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Lagrange’s Equation
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Lagrange’s Method
• Newton’s method of developing equations of
motion requires taking elements apart
• When forces at interconnections are not of
primary interest, more advantageous to derive
equations of motion by considering energies in
the system
• Lagrange’s equations:
– Indirect approach that can be applied for other types
of systems (other than mechanical)
– Based on calculus of variations – finding extremums
of quantifies expressible as integrals
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Lagrange’s Equations
d  T  T R U

 


 Qi
dt  qi  qi qi qi
where :
qi : independent coordinates necessary to
describe system's motion at any instant
Q i : corresponding loading in each coordinate
U  f1 (qi ) : potential energy in terms of coordinates
T  f 2 (qi2 ) : kinetic energy in terms of systemmasses,
mass inertias, linear/ang ular veloc ities
R  f 3 (qi2 ) : energy dissipation due to viscous friction
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Deriving Equations of Motion via
Lagrange’s Method
1. Select a complete and independent set of
coordinates qi’s
2. Identify loading Qi in each coordinate
3. Derive T, U, R
4. Substitute the results from 1,2, and 3 into
the Lagrange’s equation.
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Example 11: Spring-Mass-Damper
System
Independent coordinate: q = x
Substitute into Lagrange’s equation:
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Example 12: Pair-Share:
Restrained Plane Pendulum
• A plane pendulum (length
l and mass m), restrained
by a linear spring of
spring constant k and a
linear dashpot of dashpot
constant c, is shown on
the right. The upper end
of the rigid massless link
is supported by a
frictionless joint. Derive
the equation of motion.
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Example 12: Pair-Share:
Restrained Plane Pendulum
q   , Q  0,
1
R  c(a) 2
2
1
1
T  mv 2  m(l) 2
2
2
1
U  mgl  mgl cos   k (b ) 2
2
d T
T R U
(  )
 
Q
dt 
  
ml 2  ca 2  mgl sin   kb2  0
Linearize sin    for small 
ml 2  ca 2  mgl  kb2  0
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Example 13: Bead on a Spinning Wire Hoop
•
•
•
•
One degree of freedom, q1=angle  as an independent coordinate
Velocity of bead: R . ; Velocity of hoop:   R sin  
1 2 1
2
Kinetic energy:
2
T

mv  m R   R sin  
2
2

Potential energy relative to its position at the bottom of
the hoop (when the hoop is not rotating and  = 0), is
U  mgR  mgR cos 
•
•

R = 0, Q = 0
Substitute into Lagrange’s equation:
Solving for the angular acceleration:
R

d T
T R U
(  )
 
Q
dt 
  
mR2  mR2 2 sin  cos   mgR sin   0
•

g

    2 cos    sin  .
R

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Example 14: Pair-Share: Copying machine
•
Use Lagrange’s equation to derive
the equations of motion for the
copying machine example,
assuming potential energy due to
gravity is negligible.
q1=y, q2= θ
Q1 = F, Q2 = 0
θ
y
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Example 14: Pair-Share: Copying machine
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Example 15: Mass Spring Dashpot
Subsystem in Falling Container
• A mass spring dashpot subsystem
in a falling container of mass m1 is
shown. The system is subject to
constraints (not shown) that
confine its motion to the vertical
direction only. The mass m2,
linear spring of undeformed length
l0 and spring constant k, and the
linear dashpot of dashpot constant
c of the internal subsystem are
also shown.
• Derive equation(s) of motion for
the system using
– x1 and x2 as independent coordinates
– y1 and y2 as independent coordinates
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Example 15: Mass Spring Dashpot
Subsystem in Falling Container
Let q1  x1 , q2  x2 be positions of m1 and m 2 with respect to horizontal ground .
Q1, 2  0
1
1
m1 x12  m2 x 22
2
2
1
U  k ( x1  x2 ) 2  m1 gx1  m2 gx2
2
1
R  c ( x1  x 2 ) 2
2
Substitute into Lagrange' s Equation :
T 
d T
T
R
U
(
)


 Q1
dt x1
x1
x1
x1
m1 x1  c ( x1  x 2 )  k ( x1  x2 )  m1 g  0
d T
T
R
U
(
)


 Q2
dt x 2
x2
x 2
x2
m1 x2  c ( x 2  x1 )  k ( x2  x1 )  m2 g  0
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Example 15: Mass Spring Dashpot
Subsystem in Falling Container
Let q1  y1 be inertial position of m1 with respect to horizontal ground
q2  y2 be position of m2 relative to m1 and from unstretche d length of spring
Q1, 2  0
1
1
m1 y12  m2 ( y1  y 2 ) 2
2
2
1 2
U  ky2  m1 g y1  m2 g ( y1  y2 )
2
1 2
R  cy 2
2
Substitute into Lagrange' s Equation :
T
d T
T R U
(
)


 Q1
dt y1
y1 y1 y1
(m1  m2 ) y1  m2 y2  m1 g  m2 g  0
d T
T
R U
(
)


 Q2
dt y 2
y2 y 2 y2
m2 y1  m2 y2  cy 2  ky2  m2 g  0
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Example 16: A Block Sliding on a Wedge
•
•
•
Block mass m sliding down a wedge mass M
Independent coordinates, q1 and q2, are shown, q1 is along the plane and
is measured relative to the (moving) wedge.
q1
Velocity of the wedge is q2 , but
velocity of the block has components from both
q2
α a
q1 and q2:
2
2
2
vblock  vx2  v y2  q2  q1 cos a   q1 sin a 


•
Total kinetic energy is
•
The potential energy of the wedge can be taken as zero, and the block is
1
1
2
2
2

T  Mq2  m q2  q1 cos a   q1 sin a 
2
2
U  mgq1 sin a
•
Substitute into Lagrange’s equations and differentiate wrt to q1 and q2
q1 
g sin a
.
m
1
cos 2 a
M m
q2  
m
q1 cos a .
M m
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Example 17: Pair-Share:
Mass Pendulum Dynamic System
•
A simple plane pendulum of mass m0 and
length l is suspended from a cart of mass m as
sketched in the figure. The motion of the cart
is restrained by a spring of spring constant k
and a dashpot constant c; and the angle of the
pendulum is restrained by a torsional spring of
spring constant k, and a torsional dashpot of
dashpot constant ct. Note that the constitutive
relations for the torsional spring and torsional .
dashpot are linear expressions τ=ktθ and τ=ctθ,
respectively, where τ is the torque and θ is the
change in the angle across the element from its
undeformed configuration.
•
The torsional spring is undeformed when the pendulum is in its downwardhanging equilibrium position. Also, m0 is acted upon by a known dynamic
force F0sinωt, which remains horizontal, regardless of the angle θ and the
motion of the cart. The system is constrained to remain in the plane of the
sketch and the cart remains on the bed throughout its motion. Derive the
equation(s) of motion for the system.
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Example 16: Pair-Share:
Mass Pendulum Dynamic System
Two independent coordinates : q1  x, q2  
Q1  F0 sin t and Q2  ( F0 sin t )l cos 
Rx 
1 2
1
cx and R  ct 2
2
2
1 2 1
kx  kt 2  m0 gl (1  cos  )
2
2
T  Tcart  T pendulum
U
1 2
mx
2
1
1
T pendulum  m0 ( x  l cos  ) 2  (l sin  ) 2  m0 x 2  (l) 2  2 xl cos 
2
2
Substitute into Lagrange' s Equations :
(m  m ) x  cx  kx  m l ( cos    2 sin  )  F sin t
Tcart 

0

0

0
m0l  ct  kt  m0 gl sin   m0 xl cos   ( F0 sin t )l cos 

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Comparison of Newton’s and
Lagrange’s Methods
Newton’s (Direct Approach)
Lagrange’s (Indirect Approach)
Accelerations required
Velocities required
Generally vectors required
Generally scalars required
Free-body Diagrams useful
Free-body diagrams not useful
All forces considered
Workless forces (constraints) forces
not considered
All forces handled via same
expression
Conservative and non-conservative
forces handled separately
Intermediate forces more readily
available
Intermediate forces less readily
available
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Case Study:
Feasibility Study of a
Mobile Robot Design
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Preliminary Design Model
• Read Handout 1
• Develop a simplified model of the base
and the arm by neglecting the reaction
forces that occur at the pivot
• When do the reaction forces become
significant?
=> at high accelerations and speeds of the
base and the arm
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Preliminary Design Model
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Preliminary Design Model
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Motion Profiles
• Read Handout 2
• Plot the motor force versus time, and
motor torque versus time, and determine
whether the motor is powerful enough
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Motion Profiles
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A More Detailed Model
• The simplified model appears feasible
• Develop a detailed model, including the
reaction forces to see whether they
significantly change the results
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A More Detailed Model
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A More Detailed Model
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A More Detailed Model
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Matlab Simulation Example:
Solving Systems of Equations
•
From Matlab website: http://www.mathworks.com/support/tech-notes/1500/1510.html#time
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Matlab Simulation Example:
Solving Systems of Equations
•
From Matlab website: http://www.mathworks.com/support/tech-notes/1500/1510.html#time
•
More on ode45 commands: http://www.mathworks.com/access/helpdesk/help/techdoc/ref/ode45.html
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Homework 3: chapter 3
• 3.14, 3.21,3.24,3.25,3.31,3.33,3.36
• Case study simulation:
– Simulate the detailed model for 10 seconds with initial
conditions x0  0, x0  0, 0  100 , 0  0 and for three
cases:
1. T = 0, f =0
2. T=370 Nm, f = 0
3. T=0, f = 263 N
– For each case, plot state vector versus time
– Describe the behavior of the system for each case
and discuss the stability of the system
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Case Study Simulation Model
I p  mLx cos   T  mgl sin 
or
a  bx  c
where
a  I p ; b   mL cos  ; c  T  mgl sin 
(m  M ) x  mL cos   mL sin  2  f
or
dx  e  g
where d  (m  M ); e   mL cos  ; g   mL sin  2  f
Solve for x and 


2
ag  ec I p  mL sin   f  mL cos  T  mgl sin  
x 

ad  eb
I p m  M   mL cos   mL cos  
c b  ag  ec  T  mgl sin  mL cos 
   


a a  ad  eb 
Ip
Ip


 I p  mL sin  2  f  mL cos  T  mgl sin   






I
m

M

mL
cos


mL
cos



p
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Matlab Simulation
tspan  [0 3]; % simulation time vector
% State vector  y  [ x x   ]'
y 0  [0 0 10 * pi / 180 0]' ;
[t , y ]  ode 45(@ secondode, tspan, y 0);
% function to be integrated
function dy  secondode (t , y )
dy  zeros( 4,1);
x  y (:,1); xdot  y (:,2); phi  y (:,3); phidot  y (:,4);
% Plot state vector against time
figure
subplot ( 221)
plot (t , x )
ylabel (' x, [ m]' )
xlabel (' t , [ s ]' )
% Define parameters
I p  47.5; % kg  m ^ 2
subplot ( 222)
plot (t , xdot )
ylabel (' xdot , [ m / s ]' )
xlabel (' t , [ s ]' )
subplot ( 223)
plot (t , phi )
ylabel (' phi, [ m]' )
xlabel (' t , [ s ]' )
subplot ( 224)
plot (t , phidot )
ylabel (' phidot , [ rad / s ]' )
xlabel (' t , [ s ]' )
L  7 / 8; % m
m  40; % kg
M  100 ; % kg
g  9.81; % m / s ^ 2
T  0;
f  0;
a  I p;
b   m * L * cos( y (3));
c  T  m * g * l * sin( y (3));
d  ( m  M );
e   m * L * cos( y (3));
g   m * L * sin( y (3)) * y ( 4) * y ( 4)  f ;
% State vector  y  [ x x   ]'  [ y (1) y ( 2) y (3) y ( 4)]'
dy (1)  y ( 2);
dy ( 2)  ( a * g  e * c ) /( a * d  e * b);
dy (3)  y ( 4);
dy ( 4)  c / a  (b / a ) * ( a * g  e * c ) /( a * d  e * b);
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References
• Woods, R. L., and Lawrence, K., Modeling and
Simulation of Dynamic Systems, Prentice Hall, 1997.
• Williams, J. H., Fundamentals of Applied Mechanics,
Wiley, 1996
• Close, C. M., Frederick, D. H., Newell, J. C., Modeling
and Analysis of Dynamic Systems, Third Edition, Wiley,
2002
• Lecture notes: web.njit.edu/~gary/430/
• Palm, W. J., Modeling, Analysis, and Control of Dynamic
Systems