Transcript Document

Objective
• To investigate particle motion along a curved path “Curvilinear
Motion” using three coordinate systems
– Rectangular Components
• Position vector r = x i + y j + z k
• Velocity
v = vx i + vy j + vz k
• Acceleration a = ax i + ay j +az k
(tangent to path)
(tangent to hodograph)
– Normal and Tangential Components
• Position (particle itself)
• Velocity
v = u ut
u2
a  u u t  u n
• Acceleration

– Polar & Cylindrical Components
1  (dy / dx) 

2 3/ 2
d 2 y / dx 2
(tangent to path)
(normal & tangent)
Curvilinear Motion: Cylindrical
Components
• Section 12.8
• Observed and/or guided from origin or from
the center
r , , and z
• Cylindrical component
r and 
• Polar component “plane motion”
Application: Circular motion but observed
and/or controlled from the center
Polar Coordinates
•
•
•
•
•
•
Radial coordinate r
Transverse coordinate
 and r are perpendicular
Theta  in radians
1 rad = 180o/p
Direction ur and u
Position
• Position vector
• r = r ur
Velocity
• Instantaneous velocity
= time derivative of r
r  r ur
v  r  r u r  r u r
u   u
r

v  r u r  r  u
v  ur u r  u u
• Where
ur  r and u  r 
v r radial velocity
v transvers e velocity
Velocity (con.)
• Magnitude of velocity
u  (r) 2  (r) 2
• Angular velocity 
• Tangent to the path
• Angle =   d
1 u
d  tan ( )
ur
d
Acceleration
• Instantaneous acceleration = time derivative of v
v  r u r  r u
a  v  r u r  r u r  r u  r u  r u 
u r   u
u    u r
a  (r  r 2 )u r  (r   2 r )u
ar  r  r 2
a  r   2 r 
a  ar u r  a u
Acceleration (con.)
•
Angular acceleration
  d 2 / dt 2  d / dt (d / dt )
•
Magnitude
a  (r  r 2 ) 2  (r  2r) 2
•
•
Direction “Not tangent”
Angle   f
a
f  tan ( )
ar
1
f
Cylindrical Coordinates
• For spiral motion
cylindrical coordinates
is used r, , and z.
• Position
rp  r u r  z u z
• Velocity
v  r u r  r u  z u z
• Acceleration
a  (r  r 2 ) u r  (r  2r) u  z u z
Time Derivative to evaluate r, r, , and 
• If r = r(t) and   (t)
r  4t
2
  (8t  6)
3
r  8t
  24 t 2
r  8
  48 t
• If r = f() use chain rule
r  5
2
r  10 
r  10[ ()   () ]
r  10  2  10 
Problem
• The slotted fork is rotating about O at a constant
rate of 3 rad/s. Determine the radial and
transverse components of velocity and
acceleration of the pin A at the instant  = 360o.
The path is defined by the spiral groove r =
(5+/p) in., where  is in radians.
  360o  2p rad

r  5
p
vr  r 
 7 in
  2p
3
p
  3 rad/s
 3
r   in/s
p p
 0.955 in/s
v  r  7(3)  21 in/s
  0 rad/s 2
r  p  0 in/s 2
ar  r  r 2  0  7(3) 2  63 in/s 2
3
a  r  2r  0  2( )(3)  5.73 in/s 2
p
Example 12-20
r  0.5(1  cos  ) ft
v  4 ft/s
a  30 ft/s 2
find  and 
at   180o
r  0.5(1  cos  )
r  0.5( sin  )
r  0.5(cos  )()  0.5( sin  )
at   180o
r  1 ft
r  0
r  -0.5θ 2
u  (r) 2  (r) 2  (0) 2  (1) 2  4
   4 rad/s
a  (r  r 2 ) 2  (r  2r) 2  [0.5(4) 2  1(4) 2 ]2  [1  2(0)( 4)]2  30
  18 rad/s 2
Problem
A collar slides along the smooth vertical spiral rod, r = (2)
m, where  is in radians. If its angular rate of rotation is
constant and equal 4 rad/s, at the instant  = 90o. Determine
- The collar radial and transverse component of velocity
- The collar radial and transverse component of acceleration.
- The magnitude of velocity and acceleration

p
2
rad
r  2
r  2*
p
2
p m
  4 rad / s
r  2
r  2
r  2 * 4  8 m / s
vr  r  8 m/s
v  r  p (4)  12.56 m/s
u  (8) 2  (12.56) 2 
  0 rad / s 2
r  0
ar  r  r 2  0  p (4) 2  50.24 m/s 2
a  r  2r  0  2(8)( 4)  64 m/s 2
a  (50.24) 2  (64) 2 