# Plane Curvilinear Motion by b0VT2qH

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```									Plane Curvilinear Motion

ENGR 221
April 7, 2003
Lecture Goals

• Rectangular Coordinates
• Polar Coordinates
• Normal and Tangential Coordinates
Rectangular Coordinates

Using 2-D rectangular coordinates, the location, velocity
and acceleration can be represented in Cartesian form of
equation.
r t   x t  i  y t  j
A change in the vector, displacement is represented as

 r  t   r  t2   r  t1 
  x2  x1  i   y2  y1  j
Rectangular Coordinates
The velocity component is represented by
v  t   vx  t  i  v y  t  j
 x t  i  y t  j
The acceleration by

a  t   ax  t  i  a y  t  j
 vx  t  i  vy  t  j
 x t  i  y t  j
Example Problem – Rectangular
systems
A projectile is fired with
an initial velocity of 800
ft/s at a target B located
2000 ft above the gun A
and at a horizontal
distance of 12000 ft.
Neglect air resistance,
determine the value of
the firing angle a.
Example Problem – Rectangular
systems
Break the problem into
horizontal and vertical
components. The
horizontal velocity is

vx   800 ft/s  cos a 

x   800 ft/s  cos a  t  x0
Example Problem – Rectangular
systems
The distance traveled will
be 12000 ft. So the time
can be found

x  x0   800 ft/s  cos a  t
x  x0             12000 ft
t                      
800 ft/s  cos a  800 ft/s  cos a 
15 s
t
cos a 
Example Problem – Rectangular
systems
The vertical components
of the equation are

vy   800 ft/s  sin a      ay  32.2 ft/s    2

1 2
 y  t   vy t  a y t
2
 y  t    800 ft/s  sin a  t  16.1 ft/s 2  t 2
Example Problem – Rectangular
systems
Substitute for the time and
the location of 2000 ft.

2
 15 s                      15 s 
 cos a    16.1 ft/s   cos a  
2000 ft   800 ft/s  sin a                          2
                          
                                    

Use the substitute that 1/cos2 a = sec2 a = 1+ tan2 a
Example Problem – Rectangular
systems
Rearrange the equation

2000 ft  12000 ft  tan a    3622 ft  1  tan 2 a  
0   3622 ft  tan 2 a   12000 ft  tan a   5622 ft

Solve the quadratic tan a = 0.565 and tan a = 2.75
Example Problem – Rectangular
systems
The solution will be a = 29.5o and a = 70.0o, either
angle will reach the target
Class Problem – Rectangular
systems
A player throws a ball with an initial velocity v0 of 50
ft/s from a point A located 5 ft above the floor.
Knowing that the ceiling of the gymnasium is 20 ft
high, determine the highest point B at which the ball
can strike the wall 60 ft. away.
Polar Coordinates
Using 2-D rectangular coordinates, the location, velocity
and acceleration can be related to point with angle and
radius. So the particle location is given as

rP/O  t   r  t  er

where, O is the location of the reference point.
Polar Coordinates
The velocity component is represented by
d er
r  t   r  t  er  r  t 
dt
The vector displacement can be represented as

d er d er d

dt   d dt
Polar Coordinates
The vector component can be represented as

d er
 e
d
Back substitute

d er d er d

dt    d dt
  e
Polar Coordinates
The velocity component are written as

v  t   r  t  er  r t  e

Alternative representation of the vectors is using
Cartesian coordinates.

er  cos  i  sin  j
e   sin  i  cos  j
Polar Coordinates
Take the derivative of the vector components with
respect to 

d er
  sin  i  cos  j  e
d
d e
  cos i  sin  j  er
d
Polar Coordinates
The acceleration can be written as

d er
r  t   r  t  er  r  t 
dt
d e
 r  t  e  r  t  e  r  t 
dt
 r  t  er  r  t  e
 r  t  e  r  t  e  r  t  er
2
Polar Coordinates
The rearrange the acceleration


a  t   r  t   r  t     2
e r

  r  t    2r  t    e

if r is constant

a  t   r er  r e
2
Example Problem – Polar
systems
The rotation of 3 ft arm OA about O
is defined by the relation  = 0.15t2,
where  is expressed in radians and t
in seconds. Block B slides along the
arm in such a way that it distance
from O is r = 3 – 0.4 t2, where r is
expressed in ft and t in seconds.
Determine the total velocity and total
acceleration of block B after the arm
OA has rotated 30o
Example Problem – Polar
systems
Find out when  = 30o occurs

30
       0.524 rad
180
 0.15t  t  1.869 s
2
Example Problem – Polar
systems
Take the time derivatives of r and 

r  3  0.4t   2
  0.15t 2
r  0.8t            0.3t
r  0.8t            0.3
Example Problem – Polar
systems
The velocity components are

v  t   r  t  er  r t  e

r  0.8 1.869 s   1.495 ft/s

r  3  0.4 1.869 s   1.603 ft
2

  0.31.869 s   0.561 rad/s
Example Problem – Polar
systems
The velocity components are

v 1.869 s   1.495 ft/s er  1.603 ft  0.561 rad/s  e
 1.495 ft/s er  0.899 ft/s e

The v = 1.744 ft/s and b = 31.0o
Example Problem – Polar
systems
The acceleration components are

r  0.8 1.869 s   1.495 ft/s

r  3  0.4 1.869 s   1.603 ft
2

  0.31.869 s   0.561 rad/s
Example Problem – Polar
systems
The velocity components are


a  t   r  t   r  t   e
2
r

  r  t    2r  t    e  
Example Problem – Polar
systems
The velocity components are


a 1.869 s   0.8 ft/s  1.603 ft  0.561 rad/s  er
2                               2

 1.603 ft   0.3 rad/s 2        
                                    e
 2  1.495 ft/s  0.561 rad/s  
                                   
 1.304 ft/s 2 er  1.196 ft/s 2 e
The a = 1.770 ft/s2 and g = 42.5o
Class Problem

Determine the velocity of the rocket
and acceleration in terms of

b, , and  .
Tangential Coordinates
Using 2-D rectangular coordinates, the location, velocity
and acceleration can be related to the path of the particle
radius. So the particle path is s(t), the velocity along the
path is tangent to the instant in time.

s  t   s  t  et
Tangential Coordinates
The acceleration component is represented by
d et
v  t   s  t  et  v  t 
dt
The vector can be represented as

d et d et ds

dt    ds dt
Polar Coordinates
The vector component can be represented as
d et 
    en
ds s
s  

d et d et ds 
          en v
dt    ds dt 
v
 en

Tangential Coordinates
The acceleration component is represented by

v t 
2

a  t   s  t  et             en

Polar Coordinates
The rearrange the acceleration


a  t   r  t   r  t     2
e r

  r  t    2r  t    e

if r is constant

a  t   r er  r e
2
Homework (Due 4/14/03)
Problems:
13-103, 13-109,13-106, 13-112a&b (not c)

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