Transcript Presentation

Introduction to Analytical
Mechanics
Lagrange’s Equations
Siamak G. Faal
1/1
07/29/2016
1
Rigid body dynamics
Dynamic equations:
• Provide a relation between forces and system accelerations
• Explain rigid body motion and trajectories
Image sources (left to right):
https://www.etsy.com/listing/96803935/funnel-shaped-spinning-top-natural-wood
https://spaceflightnow.com/2016/06/30/juno-switched-to-autopilot-mode-for-jupiter-final-approach/
https://upload.wikimedia.org/wikipedia/commons/d/d8/NASA_Mars_Rover.jpg
http://www.popularmechanics.com/cars/a14665/why-car-suspensions-are-better-than-ever/
2
Kinematics vs Kinetics
Kinematics
Focuses on pure motion of
particles and bodies (without
considering masses and
forces)
Image sources (left to right):
https://www.youtube.com/watch?v=PAHjo4DXOjc
http://www.timelinecoverbanner.com/facebook-covers/billiard.jpg
Kinetics
Relationship between the
motion of bodies and forces
and torques
3
Dynamic Forces and Reactions
Dynamic analysis are essential to determine reaction forces
that are cussed by dynamic forces (e.g. Accelerations)
Image sources (left to right):
http://r.hswstatic.com/w_404/gif/worlds-greatest-roller-coasters-149508746.jpg
http://www.murraymitchell.com/wp-content/uploads/2011/04/red_aircraft_performing_aerobatics.jpg
4
Differential Equations of Motion
provide a set of differential equations which allow
modeling, simulation and control of dynamic
mechanical systems
Why humanoid
robots do not walk
or act like humans
!?
5
Background Information
6
Notations
Throughout this presentation:
Parameters
Definitions
• Vectors are noted with overlines: 𝑎
subscript G
(e.g. 𝑎𝐺 )
The quantity related to
center of mass
• The first derivative with respect to
𝑣
Linear velocity
𝜔
Angular velocity
𝑎
Linear Acceleration
time: 𝑥 =
𝑑𝑥
𝑑𝑡
• The second derivative with respect to
time: 𝑥 =
𝑑2 𝑥
𝑑𝑡 2
Σ𝐹
Resultant force vector
𝑔
Gravitational acceleration
𝑞
Generalized coordinate
7
Time Derivative of a Vector
Time derivative of a vector in rotating coordinate frame is:
𝑑
𝐴 = 𝐴𝑟𝑒𝑙 + 𝜔 × 𝐴
𝑑𝑡
Angular velocity of the
frame
Relative rate of change of
the vector
𝐴
𝜔
8
Newtonian Dynamics
Newton laws of motion:
I.
The velocity of a particle can be changed by the application of a force
II.
The resultant force is proportional to the acceleration of particle with
the factor of mass
Σ𝐹 = 𝑚 𝑎
III.
The forces acting on a body result from an interaction with another
body. The action and reaction forces have the same magnitude, the
same line of action but opposite directions
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Newtonian Equations of Motion
For each rigid body:
Linear momentum of
the center of mass:
𝑑
Σ𝐹 = 𝑝𝐺
𝑑𝑡
𝑝𝐺 = 𝑚𝑣𝐺
Σ𝑀𝐴 = 𝑚𝑟𝐺/𝐴 × 𝑎𝐴 + 𝐻𝐴
Vector from arbitrary point
A to G (center of mass)
Time derivative of the
angular momentum of
the object about point A
Acceleration of the
arbitrary point A
𝐻𝐴 = 𝐼𝐴 𝜔 + 𝜔 × 𝐼𝐴 𝜔
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Newtonian Equations of Motion
A simplified form of the equations are achieved by considering the
following assumptions:
 Mass of the object is constant
 Moment equations are calculated at the center of mass
Σ𝐹 = 𝑚 𝑎𝐺
Σ𝑀𝐺 = 𝐼𝐺 𝜔 + 𝜔 × 𝐼 𝜔
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Energy
Kinetic energy of a rigid body:
1
1
𝑇 = 𝑚 𝑣𝐺 , 𝑣𝐺 + 𝜔, 𝐼𝜔
2
2
Potential energy of an object with mass 𝑚 located at altitude ℎ due to
gravity:
𝑉 = 𝑚𝑔ℎ
Potential energy stored in an spring with stiffness 𝑘:
1
𝑉 = 𝑘Δ𝑥
2
12
Degrees of Freedom
The number of independent parameters that define
system configuration
𝜃
𝜓
In planar motion
In 3D space
𝜙
A free object has
A free object has
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DOF
𝑦
𝑧
𝑥
𝑦
𝜃
3
DOF
𝑥
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Mechanical Joints
Revolute joint
Adds 5 constraints
 Relative DOF: 1
Cylindrical joint
Adds 4 constraints
 Relative DOF: 2
Planar joint
Adds 3 constraints
 Relative DOF: 3
Prismatic joint
Adds 5 constraints
 Relative DOF: 1
Spherical joint
Adds 3 constraints
 Relative DOF: 3
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Generalized Coordinates
A set of geometrical parameters which uniquely defines the
position / configuration of the system
The minimum number of generalized coordinates that required
to specify the position of the system is equal to the number of
DOF of the system (unconstrained or independent coordinates)
Generalized coordinates do not form a unique set!
Generalized coordinates are commonly notes with 𝑞
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Generalized Coordinates
Q) What can be a set of generalized coordinates for a simple
rod in on a plane.
A)
𝑦
𝐵
𝑦𝐴
𝐿
𝜃
𝐴
𝑂
𝑥𝐴
𝑥
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Generalized Coordinates
Q) Can you think about any other generalized coordinates set
for the same system?
A)
𝑦
𝑦𝐵
𝐵
𝜃
𝐿
𝑥𝐵
𝐴
𝑂
𝑥
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Generalized Coordinates
Q) How about another one?
A)
𝑦
𝐵
𝑦𝐴
This is an
ambiguous set of
coordinates!
𝐿
𝑥𝐵
𝐴
𝑂
𝑥𝐴
𝑥
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Generalized Coordinates
Here we have more coordinates than the digress
of freedom of the system, Thus the coordinates
can not have any arbitrary value and they need
to satisfy two constrain equations:
Q) Let’s do one more
𝑥𝐵 − 𝑥𝐴 = 𝐿 cos 𝜃
𝑦𝐵 − 𝑦𝐴 = 𝐿 sin 𝜃
A)
𝑦
𝑦𝐵
𝐵
𝑦𝐴
𝐿
𝜃
𝐴
𝑂
𝑥𝐵
𝑥𝐴
𝑥
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Constraints
Any generalized coordinate which can not have any
arbitrary value is a constrained coordinate
Relation between constrained generalized coordinates
is called constraint equation
𝑀 − 𝑁 : Number of constraint equations
Number of constrained
generalized coordinates
Number of degrees
of freedom
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Virtual Displacement
𝑞3
In a virtual movement, the
generalized coordinates of the
system are considered to be
incremented by infinitesimal amount
δ𝑟
δ𝑞𝑗 from the values they have at an
arbitrary instant, with time held
constant.
𝑞2
𝑀
δ𝑟 =
δ𝑞𝑗 𝑒𝑗
𝑗=1
𝑞1
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Virtual Displacement
𝑥
Example:
Generalized coordinate: 𝑥
𝑙
𝑙
C
Q)
What are the virtual displacements
of pin F and virtual rotation of bar
EF resulting from δ𝑥
B
A
𝑙
𝑙
D
E
𝑙
F
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Virtual Displacement
𝑥
Solution:
𝑥
A
𝜃
𝜃 = cos−1
𝑟𝐹/𝐴
𝑥
2𝑙
𝑙
𝑥
3
= 𝑖 + 4𝑙2 − 𝑥 2
2
2
𝑑
𝑥
δ𝜃 =
cos−1
𝑑𝑥
2𝑙
δ𝑟𝐹/𝐴 =
B
𝜃
𝑙
C
1/2
𝑗
δ𝑥
δ𝑥 = − 2
4𝑙 − 𝑥 2
𝑙
𝑙
D
E
1/2
𝑑
1
3𝑥
𝑟𝐹/𝐴 δ𝑥 = 𝑖 −
𝑑𝑥
2
2 4𝑙2 − 𝑥 2
1/2
𝑙
𝑗 𝛿𝑥
F
𝑦
23
Virtual Work
When a particle is given a virtual displacement δ𝑟, the forces acting on the
particle do virtual work δ𝑊
Since δ𝑟 is infinitesimal, δ𝑊 is infinitesimal as well
Because the change is virtual, time is held constant at an arbitrary value
The virtual work done by constraint or internal forces is zero!
24
Generalized Forces
δ𝑊 =
𝐹 ⋅ δ𝑟
𝑀
→ δ𝑊 =
𝐹⋅
𝑗=1
𝑀
δ𝑊 =
𝑗=1
𝜕𝑟
δ𝑞𝑗
𝜕𝑞𝑗
𝜕𝑟
𝐹⋅
δ𝑞𝑗
𝜕𝑞𝑗
𝑄𝑗 Generalized force
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Joseph-Louis Lagrange
25 January 1736 – 10 April 1813,
Italian Enlightenment Era mathematician
and astronomer. Lagrange made
significant contributions to the fields of
analysis, number theory, and both
classical and celestial mechanics.
Lagrange’s Equations
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Lagrange’s Equations of Motion
𝑑 𝜕𝑇
𝜕𝑇
𝜕𝑉
−
+
= 𝑄𝑗 ,
𝑑𝑡 𝜕𝑞𝑗
𝜕𝑞𝑗 𝜕𝑞𝑗
𝑗 = 1, 2, … , 𝑀
Alternative (common) form:
ℒ =𝑇−𝑉
𝑑 𝜕ℒ
𝜕ℒ
−
= 𝑄𝑗 ,
𝑑𝑡 𝜕𝑞𝑗
𝜕𝑞𝑗
𝑗 = 1, 2, … , 𝑀
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Lagrange’s Equations of Motion
 Directly provides the differential equations of motion
without any need for solving a system of equations.
 Provides a much easier method to solve multi-body systems
 It is very systematic and can be used to automatically
generate differential equations of motion for different
systems (As long as the derivatives are computed correctly)
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Example 1
Find differential equations of motion for the following system
using both Newtonian mechanics and Lagrange’s equations
𝑥
𝑘
𝑚
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Example 1
Newtonian approach:
𝑥
Spring force is equal to: 𝐹𝑠 = −𝑘 𝑥 − 𝑥0
𝑘
𝑚
Assume 𝑥0 = 0
→ 𝐹𝑠 = −𝑘𝑥
Newton’s second law for x-axis:
Free-body diagram:
Σ𝐹 = 𝑚𝑥
Since the only force applied to the system is the spring
force:
𝑥
𝐹𝑠
𝑚
Σ𝐹 = 𝐹𝑠 = −𝑘𝑥
So the differential equation of motion for the system is:
𝑥 + 𝑘𝑥 = 0
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Example 1
Lagrangian approach:
𝑥
1
2
The kinetic energy of the system is: 𝑇 = 𝑚𝑥 2
𝑘
𝑚
1
2
The potential energy of the system is: 𝑉 = 𝑘𝑥 2
Since there is no force acting on the system, the generalized
Calculations:
force is zero: 𝑄 = 0
Lagrangian of the system is:
1
1
ℒ = 𝑚𝑥 2 − 𝑘𝑥 2
2
2
𝜕ℒ
= 𝑚𝑥
𝜕𝑞
𝑑 𝜕ℒ
= 𝑚𝑥
𝑑𝑡 𝜕 𝑞
Thus, the differential equation of motion is:
𝑑 𝜕ℒ
𝜕ℒ
−
= 0 → 𝑚𝑥 − 𝑘𝑥 = 0
𝑑𝑡 𝜕 𝑞𝑗
𝜕𝑞𝑗
𝜕ℒ
= −𝑘𝑥
𝜕𝑞
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Example 2
Find differential equations of motion for the following system
using both Newtonian mechanics and Lagrange’s equations
𝑥
𝑘
𝑚
𝑢
32
Example 2
Newtonian approach:
𝑥
𝑘
Spring force is equal to: 𝐹𝑠 = −𝑘 𝑥 − 𝑥0
𝑚
𝑢
Assume 𝑥0 = 0
→ 𝐹𝑠 = −𝑘𝑥
Newton’s second law for x-axis:
Free-body diagram:
Σ𝐹 = 𝑚𝑥
The summation of forces along x axis is:
𝑥
𝐹𝑠
𝑚
𝑢
Σ𝐹 = 𝑢 − 𝑘𝑥
So the differential equation of motion for the system is:
𝑥 + 𝑘𝑥 = 𝑢
33
Example 2
Lagrangian approach:
𝑥
1
The kinetic energy of the system is: 𝑇 = 𝑚𝑥
2
𝑘
2
𝑚
1
The potential energy of the system is: 𝑉 = 𝑘𝑥
2
2
The generalized force acting on the system is:
𝜕𝑟
𝜕𝑥
Σ𝐹 ⋅
=𝑢⋅
=𝑢
𝜕𝑞𝑗
𝜕𝑥
Lagrangian of the system is:
1
1
ℒ = 𝑚𝑥 2 − 𝑘𝑥 2
2
2
Thus, the differential equation of motion is:
𝑑 𝜕ℒ
𝜕ℒ
−
= 𝑢 → 𝑚𝑥 + 𝑘𝑥 = 𝑢
𝑑𝑡 𝜕𝑞𝑗
𝜕𝑞𝑗
Calculations:
𝜕ℒ
= 𝑚𝑥
𝜕𝑞
𝑑 𝜕ℒ
= 𝑚𝑥
𝑑𝑡 𝜕 𝑞
𝜕ℒ
= −𝑘𝑥
𝜕𝑞
34
Example 3
Find differential equations of motion for the following system
using both Newtonian mechanics and Lagrange’s equations
𝑦
𝑙
𝑔
𝑚, 𝐼
𝜏
𝜃
𝑥
35
Example 3
𝑔
𝑚, 𝐼
𝜏
Moment of inertia of a rod:
1 2
𝑚𝑙
3
𝑙
𝑦
Newtonian approach:
𝜃
1
𝑚𝑙 2
12
𝑥
Free-body diagram:
𝑦
𝑚𝑔
Moment equation around point O gives us:
1 2
𝑙
𝑚𝑙 𝜃 = 𝜏 − 𝑚𝑔 cos 𝜃
3
2
𝐹𝑦
𝑙
2
𝜃
𝑥
𝑂
𝐹𝑥
36
Example 3
𝑦
Lagrangian approach:
Center of mass position in terms of generalized coordinates
𝑙
𝑙
cos 𝜃 𝑖 + sin 𝜃 𝑗 →
2
2
𝑙
𝑙
= − 𝜃 sin 𝜃 𝑖 + 𝜃 cos 𝜃 𝑗
2
2
𝑟𝑐𝑚 =
𝑟𝑐𝑚
𝑙
𝑔
𝑚, 𝐼
𝜃
𝑥
The kinetic energy of the system is:
𝑇=
=
1
1
1
1
𝑚 𝑟𝑐𝑚 , 𝑟𝑐𝑚 + 𝜔 , 𝐼 𝜔 = 𝑚𝑙 2 𝜃 2 +
𝑚𝑙2 𝜃 2
2
2
8
24
1 2 2
𝑚𝑙 𝜃
6
The potential energy of the system is:
𝑙
𝑉 = 𝑚𝑔 𝑟𝑐𝑚 ⋅ 𝑗 = 𝑚𝑔 sin 𝜃
2
37
Example 3
𝑦
Lagrangian approach (cont.):
The generalized force is
𝑄 = Σ𝐹 ⋅
𝑙
𝑔
𝑚, 𝐼
𝜃
𝜕𝑟
𝜕𝜃
=𝜏
=𝜏
𝜕𝑞𝑗
𝜕𝜃
𝑥
Lagrangian of the system is:
Calculations:
1
𝑙
ℒ = 𝑚𝑙 2 𝜃 2 + 𝑚𝑔 sin 𝜃
6
2
𝜕ℒ 1 2
= 𝑚𝑙 𝜃
𝜕𝑞 3
Thus, the differential equation of motion is:
𝑑 𝜕ℒ
𝜕ℒ
−
=𝜏
𝑑𝑡 𝜕𝑞𝑗
𝜕𝑞𝑗
1 2
𝑙
→ 𝑚𝑙 − 𝑚𝑔 cos 𝜃 = 𝜏
3
2
𝑑 𝜕ℒ
1
= 𝑚𝑙2 𝜃
𝑑𝑡 𝜕 𝑞
3
𝜕ℒ
𝑙
= 𝑚𝑔 cos 𝜃
𝜕𝑞
2
38
References
[1] Ginsberg, Jerry H. Advanced engineering dynamics. Cambridge
University Press, 1998.
[2] Greenwood, Donald T. Advanced dynamics. Cambridge University
Press, 2006.
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