PHYS 705: Classical Mechanics. Examples: Lagrange Equations and Constraints

Transcription

1 1 PHYS 705: Classical Mechanics Examples: Lagrange Equations and Constraints

2 x R Note: an object rolls because of friction but static friction does no work this is different from our previous case with a disk rolling on a D plane. This has 1 less dof Pick the coordinates as shown. The constraint eq (rolling without slipping) is: x, xr 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.

3 3 x R m l 1 1 T mx I I( hoop) mr 1 1 T mx mr R x (constraint) T mx Now, pick U=0 to be at where the hoop is at the bottom of the incline plane, So, we then have, U mg( lx)sin Lmx mg( lx)sin Lagrange Equation gives, mx mg sin 0 x g sin (Correct acceleration for a hoop rolling down an incline plane)

4 4 y In this case, we need to go back to Newtonian mechanics to mg sinxˆ get the constraint force: F c x The constraint force is the static friction F c needed to keep the hoop rolling without slipping. Newton nd law gives, mg sin F mx F mg sin mx Plug in our result for and we get, x c c F c mg sin mg sin mgsin F c mg sin x ˆ

5 5 x R Using both coordinates : x and We have one holonomic constraint and we will have one Lagrange multiplier. g x m l (for x eq) #: Now without explicitly eliminating one of the coordinates using the constraint equation, we will use Lagrange Equation with Lagrange multipliers to get both the EOM and the magnitude of the constraint force. g x, xr 0 The relevant terms to be included in the Lagrange equation are: and g R (for eq)

6 6 1 1 T mx mr U mg( lx)sin 1 1 L mx mr mg lx ( )sin The EOM are: x d L L g 0 dt x x x mx mg sin (1) d L L g dt mr R mr () 0 x,, and We have three unknowns: to be solved here. xr 0 (3) Together with the constraint equation these system of equations can be solved. (Note: Constraint Eq is applied after EOM is obtained! )

7 7 Combining Eqs (1) and () 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 g sin x (same EOM for x as previously) Now, we can substitute this back into Eq (1) to solve for, mg sin mg sin mx mg sin mg sin

8 8 The magnitude of the force of constraint corresponding to the x-eom is given by: Q x g x mgsin By the way, we can also get the EOM for the variable, mr g sin R mg sin Notice that there is another force of constraint (the normal force : ). 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. F mgcos N

9 9 Mass Rolling off from a Hemispheric Surface Problem: A mass sits on top of a smooth fixed a r U 0 hemisphere with radius a. Find the force of constraint and the angle at which it flies off the sphere. r and Use coordinates: and constraint: gr (, ) ra0 1 m T mv r r U mgrcos m LT U r r mgr note: v r rˆ r θˆ cos g r g 0

10 Mass Rolling off from a Hemispheric Surface m L r r mgr 10 cos r d L L g 0 dt r r r d mr mr mg cos 0 dt cos (1) mr mr mg d L L g 0 dt d mr mgr sin 0 dt mr mrr mgr sin 0 r r gsin 0 () r a and r r 0 Inserting constraint:, we have ma mg cos (1') a g sin 0 g sin (') a NOTE: To find force of constraint, insert constraint conditions AFTER you have gotten the E-L equation (with the multiplier) already.

11 11 Mass Rolling off from a Hemispheric Surface Note that d dt Substituting from Eq ( ) into the above equation, we have, d ( ) g sin g sin g dt a a a g d( ) dcos a Integrating both sides, we arrive at the EOM for, g cos C a g 1cos a g a d cos dt C is an integration constant. Assuming initial condition with, we have sin C g a (0) (0) 0

12 1 Mass Rolling off from a Hemispheric Surface Plugging the last expression into Eq (1 ), we have g ma 1 cos mg cos a mg mg cos mg cos ma mg cos (1') mg(3cos ) (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 ) 0 c c 1 cos 3 48.