Lecture 2_Motion in two dimensions_physics.ppt

Introduction:
In this chapter, we derive the kinematic equations from fundamental
definitions of displacement, velocity, and acceleration vectors;
In two dimensions, two special cases are considered: motion with,
constant acceleration , , and UCM (uniform circular motion).
Displacement, velocity, and acceleration vectors:
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Motion in two dimensions
st
a C



r


x
y
v0
P, ti
O
Q, tf
i
r

f
r

In the time interval
, the position
vector changes from
to
f i
t t t
  
i
r

f
r


Displacement, velocity, and acceleration vectors (Cont’d)
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 From the previous figure we can find that ; so the displacement vector is
expressed as:
(4.1)
 The average velocity is defined as:
(4.2)
 The instantaneous velocity (along the tangent) is expressed like:
(4.3)
 The average acceleration (along ) has as expression:
(4.4)
 Finally, the instantaneous acceleration will have as expression:
(4.5)
f i
r r r
  
  
f i
r r r
  
  
r
v
t



















0
lim
t
r d r
v
t dt
 

 

 

v


f i
f i
v v v
a
t t t
 
 
 
  















2
2
0
lim =
t
v dv d r
a
t dt dt
 

 

  


Motion in two dimensions with constant acceleration
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 In the xy-plane, the position vector is written:
(4.6)
where change with time as the particle move;
 By definition, the velocity vector is given by:
(4.7)
 Since ; therefore,
the components of the velocity vector can be written (chapter 3) as:
 Equation (4.7) is then expressed as:
r

r xi y j
 
 

, and
x y r

x y
d r d x di d y dj
v i x j y v v i v j
dt dt dt dt dt
       
 
    
 
its components are also constants
x
st
y
a
a C
a


  


















0 0
and
x x x y y y
v v a t v v a t
   
       
0 0 0 0
0 (4.8)
x x y y x y x y
v v a t i v a t j v i v j a i a j t
v v at
       
  
     

  

Motion in two dimensions with constant acceleration (Cont’d)
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 From kinematics and for this case of constant acceleration, the coordinates of a
particle are:
 These expressions into Equation (4.6) give:
 Considering , the component forms for (4.8) and (4.9) are:
2
0 0
2
0 0
1
2
1
2
x x
y y
x x v t a t
y y v t a t

  



   


     
2 2
0 0 0 0
2 2
0 0 0 0 0 0
1 1
2 2
1 1
(4.9)
2 2
x x y y
x y x y
r x v t a t i y v t a t j
x i y j v i v j t a i a j t r r v t at
   
     
   
   
         
 

         
0 0
r 


2
0
0 2
0 0
2
0
0
1
1 2
For ; and for
1
2
2
x x
x x x
y y y
y y
x v t a t
v v a t
v v at r v t at
v v a t
y v t a t

 

 

 
     
 
 

   


     

Projectile motion
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 In this motion:
 If
is constant and downward directed;
air resistance is neglected;
trajectory is always a parabola;
0
x
y
g
a
a g












 
 
0 0 0 0 0
0 0 0
0 0
0 , at 0 cos
then
sin
,
x
y
x y t v v
v v
Ox v



   




 
















 
 
x
y
O(0, 0)
1
0;
y
v t

0
v

0

0
x
v
0
y
v

0
x x
v v

y
v
v

0
 

0
v v

 
R 0
y y
v v

0
x x
v v

0
x x
v v

h
0 0
t t
 
2 1
2
t t t
 
2
R

Projectile motion (Cont’d 1)
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 Substituting those accelerations into the components of equations (4.8) and
(4.9), we get:
 Solving Equation (4.12) for time and substituting its value into equation
(4.13), we get the equation of a parabola:
whose form is:
0 0 0
0 0 0
0 0
cos (4.10);
sin (4.11);
and
st
x x
y y
x
v v v C
v v gt v gt
x v t v


   


   


  
 
0
2 2
0 0 0
cos (4.12);
1 1
sin (4.13).
2 2
y
t
y v t gt v t gt





   


t
  2
0 2 2
0 0
tan (4.14)
2 cos
g
y x x
v


 
   
 
2
y ax bx
 

Projectile motion (Cont’d 2)
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 At any time it is possible to determine respectively the magnitude and
direction of particle’s velocity, :
Horizontal range, , and the maximum height, :
 We have to find and in terms of :
v

   
2 2
1
(4.15)
, horizontal direction, tan (4.16)
x y
y
x
v v v v
v
Ox v v
v
 
   


  
  
  
  
















  
 
R h
h R 0 0
, and
v g

0 0
1
2 2
0 0
1
2
0 0 0
1
0 0 0
sin
At the peak point: 0 Equation (4.11) gives
sin
into (4.13) gives:
2
2 sin cos
The range horizontal distance travelled in a time 2
Since sin 2 2sin cos ,
y
v
v t
g
v
t h
g
v
R t t R
g


 
  
  

   





2
0 0
2
0 0
0 max
sin 2
then (4.18)
The maximum range is obtained when 45 : .
v
R
g
v
R
g



 


Uniform circular motion
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 This is the motion for an object moving in a circular path with constant
speed, ;
 In a circular motion, there are two ways to produce accelerations:
Tangential and radial accelerations in a curvilinear motion:
 This is the motion of a particle moving along a curved path: here the
velocity vector changes both in direction and magnitude:
v v


2
from the of tangential acceleration:
from the of centripetal or radial acceleration: (4.19)
t
c r
d v dv
change in magnitude v a
dt dt
v
change in direction v a a
r
  
  





v

P
Q
r t
a a a
 
  
' ' '
r t
a a a
 
  

Motion in two dimensions (Cont’d)
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Tangential and radial accelerations in curvilinear motion (Cont’d):
 From point to point, the total acceleration (previous figure) changes and
can be written as:
(4.20)
where , , and representing the radius of
curvature at a given point;
 Pythagoras theorem gives the magnitude of :
Relative velocity and relative acceleration:
 In this section, we see how observations made by different observers in
different frames of reference are related to each other;
r t
a a a
 
  
a

t
d v dv
a
dt dt
 
 2
c r
v
a a
r
  r
a
 2 2
r t
a a a
 

Relative velocity and relative acceleration (Cont’d)
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 Example:
More general situation:
 Consider the picture of particle’s motion.At a given time , the particle
will be located at a point P relative to two reference frames
;
 The motion can be described by two observers: (1) related to a fixed
reference frame (with respect to the Earth); and
Observer A
Path seen by A
is a vertical
straight line
Experimenter A throws an object
Observer B
Path
seen by
B is a
parabola
Frame of reference in motion (Observer A) Fixed frame of reference (Observer B)
u

v
 V u v
 
  
u

u

0
t t

'
and
S S
S

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(2) related to the moving reference frame relatively to at a constant
velocity .
 We label by:
 The two position vectors are related to each other by:
or (4.23)
 Differentiating (4.23) with respect to time, we get:
'
S S
u

' '
: the position vector in , and
: the position vector in .
r S
r S




'
'
At 0
At
i
f i
t O O
t t t OO ut
  
   















 

u

'
OO















r

'
r

S '
S
O
'
O
P
'
r r ut
 
   '
r r ut
 
  

More general situation (Cont’d 2)
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 Equations (4.23) and (4.24) are called “Galilean transformation
equations”;
 Although observers in and measure different displacements and
velocities, they find the same acceleration:
'
'
(4.24)
dr d r
u v v u
dt dt
    
 
   
S '
S
'
'
, since 0.
dv dv du du
a a
dt dt dt dt
    
    
 

2017-2018 IN CLASS EXERCISES_Problem 3
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Statement:
If one throws a ball at 21 m / s at an angle of elevation of 30 ° from a roof
top located at 16 m from the ground (Figure 1).
Find:
a) the duration for the trajectory of the ball;
b) its horizontal range;
c) the maximum height;
d) the impact angle of the ball on the ground;
e) its speed at 2 m above the roof top; and
f) its speed at 2 m above the ground.

2017-2018 IN CLASS EXERCISES_Problem 3 (Cont’d 1)
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y
x
16 m
v0
= 21 m/s
R
H
Figure 1
0
0 30
 

v

3
v

1
v

2
v

2 m
2 m
 
0,0
O

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Problem solving:
 a) Data: ;
Equations of motion:
 To be found: Duration of the trajectory,
 Equation to be used: ; knowing that at the end of the
trajectory we have
 Solving:
1 0 2
0 0 0 0
0; 16 ; 21 ; 30 ; 9.80
x y m v m s g m s

 
      
 
 
2
18.2 i
16 10.5 4.90 ii
x t
y t t




  


0 1
0 0 0
0 1
0 0 0
cos 21 cos30 18.2
sin 21 sin30 10.5
x
y
v v m s
v v m s




     

 
    


?
t 
2
16 10.5 4.90
y t t
  
0
y 
2 2 2
0 16 10.5 4.90 4.90 -10.5 -16 0 0
t t t t at bt c
        
   
2
2
2
10.5 10.5 4 4.90 16
4
4
2 2 2 4.90
b b b ac
b ac t
a a
     
     
      


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 b) Data: ;
To be found: Horizontal range, ;
Equation to be used: (i) ;
Solving:
 c) Data: ;
To be found: Maximum height, ;
Equation to be used: ;
Solving:
10.5 110.25 313.64 10.5 20.59
3.17 to be kept
9.80 9.80
10.5 110.25 313.64 10.5 20.59
1.03 to be rejected: before the motion
9.80 9.80
s
t
s
   
 


 
  
   




3.17
t s

?
R 
18.2
x t

18.2 3.17 57.7
R R m
   
1 2
0 0
10.5 ; 0; 9.80 ; 16
y y
v m s v g m s y m
 
     
?
H 
 
2 2
0 0
2
y y
v v g H y
  
   
2
2 110.25
0 10.5 2 9.80 16 16
19.6
at 21.6 from the ground level, or
21.6
at 5.6 from the roof top.
H H
m
H m
m
       

 



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 d) Data: ;
To be found: Impact angle ;
Equations to be used:
Solving:
 e) Data:
To be found: speed at 2 m above the roof top,
Equations to be used:
1
0
3.17 ; 18.2
x x
t s v v m s
   
?
 
 
0
1
10.5 9.80 ;
, tan
y y x y
y
x
v v gt t v v i v j
v
Ox v
v
 
      

  
   

 

 















 

 
1 1
1 0 0
10.5 9.80 3.17 20.6 ; 18.2 20.6
20.6
tan 48.5 or 48.5 under the horizontal direction
18.2
y
v m s v i j m s
 
 

        

 
 
 
  
 

 

 
1 1 2
0 0 0 0
18.2 ; 10.5 ; 9.80 ; 2 18 ; 16
x x y
v v m s v m s g m s y y m y m
  
          
?
v v
 

 
2 2
0 0
2 2
2 ;
y y x y
x y
v v g y y v v i v j
v v v v
     


  


 



Solving:
f) Data: ;
To be found: speed at 2 m above the ground level,
Equation to be used:
Solving:
   
1
1
2
2 2 2
1
2
8.43
10.5 2 9.80 18 16 110.25 19.6 2 71.05
8 43
y
y
y
v m s
v m s
v m s



  

           
  




 
 
       
1
1 1 1
1
2 2 2
2 2 2 2
1 2
(18.2 8.43 ) , for the ascending motion 0
(18.2 8.43 ) , for the descending motion 0 .
18.2 8.43 18.2 8.43
331.24 71.07 20.06
x y y
x y y
v v i v j i j m s v
v v i v j i j m s v
v v v
v v


      

 
     


       
   
   

   

 
 1
m s

1 1 2
0 0 0
18.2 ; 10.5 ; 9.80 ; 2 ; 16
x x y
v v m s v m s g m s y m y m
  
        
?
v v
 

 
2 2
0 0
2 2
2 ;
y y x y
x y
v v g y y v v i v j
v v v v
     


  


 


     
1
4
2
2 2 2
1
3
19.6
10.5 2 9.80 2 16 110.25 19.6 14 384.65
19.6
y
y
y
v m s
v m s
v m s



  

            
 





05/20/25
Abdou Safari Kagabo
19
From both solutions, we keep only because it is the sole
to correspond to the downward motion .
Individual problem: Solve yourself the same problem by considering the
origin of coordinates O(0, 0) located at the ball launching point.
1
3 19.6
y
v m s
 

   
1
3 3 3
18.2 19.6 , for the descending motion 0
x y y
v v i v j i j m s v

     
   

   
2 2 1
3 3 18.2 19.6 26.75
v v v v m s
       



Lecture 2_Motion in two dimensions_physics.ppt

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