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PHYS 705: Classical Mechanics Examples: Lagrange Equations and Constraints

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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

10 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