t v v t v v t v v v               r Analyzing the motion cos sinp px r t y r t   p p x y dx dy v v dt dt   [remember, ] v v r r   
  • 7. Acceleration ˆ ˆi j ˆ ˆ= i j yx x y dvdvdv a dt dt dt a a               r r     2 2 2 22 2 cos sinx y v v a a a v r r         sin cos cos sin x y x y v v t v v t a v t a v t              recall : cos sinp px r t y r t     2 2 cos sinx y v v a t a t r r      v r   
  • 8. Relative motion PA PB BAr r r  r r r Position: PA PB BA PA PB BA PB R dr dr dr v v v v v dt dt dt        r r r r r r r r Velocity: PA PB R PA PB dv dv dv a a dt dt dt     r r r r r Acceleration (if frame B is “inertial”):