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 Review of Planar Kinematics and Kinetics
 General Features of Planar (2-D) motion of a rigid body
1. Translation (No rotation)

 
r , v , and a : 3-D vectors but only 2 components change
2. Rotation
Motion of any point P in a rigid body: Restricted on a circle


Directions of  and  : Fixed (Normal to the plane of rotation)
3. General motion = Translation + Rotation
 Kinematics

r  xiˆ  yˆj  zkˆ
Define the object’s position:
 dx
dy ˆ dz ˆ
 Find the velocity (Time derivative of displacement) v  iˆ 
j k
dt
dt
dt
 Find the acceleration (Time derivative of velocity)
dv y
 dv
ˆj  dvz kˆ
a  x iˆ 
dt
dt
dt
 Kinetics
Find all forces acting on the object.
 These forces generate the acceleration along the direction of force
Fx  ma x
Fy  ma y
Fz  ma z
or
M x  I x
M y  I y
M z  I z
1. 2D Kinematics of a rigid body
- How to determine Velocity and Accel. of a point in the body
 Translation
A
Rectilinear
B
Curvilinear
Position
Velocity

 
rB  rA  rB / A
 
- rA (rB ) = Position vector of point A (B) in the body

- rB / A = Relative-position vector of B with respect to A



drB d  
drA drB / A

vB 
 (rA  rB / A ) 

dt dt
dt
dt


 vB  v A
because of a Rigid body


Acceleration  aB  a A
All the points on the body have the same motion!
 Rotation about a fixed axis (Polar coordinate system)

(1) Position of a point P in the body: r  rrˆ
(2) Velocity of a point P
because of a rigid body

drˆ
 
 dr dr
v
 rˆ  r  rrˆ  rˆ  rˆ    r
dt dt
dt
d
  (angular speed)
where  
dt


Direction  // d // zˆ
(3) Acceleration of a point P
1
2



 dv d   d   dr     
a
 (  r ) 
r 
   r    (  r )
dt dt
dt
dt
d d d 2


where  
(angular acceleration)
dt
dt dt 2


Direction  //  (Acceleration) or  //   (Deceleration)

1. Tangential comp. at  zˆ  rrˆ  rˆ
2. Normal comp.
(Faster and slower rotation)
    

an    v    (  r )  r 2rˆ
(Centripetal)


Note: Velocity ( v ) & Accel. ( a ): Motion of a point mass P in the body


Angular vel. (  ) & Angular accel. (  ): Motion of a whole body
 General Plane Motion (= Translation + Rotation)
Analysis Method:
Step 1. Set a Fixed reference frame (Origin O)
Step 2. Set a Translating reference frame (Origin A in the body)
Step 3. Separate General motion of a point B of interest into
= Translation of A + Relative motion (Rotation) of B about A

 
(1) Position of B: rB (Arbitrary point in the body)  rA  rB / A



 
drB drA drB / A 



 v A    rB / A
(2) Velocity of B: vB 
dt
dt
dt
= Translation of A + Rotation of B about A


dvB dv A d  


 (  rB / A )
(3) Acceleration of B: a B 
dt
dt dt
 
  

 a A    rB / A    (  rB / A )
2. 2D Kinetics of a rigid body
- How to establish Newton’s equations of motion
 Equations of motion

(1) Translation – Effect of Forces [Mass (m) and Acceleration ( a )]


 F  ma G
: 2 equations (2D planar motion)
(2) Rotation – Effect of Moment (torque)

[Moment of inertia (I) and angular acceleration (  )]


 M  I
: 2 equations
 Finding Moment of inertia (I )
- Dependant to the Body shape & the Axis of rotation.
I   mi ri2 (Discrete)
i
or
I =  r 2 dm or  r 2 dV (Continuous)
m
V
Parallel-Axis Theorem: I  I G  md 2
where IG = Moment of inertia about the axis passing through
the mass center G
d = Perpendicular distance between two parallel axes
(See the back cover of textbook for typical examples of I.)
 Work and Energy
1
1
Kinetic energy: T  mvG2  I G 2
2
2
Potential energy: VF   F cos  ds


( = Angle between F and ds )
= Negative of Work of a Force (U F )
 Special examples
U F  FC cos  ( s2  s1 ) : Constant force
UW  Wy  mgy : Gravitational force
1
U S  ks 2
2
: Spring force
 Principle of Work and Energy
Tiniital  U initial  final  T final
U initial  final  T final  Tinitial
 T
: Total work done by all the external forces on the body
= Difference in Kinetic energy before and after applying the force.
 Conservation of (Mechanical) Energy (For a conservative force)
Ti  Vi  T f  V f
or T f  Ti  (V f  Vi )
or T  V
 Impulse (How fast does the momentum change?)
Momentum



Linear momentum: L   mi vi  mvG
i
Angular momentum:
H G  I G  (about an axis passing through G)
 Principle of Impulse and Momentum
 d

 F  dt (mvG ) →


d
M
 G  ( I G ) →
dt
t2



  Fdt (Linear impulse)  m(vG ) 2  m(vG )1
t1
t2

M
  G dt (Angular impulse)  I G2  I G1
t1
 Conservation of momentum
t2
If

F
  dt = 0 
mvG initial = mvG  final
t1
t2
If

M
  dt = 0 
IG initial
IG  final
t1
For a momentum change;
Over a short (long) time period

Large (small) force felt by a body
e.g. Egg falling on hard floor or carpet