A heavy particle is placed at at the top of of a vertical vertical hoop. Calculate the reaction reaction of the hoop on the particle by by means of of Lagrange’s undetermined undetermined multipliers multipliers and Lagrange’s Lagrange’s equations… equations… z = 0 is at the top of the ball. Potential energy and kinetic energy in polar coordinates are, 2
V ( z ) mg z mg r cos V ( r, )
d T 12 m dt ( r ) 12 m r2 ( r ) 2
[I.1]
That makes the Lagrangian,
L T V
1 2
[I.2]
r R 0 f ( r)
[I.3]
m r 2 (r )2 mgr cos
Constraint, expressed as
rR
Lagrange’s undetermined multipliers appear as, m m f d f d f f d d L d L f 1 0 0 [ I.4] 1 qk dt qk d t q q d t d t q qk dt qk 1 1 k k k
For qk r , , and using [I.3], we see that [I.4] becomes the system,
mr
2
mg cos
d dt
mr
mgr sin
d
( r R) r
mr 0 dt
2
2 m(r ) m( r g cos )
2rr r
2
gr sin
[I.5]
[I.6]
Using r R (as agreed in [I.3]), the equations of motion [I.5] and [I.6] become,
mR 2 m(0 g cos )
2 R 0 R 2 gR sin
mR 2 mg co s
R g sin
[I.7]
The second equation of [I.7] is a nonlinear equation for . However, it is not nonlinear but separable for ,
Since is a function f unction of alone (
g R
d dt
d d d dt dt
d
d
d d
[I.8]
R g sin , as in [I.7] itself), we can just integrate as we would a separable ODE,
sin d
d
g R
( cos )
2 1 2
0
[I.9]
Now we have have the angular angular velocity velocity at all all , and we know the angular velocity is zero when cos cos 0 1 and the heavy particle is just sitting at the top of the barrel ( 0 ), and so, 0
R g
2
2 Rg 1 cos
[I.10]
That we know in terms of alone means we can eliminate a time derivative from the first relation in [I.7],
mR 2 R g 1 cos mg cos
2mg 1 cos mg cos
[I.11]
The has units of force, and is, by what Goldstein says on p. 48, the force of reaction by the barrel,
mg cos 2mg cos 1 mg 3cos 2
[I.12]
st The mg cos is the ordinary normal force, and 2mg (cos 1) mR (see [I.7], 1 equation) is the “effective 2
potential” that flings the particle out a bit and acts t o reduce the normal force a bit (note that cos 1 is always negative for 0
2
).
Find the height at which the particle falls off… The particle falls off when the normal force is zero. The height is the z- coordinate. Don’t forget that R z R for this problem. Checking our answer
Consider this geometry,
[I.13]
By this geometry, note that V = 0 at top of hoop. Therefore, T = V, and we have a potential energy of,
V mgR 1 cos T 12 mv 2
[I.14]
We don’t care what v is, only when the particle falls off the hoop. Solve for the quantity
mv2 R
mv2 R
F cent in [I.14],
2mg 1 cos F cent
[I.15]
Particle falls off when centripetal force equals normal force. Normal force at any is,
F N mg cos
[I.16]
Setting [I.15] and [I.16] equal,
Fcent FN
2mg 1 cos mg cos
2 3cos
cos 1( 23 )
[I.17]
Acknowledgements: thanks to allen majewski for [I.13] to [I.17], which provided a useful check of the work up until INSERT REFERENCE.
