1
PHYS 705: Classical Mechanics
Examples: Lagrange Equations
and Constraints
2
Hoop Rolling Down an Incline Plane
R
Note: • an object rolls because of friction but static

x
friction does no work

• this is different from our previous case with a
disk rolling on a 2D plane. This has 1 less dof
Pick the coordinates x,  as shown. The constraint eq (rolling without
slipping) is: x  R  0
We will solve this problem in two ways:
#1: The problem really has one “proper” generalized coordinate x and we will
explicitly use the constraint equation to eliminate  from our analysis. The EOM
is simpler (1D) but we can’t get an expression for the constraint force.
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Hoop Rolling Down an Incline Plane
R
m

x
1 2 1 2
mx  I
I (hoop )  mR 2
2
2
1
1
T  mx 2  mR 2 2
R  x (constraint)
2
2
T  mx 2
T
l

Now, pick U=0 to be at where the hoop is at the bottom of the incline plane ,
we then have,
So,
U  mg (l  x) sin 
L  mx 2  mg (l  x) sin 
Lagrange Equation gives,
2mx  mg sin   0  
x
g sin 
2
(Correct acceleration for a
hoop rolling down an
incline plane)
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Hoop Rolling Down an Incline Plane
y
In this case, we need to go back to Newtonian mechanics to
mg sin  xˆ
Fc
get the constraint force:
x


The constraint force is the static friction Fc needed
to keep the hoop rolling without slipping.

Newton 2nd law gives, mg sin   Fc  mx
 Fc  mg sin   mx
Plug in our result for x and we get,
mg sin  mg sin 

Fc  mg sin  
2
2
Fc  
mg sin 
xˆ
2
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Hoop Rolling Down an Incline Plane
R
#2: Now without explicitly eliminating one of
m

x
the coordinates using the constraint equation,
l
we will use Lagrange Equation with Lagrange

multipliers to get both the EOM and the
magnitude of the constraint force.
Using both coordinates : x and 
We have one holonomic constraint g  x,    x  R  0 and we will
have one Lagrange multiplier .
The relevant terms to be included in the Lagrange equation are:
g

  (for x eq)
x
and

g
  R (for  eq)

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Hoop Rolling Down an Incline Plane
1 2 1
mx  mR 2 2
2
2
U  mg (l  x) sin 
T
L
1 2 1
mx  mR 2 2  mg (l  x) sin 
2
2
The EOM are:
x
d  L  L
g


0

 
dt  x  x
x
mx  mg sin    (1)

d  L  L
g



0
 
dt    

mR 2   R
mR   (2)
We have three unknowns: x,  , and  to be solved here.
Together with the constraint equation
x  R  0 (3) these system of
equations can be solved. (Note: Constraint Eq is applied after EOM is obtained! )
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Hoop Rolling Down an Incline Plane
Combining Eqs (1) and (2) by eliminating , we have,
mx  mg sin   mR
Now, from Eq (3), we have 
x  R
Substituting this into the equation above, we have,
mx  mg sin   mx

x
g sin 
2
(same EOM for x as previously)
Now, we can substitute this back into Eq (1) to solve for ,
  mx  mg sin  
mg sin 
mg sin 
 mg sin   
2
2
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Hoop Rolling Down an Incline Plane
The magnitude of the force of constraint corresponding to the x-EOM
is given by:

g mg sin 

2
x
By the way, we can also get the EOM for the  variable,
mg sin 
mR    
2
g sin 


2R
Notice that there is another force of constraint (the normal force : FN  mg cos  ).
We could get that out by introducing another “improper” coordinate y that permits
motion normal to the incline plane and imposing the constraint y=0.
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Mass Rolling off from a Hemispheric Surface
Problem: A mass sits on top of a smooth fixed
a

hemisphere with radius a. Find the force of
r
U 0
constraint and the angle at which it flies off the
sphere.
Use coordinates: r and  and constraint: g (r ,  )  r  a  0
1 2 m 2 2 2
T  mv 
r  r 
2
2


note: v  rrˆ  rθˆ

g

r

g
0

U  mgr cos 
L  T U 


m 2 2 2
r  r   mgr cos 
2
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L


m 2 2 2
r  r   mgr cos 
2
Mass Rolling off from a Hemispheric Surface
r

d  L  L
g

0
  
dt  r  r
r
d
 mr   mr2  mg cos     0
dt
mr  mr 2  mg cos    (1)
d  L  L
g


0

 
dt    

d
mr 2  mgr sin   0
dt
mr 2  2mrr  mgr sin   0
r  2r  g sin   0 (2)


r  0 , we have
Inserting constraint: r  a and r  
ma 2  mg cos    (1')
a  g sin   0
 
g
sin 
a
(2 ')
NOTE: To find force of constraint, insert constraint conditions AFTER you have
gotten the E-L equation (with the multiplier) already.
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Mass Rolling off from a Hemispheric Surface
Note that
 
d 2

  2
dt
 
g
sin 
a
Substituting  from Eq (2’) into the above equation, we have,
2 g d  cos  
d ( 2 )
g
 2g
 2  sin   
sin   
dt
a
dt
a
 a
2g
2

d ( )  
d  cos  
a
Integrating both sides, we arrive at the EOM for ,
2g
cos   C
a
2g
 2 
1  cos  
a
 2  
C is an integration constant. Assuming
initial condition with(0)   (0)  0 , we
have C  2 g a
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Mass Rolling off from a Hemispheric Surface
ma 2  mg cos    (1')
Plugging the last expression into Eq (1’), we have
 2g

m a 
1  cos     mg cos   
 a

2mg  2mg cos   mg cos   
  mg (3cos   2)
(This gives the mag of the constraint force.)
The particle flies off when the constraint force = 0. By setting =0, we have
the condition,
mg (3cos  c  2)  0
 c  cos 1  2 3  48.2