t. k. ng, hkust lecture iv. mechanics of rigid bodies
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T. K. Ng, HKUSTT. 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 (t) and (t), describing the angular orientation of the dumbbell with respect to a chosen coordinate system (rotation).
0
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
i
ii
ii masstotalmMxmM
X )(,1
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
ext
toti
ii
ii
iii
ii
iii
FFamAM
vmVM
mMxmXM
)(,
We consider motion of center of mass
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
i
iii vrmL
Newton’s Law for rotation
(1)Angular momentum & rotation
''
', )()(
ii
ii
iiii
vrmVRM
vVrRmL
It is convenient to separate the CM motion by writing ri = R + ri’
(where R = CM coordinate)
since i
iirm 0'
Newton’s Law for rotation
(1)Angular momentum & rotation
''
', )()(
ii
ii
iiii
vrmVRM
vVrRmL
It is convenient to separate the CM motion by writing ri = R + ri’
(where R = CM coordinate)
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
'
'
''
totCMtot
iii
tot
ii
ii
frFR
armARMt
L
The rate of change of angular momentum is given by
CM motion Rotation about CM
Newton’s Law for rotation
(1)Angular momentum & torque
0)(
)(
:
''
'''
ijji
j i
extiiij
exti
i
itot
exttot
i
exti
i
iCMtot
j
ijext
ii
frrif
frffr
FRfRfR
fffNotice
Force directed along the line joining the particles
(central force)
Newton’s Law for rotation
(1)Angular momentum & torque
'ext
CMextt
L
i.e., the rate of change of angular momentum is govern by external force only.
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
)0),cos(),sin((
)),sin(),cos((),,(
tirtiriviztirtiriziyixir
vrmLi
iii
Newton’s Law for rotation
Show that
i
ziiz IrmL 2
Notice:
222
sin,cos
),,,(
iii
iiiiii
iiii
yxr
ryrx
zyxx
Newton’s Law for rotation
With external torque
,extzz
zt
It
L
i.e. torque change in rate of rotation
Newton’s Law for rotation
With external torque
,extzz
zt
It
L
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?
i
iiz rmI 2
Examples
What is the moment of inertia for the following objects?
(a) a sphere with uniform density
223
2
0
2
0
22
5
2
3
4
5
2
)(sinsin)(2
MRRR
ddrrrIR
z
Examples
What is the moment of inertia for the following objects?
(b)a cylinder with uniform density
About z-axis
z-axis
length = l
radius=R
Examples
(b)a cylinder with uniform density
About z-axis
222
00
2
2
1
2
1
)(2
MRRlR
dzdrrrIlR
z
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
2
22
2
1
)(2
1
2
1..
z
i iiiii
I
rmvmEK
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
)0(
)(
'2
2'2
i
iiCMCM
i
i
CMi
i
ii
rmIMR
rRmrmI
CM
RCM
I
Example (Physical pendulum):
Newton’s Law for rotation
sin||)(
2
2
CMCM RMgzMgRdt
dI
CM
I
Example (Physical pendulum):
Small angle of oscillation:
CMCM MgRRMg
dt
dI ||
2
2
CM
IOscillation frequency
CMCM
CMCM
IMR
MgR
I
MgR
2
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
22
22
2
1
2
1
mrMR
mvr
I
mvr
mrMRmvr
Solution (1): Let’s check energy change
2
22
22
222
2
1
212
1
2
1
2
1
mvmrMR
mrmv
mrMRE f
Solution (1): the torque due to friction is
gmr
gMRR
rdrr
P
R
MGA
PMGAtot
3
2
3
2)2(
3
0
Newton’s Law implies.
Idt
d tot
Therefore, time taken for MGA to slow down is
.)
3
2(
)0(
mrMRg
mvrIt
tot
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:
.
)0(
rMm
mMm
MmrRCM
Solution (2): conservation of momentum (translational motion of CM)
vMm
mv
vmMmv
f
f
)(
Final velocity
Solution (2): conservation of angular momentum (rotation about CM)
22
222
2
1
2
1)(
)(
rmM
MmMR
MRMRmM
MrmI
ImM
MrmvRrmv
CMCM
CMCM
Final frequency of
rotation about CM
Solution (2): conservation of angular momentum (rotation about CM)
mM
Mm
rMR
v
ImM
mMv
CM
22
2
1)(
Solution (2): conservation of angular momentum (rotation about CM)
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 .
End
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