Download Lecture Mechanics Rigid Body ppt

T. K. Ng, HKUST
Lecture IV. Mechanics of rigid bodies
Mechanics of Rigid Bodies
• 1) Mathematics description of rigid bodies
• 2)Rigid Body in static equilibrium
• 3)Dynamics of rigid bodies
1). Mathematical Description of
Rigid Bodies
Before discussing the laws of motion for
rigid bodies, first we have to understand how
many ways can a rigid object move?
Example:
how many ways a dumbbell can
move (assuming the solids at the two ends of
the dumbbell can be regarded as point
particles) and how can we describe its motion?
Dumbbell can
(i) move as a whole, and
(ii) rotate upon its center
To describe motion as a whole, need
(i) x (t) (x = position of center for example), and
(ii) angles q (t) and f (t), describing the angular
orientation of the dumbbell with respect to a
chosen coordinate system (rotation).
q
0
f
Exercise!! How can we describe
mathematically the motion of the following
objects:
(i) bowling ball rolling on floor,
(ii) rigid triangle moving in space,
(iii) A piece of rock
(iv) A piece of rubber
Question: how many ways can a general rigid
body move? (or how many mathematical
variables do we need to describe it’s motion)
Answer: Translation (3-degrees of freedom)
+ Rotation (3-degrees of freedom)
(why?)
Therefore, to understand motion of rigid
bodies, we just need to know how Newton’s
Law governs Translation & Rotation!
To describe translation and rotation, we
introduce the notion of center of mass
Recall: center of mass
• The center of mass is a special point in a
rigid body with position defined by

1
X 
M

 mi xi ,
i
(M 
 mi  total mass )
i
This point stays at rest or with uniform
motion when there is no net force acting on
the body
Center of mass (CM)
• For rigid bodies the rest of the body may be
rotating upon this point which acts as the
center of rotation.
• i.e., the motion of a rigid body can be
described as motion of CM (translation) +
rotation of the rest of the body upon CM.
Proof of the above statements
Assumption: We imagine a rigid body as
composed of many small point masses.
mi
ri
Proof of the above statements
Therefore, A rigid body is characterized
by the set {mi,ri}.
mi
ri
Newton’s Law for translation of rigid body
We consider motion of center of mass


MX   mi xi ,
( M   mi )
i
i


MV   mi vi
i


 ext

MA   mi ai   Fi  Ftot
i
i
Newton’s Law for CM motion
Newton’s Law for rotation
(1)Angular momentum & rotation
The angular momentum of a rigid body
defined by a set {mi,ri} is

 
L   mi ri  vi
i
Newton’s Law for rotation
(1)Angular momentum & rotation
It is convenient to separate the CM motion by
writing ri = R + ri’
(where R = CM coordinate)

 ,
 '
 L   mi ( R  ri )  (V  vi )
i
 
' '
 MR  V   mi ri  vi
i
since
'
 mi ri  0
i
Newton’s Law for rotation
(1)Angular momentum & rotation
It is convenient to separate the CM motion by
writing ri = R + ri’
(where R = CM coordinate)

 ,
 '
 L   mi ( R  ri )  (V  vi )
i
 
' '
 MR  V   mi ri  vi
i
CM motion
Rotation about CM
Newton’s Law for rotation
(1)Angular momentum & rotation
CM motion
Rotation about CM
Newton’s Law for rotation
(1)Angular momentum & torque
The rate of change of angular momentum is
given by



'
'
L

 MR  A   mi ri  ai
t
i


 tot
'
 R F
  ri  f i
 CM
  tot
CM motion
'
i
  tot
Rotation about CM
Newton’s Law for rotation
(1)Angular momentum & torque

 ext
fi  fi

Notice :
 CM
 tot
'
 tot 

 R

i
if

i

j


fi  R 
 ext
'
ri ( f i

f ij
 ext

 ext
fi
 R  Ftot



  f )  r

'
'
( ri  r j )  f ij  0
i
ij
j
i
'
 ext
 fi
i
Force directed along the line joining the particles
(central force)
Newton’s Law for rotation
(1)Angular momentum & torque
i.e., the rate of change of angular momentum is
govern by external force only.

 CM
'
L

  ext   ext
t
Notice: internal force is needed to maintain
rigidity of the body
Necessary conditions for a rigid body in
static equilibrium
(a)net force acting on the body=0
(b)net torque acting on the body=0
Q. Are these conditions sufficient?
Necessary conditions for a rigid body in
static equilibrium
Notice: A rigid body can
translate with a uniform velocity +
rotate with a uniform speed about CM
even if external force = external torque = 0
Rigid body in static equilibrium:
stability problem
• Question: Which configuration shown below is
stable? And why?
a
b
Rigid body in static equilibrium:
stability problem
• a is unstable because the orange block will fall if
displaced a little bit away from equilibrium.
a
b
Rigid body in static equilibrium:
stability problem
• Question: A piece of wood with uniform density is
put on the table in two ways (a) and (b). Which
way is more stable? And why?
a
b
Rigid body in static equilibrium:
stability problem
• Ans: (b) is more stable because its center of mass
is lower, and is more difficult to be turned over by
a force. (or its potential energy is lower)
a
b
Sufficient conditions for a rigid body in static
equilibrium :
When the rigid body is displaced
a little bit away from its
equilibrium position, the force it
felt pushes the body back to its
equilibrium position
Sufficient conditions for a rigid body in static
equilibrium :
A rigid body is more stable if it
has a configuration with lower
potential energy.
Examples:
Questions (2) & (3) in
assignment I. Questions (6) & (7)
in last year assignment II + ….
Dynamics of rigid bodies
• a) Rotation about a single axis passing
through CM.
• b) Rotation ….. + translational motion of
CM.
Rotation about a single axis
Example: rotation of an object about a
given axis (z) passing through CM.
Rotation about a single axis
Let the angular speed of rotation be.
Therefore

 
L   mi ri  vi
i

ri  ( xi , yi , zi )  (ri cos(t ), ri sin(t ), zi )

vi  (ri sin( t ), ri cos(t ),0)
Newton’s Law for rotation
Show that
Lz   mi ri2  I z
i
 Notice:
xi  ( xi , yi , zi ),
xi  ri cos i ,
yi  ri sin  i

ri2  xi2  yi2
Newton’s Law for rotation
With external torque
Lz

 Iz
  zext ,
t
t
i.e. torque change in
rate of rotation
Newton’s Law for rotation
With external torque
Lz

 Iz
  zext ,
t
t
Notice, IZ acts as “mass”
(inertia) in Newton’s Law
of translation
Examples
What is the moment of inertia for the
following objects?
(a) a sphere with uniform density?
(b)a cylinder with uniform density?
Iz  
i
2
mi ri
Examples
What is the moment of inertia for the
following objects?
(a) a sphere with uniform density
R
2
0
0
I z  2  r 2 ( r 2 dr )
2
sin
 (sin d )

2  4 3  2
2
 
R   R  MR 2
5 3
5

Examples
What is the moment of inertia for the
following objects?
(b)a cylinder with uniform density
About z-axis
length = l
radius=R
z-axis
Examples
(b)a cylinder with uniform density
About z-axis
R
l
0
0
I z  2  r 2 (rdr )  dz


1
1
2
2

R l R  MR 2
2
2
Question:
For a cylinder with uniform density under
a constant torque. Will it’s angular
acceleration larger if the torque is acting
upon x- or z- directions?
Rotational kinetic energy
1
1
2
K .E.   mi vi   mi (ri ) 2
2 i
2 i
1
 I z 2
2
Example (Physical pendulum):
What determines the oscillation
frequency of a rigid body suspended and
free to swing under it’s own weight
about a fixed axis of rotation?
I
CM
Example (Physical pendulum):
Moment of inertial about fixed point I
I 

2
mi ri 
i

2
MRCM


' 2
mi ( RCM  ri )
i
 I CM
(

'
mi ri  0)
i
I
RCM
CM
Example (Physical pendulum):
Newton’s Law for rotation
I
d 2
dt 2



 RCM  (Mgz )  Mg | RCM | sin 
I
q
CM
Example (Physical pendulum):
Small angle of oscillation:
I
d 2
dt 2

 Mg | RCM |   MgRCM 
Oscillation frequency

MgRCM

I
I
MgRCM
2
MRCM
 I CM
q
CM
Example 1: A person moving with velocity v
and mass m jumps on a merry-go-around
(MGA) which is initially at rest. The mass of
the MGA is M and its radius is R. The person
lands on the MGA at a point x with distance r
from the origin. The velocity v is
perpendicular to the line joining the origin
and x. What is the final angular velocity of the
MGA.
Example 1: A person moving with velocity v
and mass m jumps on a merry-go-around
(MGA) which is initially at rest. The mass of
the MGA is M and its radius is R. The person
lands on the MGA at a point x with distance r
from the origin. The velocity v is
perpendicular to the line joining the origin
and x. What is the final angular velocity of the
MGA. How long will it take for the MGA to
stop if the friction coefficient between ground
and MGA is ?
Solution:
What are the conservation laws that hold in
example (1)?
Energy? Momentum?Angular Momentum?
(N)
(N)
(Y)
Solution (1):
Conservation of angular momentum
1
mvr  MR 2  mr 2
2
mvr
mvr
 

1
I
MR 2  mr 2
2
Solution (1): Let’s check energy change
11
2
2 2
E f   MR  mr 
22


2

1
mr
 mv 2 
2
 1 MR 2  mr 2

2

 1
  mv2
 2


Solution (1): the torque due to friction is
 tot   MGA   P
R
 MGA 

0
2R 3
2
 (2r )rdr 
 gMR
3
3
 P  gmr
Newton’s Law implies
d  tot

.
dt
I
Therefore, time taken for MGA to slow
down is
t 
(  0) I
 tot

mvr
2
g ( MR  mr )
3
.
Example 2: A person moving with velocity v
and mass m jumps on a merry-go-around
(MGA) which is initially at rest. The mass of
the MGA is M and its radius is R. The person
lands on the MGA at a point x with distance r
from the origin. The velocity v is
perpendicular to the line joining the origin
and x. What is the final angular velocity of the
MGA. What happens if the MGA is replaced
by a disc of same mass and is resting on a
frictionless ground?
This is an example of translational motion of
CM + rotation about CM (one axis).
Solution:
We decompose the motion into
(i) Translational motion of CM
(ii)Rotation of the body above CM.
This is an example of translational motion of
CM + rotation about CM (one axis).
Conservation Laws:
Translational motion of CM – conservation
of momentum
Rotation about CM – conservation of
angular momentum
CM position:
RCM
mr  M (0)

mM
m

r.
mM
Solution (2): conservation of momentum
(translational motion of CM)
mv  ( M  m)v f
 vf
m

v
mM
Final velocity
Solution (2): conservation of
momentum (rotation about CM)
angular
 Mr 
mv(r  RCM )  mv
  I CM 
M m
I CM
Mr 2 1
2
 m(
)  MR 2  MRCM
M m
2
1
Mm 2
2
 MR 
r
2
M m
Final
frequency of
rotation about
CM
Solution (2): conservation of
momentum (rotation about CM)
angular
mMv
v
 

1
( M  m) I CM
MR 2  r 2
2
Mm 

 

M m

Solution (2): conservation of
momentum (rotation about CM)
angular
Notice that the system is rotating
about CM, NOT about center of
MGA!
CM
Center of MGA
Example: A dumbbell is formed by 2 equal
masses m joint by a light rod of length l.
Discuss the system’s motion if a force F acts
on the first mass for a very short time t. The
angle between the force and the rod joining
the masses is q.
q
End