# Lec5

```Relations for 2-D motion
Motions that occur in 2-dimensions are more complicated since
now the direction of the acceleration, velocity, force, etc. need to
be taken into account.


F  ma


These vectors, F , and a , are still proportional to each other by
the proportionality constant, mass, and the two vectors are in the
same direction.



F  ma  mr
Another important difference is that instead of taking time
derivatives of a single valued function x, (i.e. a scalar), we now

have to take successive time derivatives of a vector r .
5-1
How do we take time derivatives of a
vector?

 dr
r
dt
The answer is that it depends….. It depends on which
coordinate system we choose as convenient for a specific
problem. For 2-D motion, we will work in Cartesian and Polar
coordinates. First, let’s consider how to write functions and take
derivatives in Cartesian coordinates
FORCE
ACCELERATION
xˆ yˆ

F  Fx xˆ  Fy yˆ

a  ax xˆ  ay yˆ
Are unit vectors in the x- and y- directions (i.e. having
magnitude of 1 and whose purpose is only to indicate direction)
5-2
Time derivatives in Cartesian
Coordinates
Why is it so convenient to work in Cartesian coordinates? The
answer is that the unit vectors do not change in time. They
always point in the same direction.
Consider the vector:

r  xxˆ  yyˆ

  dr
v r 
 xxˆ  xxˆ  yyˆ  yyˆ  xxˆ  yyˆ
dt
  d xxˆ  yyˆ 
ar 
 xxˆ  xxˆ  yyˆ  yyˆ  xxˆ  yyˆ
dt
Thus, if the force is only in the x- or y- direction, then the trajectory
of the object can be computed the same as for 1-dimensional motion.
5-3
Polar Coordinates
In certain types of motion, such as revolution about an axis, it is sometimes
more convenient to use polar coordinates. The main difficulty in polar
coordinates is that the unit vectors can change direction as a function of
position or time.
y ˆ
y  r sin 

rˆ

r  x, y   r, 
x  r cos 
x
The transformation between Cartesian and Polar unit vectors can be
written as follows:
rˆ  xˆ cos   yˆ sin 
ˆ   xˆ sin   yˆ cos 
 rˆ   cos  sin    xˆ 
ˆ   sin   cos   yˆ 
  
 
5-4
Converting back to Cartesian from Polar
Coordinates
 rˆ   cos  sin    xˆ 
Given
ˆ   sin   cos   yˆ 
  
 
The transformation between Polar and Cartesian unit vectors can be
written as follows:
 cos   sin    rˆ   cos   sin    cos   sin    xˆ   xˆ 
 sin   cos   ˆ   sin   cos    sin   cos    yˆ    yˆ 

   
 
   
1
1
 xˆ  cos   sin    rˆ 
 yˆ    sin   cos   ˆ
  
 
Given these forward and reverse transformations between Cartesian and
polar coordinates, the next question is how to take time derivatives of
vectors written in polar coordinates? Suppose there is a vector, written as
follows:


What is r ?
r  rrˆ
5-5
Time derivatives in Polar Coordinates: I

r  rrˆ  rrˆ

d
r  rrˆ  r xˆ cos   yˆ sin  
dt



r  rrˆ  r   xˆ sin    yˆ cos  


 
ˆ



r  rrˆ  rˆ
Hence the velocity in polar coordinates can be given in terms of
an object’s radial position, radial velocity, and rotation frequency.
5-6
Time derivatives in Polar Coordinates:
II

r  d rrˆ  rˆ
dt

r  rrˆ  rrˆ  rˆ  rˆ  rˆ


 rˆ

 
  r 2 rˆ

r  r  r2 rˆ  2r  r ˆ



d
 xˆ sin    yˆ cos 
dt

ˆ  xˆ cos   yˆ sin    rˆ
ˆ 
This is a very interesting result, but is it useful? Why can’t we
solve everything in the much simpler Cartesian coordinates?
It turns out that a certain class of forces, known as “central
forces” are most easily described using polar coordinates.
5-7

F  Fx xˆ  Fy yˆ
Fx  m ax  0
Fy  m ay  m g

ay  g 
dvy
dt
dy
v y   gt  v y 0   gt 
dt
a x t   0 
v x t   v x 0
dvx t 
dt
dx

dt
1 2
 dy   g  tdt  y  y0  y   2 gt
1/ 2
  2y 

t  
 g 
  2y 
 dx   v x0 dt  x  x0  x  v x0 t  v x0  g 
1/ 2
5-8
Solution
1/ 2
  2y 

x  vx 0 
 g 
g  9 .8
m
s2
1/ 2
 g 

 vx 0  x
 2y 
x  40 m
y  16m  26m  10m
vx 0  28m / s
5-9

F  Fx xˆ  Fy yˆ

v  vx xˆ  vy yˆ
where
With Initial
Conditions
Fx  m ax  0
Fy  m ay  m g
vx  v cos 
v y  v sin  
vx  0  vx  v cos 
v y   g  v y   gt  v sin  
x  v cos   xt   v cos t  x0
1 2
y   gt  v sin    y t    gt  v sin  t  y0
2
5 - 10
Given that we know the y position of the projectile for t=0 and
t=final are at the same position, we can solve for t.
1 2
 1

yt   y0  0   gt  v sin  t  t   gt  v sin  
2
 2

2v sin  
t  0; t 
g
2v 2 sin   cos 
v 2 sin 2 
xt   v cos t  x0 
 x0 
 x0
g
g
When is
xt   x0 a maximum?
Answer: It is maximal when
sin2   1;
  45
o
In other words, at this angle the projectile can travel a maximal distance. So,
what initial velocity is required for the projectile to travel 12km?
v2
xt   x0   12km  v 
g
12kmg  343ms
5 - 11
Initial Conditions:
r  70 sft2
  0.02 rad
s
2
r  30000ft   60o

 

r  r  r2 rˆ  2r  r ˆ

r  rrˆ  rˆ
Terms we
don’t know
x  const.
d
r cos  
dt
d2
x  0  2 r cos 
dt
x  0 
5 - 12
x  0 
d
r cos    r cos    r sin    0
dt
sin  

r  r
 r tan 
cos 
d2
d
x  0  2 r cos   r cos   r sin  
dt
dt
x  0  rcos   r sin   r sin   r cos 2  r sin 
r cos 
r  cos   2



2  

r sin  
r
sin  
r cos 
cos   2
2




 2 tan  

r sin  
sin  
5 - 13
Solving for velocity magnitude
 

r  r 2  r
2


  
2

r tan   r
2
 r tan2    1  r cos 

 
r  30000ft  0.02 s cos   1200 fts
3
Solving for acceleration magnitude
r 
r  r   2r  r
r  r   2r tan 
r 
r  r   rcot   r cot  
r 
2 2
2 2
2
2
2 2
r  70 ft2  30000ft 0.02 rad 2
s
s
r  67 ft2
s
 r cot   2r tan  2  r cot  2
2 2

2
 
 r  r 2 1  cot2    r  r 2 sin  
3
 
sin  
3
5 - 14
Problem 1
As observed from a ship moving due east at 8 km/h, the
wind appears to blow from the south. After the ship has
changed course, and as it is moving due north at 8 km/h,
the wind appears to blow from the southwest. Assuming
that the wind velocity is constant during the period of
observation, determine the magnitude and direction of the
true wind velocity.
5 - 15
Problem 1
As observed from a ship moving due east at 8 km/h, the
wind appears to blow from the south. After the ship has
changed course, and as it is moving due north at 8 km/h,
the wind appears to blow from the southwest. Assuming
that the wind velocity is constant during the period of
observation, determine the magnitude and direction of the
true wind velocity.
1. To solve relative motion problems:
a. Write the vector equation that relates the velocities of two
particles.
vB = vA + vB/A
vB and vA are the velocities of particles B and A relative to a fixed
frame, respectively. vB/A is the velocity of B relative to A.
5 - 16
Problem 1
As observed from a ship moving due east at 8 km/h, the
wind appears to blow from the south. After the ship has
changed course, and as it is moving due north at 8 km/h,
the wind appears to blow from the southwest. Assuming
that the wind velocity is constant during the period of
observation, determine the magnitude and direction of the
true wind velocity.
b. Express all velocity vectors in terms of their rectangular
components.
c. Scalar equations. Substitute the velocity vectors in
vB = vA + vB/A and solve the two independent sets of scalar
equations obtained.
5 - 17
Problem 1 Solution
First observation
Write the vector equation
that relates the velocities
of two particles.
y N
vs1
vw = vs1 + vw/s1
vw/s1
E
x
Express all velocity vectors in terms
of their rectangular components.
vw = vwx i + vwy j
Scalar equations.
vs1 = 8 i km/h
vwx = 8 + 0
vw/s1 = vw/s1 j
vwx = 8 km/h
vwy = 0 + vw/s1
5 - 18
Problem 1 Solution
Second observation
Write the vector equation
that relates the velocities
of two particles.
y N
vs2
vw/s2
vw = vs2 + vw/s2
o
45
E
x
Express all velocity vectors in terms
of their rectangular components.
vw = vwx i + vwy j
vs2 = 8 j km/h
vw/s2 = vw/s2 cos 45o i + vw/s2 sin 45o j
Scalar equations.
vwx = 0 + vw/s2 cos 45o
vw = 8 i +16 j km/h
vwy = 8 + vw/s2 sin 45o
N
vwy = 16 km/h
vw = 17.89 km/h
63.4o
E
5 - 19
```