Capítulo 2. Revisión de la literatura 8
2.5 Implementación del ABC/M 41
in the Lagrange-
Euler equations 660. The following examples, 616−619 involve one-sided constraints which exhibit holonomic behavior for restricted ranges of the constraint surface in coordinate space, and this range is case specific. When the forces of constraint press the object against the constraint surface, then the system is holonomic, but the holonomic range of coordinate space is limited to situations where the constraint forces are positive. When the constraint force is negative, the object flies free from the constraint surface. In addition, when the frictional force where is the static coefficient of friction, then the object slides negating any rolling constraint that assumes static friction.
6.16
Example: Mass sliding on a frictionless spherical shell
Masssliding on frictionless cylinder of radius.
Consider a mass starts from rest at the top of a frictionless fixed spherical shell of radius . The questions are what is the force of constraint and determine the angle at which the mass leaves the surface of the spherical shell. The coordinates shown are the obvious generalized coordinates to use. The constraint will not apply if the force of constraint does not hold the mass against the surface of the spherical shell, that is, it is only holonomic in a restricted domain. The Lagrangian is =1 2 ³ ˙ 2+2˙2´−cos
This Lagrangian is applicable irrespective of whether the constraint is obeyed, where the constraint is given by
( ) =−= 0
For the restricted domain where this system is holonomic, it can be solved using generalized coordinates, generalized forces, Lagrange multipliers, or Newtonian mechanics as illustrated below.
Minimal generalized coordinates:
The minimal number of generalized coordinates reduces the system to one coordinate , which does not determine the constraint force that is needed to know if the constraint applies. Thus this approach is not useful for solving this partially-holonomic system.
Generalized forces:
The radial constraint has a corresponding generalized force . The Lagrange equation Λ= gives
¨+cos−˙2= (a)
The Lagrange equationΛ== 0since there is no tangential force for this frictionless system. Therefore
2¨−sin+ 2˙˙= 0 (b) When constrained to follow the surface of the spherical shell, the system is holonomic, i.e. = and
˙
= ¨= 0. Thus the above two equations reduce to
cos−˙2 = (c) 2¨−sin = 0 That is ¨ = sin Integrate to get ˙ using the fact that
¨ = ˙ = ˙ ˙ then Z ¨ = Z ˙ ˙ = Z sin Therefore ˙ 2= 2 (1−cos) (d)
assuming that ˙ = 0at = 0Substituting equation ()into equation () gives the constraint force, which is normal to the surface, to be
==(3 cos−2)
Note that == 0 when cos= 23, that is = 482
Lagrange multipliers:
For the holonomic regime, which obeys the constraint, ( ) =−= 0 the Lagrange equation for is Λ=Since = 1 then
¨+cos−˙2= (a) The Lagrange equation for gives ∆= = 0 since = 0 Thus
2¨−sin+ 2˙˙= 0 (b) As above, when constrained to follow the surface of the spherical shell, the system is holonomic = and ˙= ¨= 0 Thus the above two equations reduce to
cos−˙2 = (c) 2¨−sin = 0 (d) That is, the answers are identical to that obtained using generalized forces, namely;
˙
2= 2
(1−cos) (d)
assuming that ˙ = 0at = 0
The force of constraint applied by the surface is =
Substituting equation () into equation () gives
==(3 cos−2)
Note that = 0when cos=23, that is = 482
Both of the above methods give identical results and give that the force of constraint is negative when 482Assuming that the surface cannot hold the mass against the surface, then the mass willfly off the
spherical shell when 482 and the system reduces to an unconstrained object falling freely in a uniform
gravitationalfield, which is holonomic, that is == 0Then the equations of motion ()and ()reduce
to
¨+cos−˙2 = 0 (e) 2¨−sin+ 2˙˙ = 0 (f)
Energy conservation:
This problem can be solved using energy conservation
1 2
2=[1
−cos]
Thus the centripetal acceleration
2
= 2[1−cos]
The normal force to the surface will cancel when the centripetal acceleration equals the gravitational acceler- ation, that is, when
2
= 2[1−cos] =cos This occurs when cos= 2
3. This is an unusual case where the Newtonian approach is the simplest.
6.17
Example: Rolling solid sphere on a spherical shell
Disk of mass, radiusrolling on a cylindrical surface of radius. This is a similar problem to the prior one with the added
complication of rolling which is assumed to move in a vertical plane making it holonomic. Here we would like to determine the forces of constraint to see when the solid sphereflies off the spherical shell and when the friction is insufficient to stop the rolling sphere from slipping.
The best generalized coordinates are the distance of the center of the sphere from the center of the spherical shell, and It is important to note that is measured with respect to the vertical, not the time-dependent vector r. That is, the direction of the radius is which is time dependent and thus is not a useful reference to use to define the angle . Let us assume that the sphere is uniform with a moment of inertia of =
2 5
2If the tangential frictional force is less than the limiting value , with 0 then the sphere will roll without slipping on the surface of the cylinder and both constraints apply.
Under these conditions the system is holonomic and the solution is solved using Lagrange multipliers and the equations of constraint are the following:
1) The center of the sphere follows the surface of the cylinder 1=−−= 0 2) The sphere rolls without slipping
The kinetic energy is = 12³˙2+2˙2´+12˙2 and the potential energy is =cosThus the Lagrangian is = 1 2 ³ ˙ 2+2˙2´+1 2˙ 2 −cos
Consider the solution using Lagrange multipliers for the holonomic regime where both constraints are satisfied and lead to the following differential constraint relations
1 = 1 1 = 0 1 = 0 2 = 0 2 = 2 =−(+) The Lagrange operator equation Λgives,
˙ − =1 1 +2 2 that is ¨+cos−˙2=1 (a) Λ gives 2¨+ 2˙˙−sin=−2(+) (b) Λ gives ¨=2 (c)
Since the center of the sphere rolling on the spherical shell must have =+ then ˙ = ¨= 0 ¨ = ¨ Substituting this into ()gives
¨
=
2
2 Insert this into equation ()gives
2= sin ¡ +22 ¢ The moment of inertia about the axis of a solid sphere is = 2
5 2 Then 2=2sin 7 But also ¨ =˙˙ = 2 2= 5 22= 5sin 7 Integrating gives Z ˙ ˙ = 5 7 Z sin That is ˙ 2= 10 7 (1−cos) assuming that ˙ = 0at = 0Inserting this into equation ()gives
−10
That is
1=
7 [17 cos−10]
Note that this equals zero when
cos=10 17
For larger angles 1 is negative implying that the solid sphere will fly off the surface of the spherical shell. The sphere will leave the surface of the cylinder when cos=1017 that is,= 5397This is a significantly larger angle than obtained for the similar problem where the mass is sliding on a frictionless cylinder because the energy stored in rotation implies that the linear velocity of the mass is lower at a given angle for the case of a rolling sphere.
The above discussion has omitted an important fact that, if ∞ the frictional force becomes insufficient to maintain the rolling constraint before = 5397 that is, the frictional force will exceed
the sliding limit . To determine when the rolling constraint fails it is necessary to determine the frictional torque
=−2
Thus
=−2
It is in the negative direction because of the direction chosen for The required coefficient of friction is given by the ratio of the frictional force to the normal force, that is
= 2
1
= 2 sin [17 cos−10]
For = 1 the disk starts to slip when = 47540 Note that the sphere starts slipping before it flies off
the cylinder since a normal force is required to support a frictional force and the difference depends on the coefficient of friction. The no-slipping constraint is not satisfied once the sphere starts slipping and the frictional force should equal 1Thus for the angles beyond 4754the problem needs to be solved with the rolling constraint changed to a sliding non-conservative frictional force. This is best handled by including the frictional force and normal forces as generalized forces. Fortunately this will be a small correction. The friction will slightly change the exact angle at which the normal force becomes zero and the system transitions to free motion of the sphere in a gravitationalfield.
6.18
Example: Solid sphere rolling plus slipping on a spherical shell
Consider the above case when the frictional force is insufficient to constrain the motion to rolling. Now the frictional force is given by
= when is positive.
This can be solved using generalized forces with the previous Lagrangian. Then ˙ − == which gives ¨+cos−˙2= Similarly Λ==−(+)gives 2¨+ 2˙˙−sin=−(+) Similarly Λ= = gives ¨=
These can be solved by substituting the relation = . The sphere flies off the spherical shell when ≤0leading to free motion discussed in example 62. The problem of a solid uniform sphere rolling inside a hollow sphere can be solved the same way.
6.19
Example: Small body held by friction on the periphery of a rolling wheel
M N F O m y x xSmall body of massheld by friction on the periphery of a rolling wheel of mass and radius. Assume that a small body of mass is bal-
anced on a rolling wheel of mass and radius as shown in the figure. The wheel rolls in a vertical plane without slipping on a horizontal surface. This example illustrates that it is possi- ble to use simultaneously a mixture of holonomic constraints, partially-holonomic constraints, and generalized forces.3
Assume that at = 0 the wheel touches the floor at = = 0 with the mass perched at the top of the wheel at = 0. Let the frictional force acting on the massbe and the reaction force of the periphery of the wheel on the mass be . Let ˙ be the angular velocity of the wheel, and ˙ the horizontal velocity of the center of the wheel. The polar coordinates of the mass are taken with measured from the center of the wheel with measured with respect to the vertical. Thus the cartesian coordinates of the small mass are (+sin +cos)with respect to the origin at == 0.
The kinetic energy is given by
=1 2˙ 2+1 2˙ 2+1 2 ∙³ ˙
+˙cos+˙sin´2+³˙cos−˙sin´2 ¸
The gravitational force can be absorbed into the scalar potential term of the Lagrangian and includes only the potential energy of the mass since the potential energy of the rolling wheel is constant.
= +(+cos)
Thus the Lagrangian is
=1 2(+)˙ 2+1 2˙ 2+1 2 h
2˙2+ 2˙˙cos+ 2˙˙sin+˙2i−(+cos)
The equations of constraints are:
1) The wheel rolls without slipping on the ground plane leading to a holonomic constraint: 1=−=˙−˙ = 0
2) The mass is touching the periphery of the wheel, that is, the normal force 0This is a one-sided restricted holonomic constraint.
2=−= 0
3) The mass does not slip on the wheel if the frictional force . When this restricted
holonomic constraint is satisfied, then
3=˙−˙ = 0
The rolling constraint is holonomic, and can be accounted for using one Lagrange multiplier plus the
differential constraint equations
1 = 1 1 = 0 1 = 1 = 0
The other two constraints are non-holonomic, and thus these constraint forces are expressed in terms of two generalized forces and that are related to the tangential force and radial reaction force . For
simplicity, assume that the wheel is a thin-walled cylinder with a moment of inertia of = 2
The Euler-Lagrange equations for the four coordinates are
−
³
(+)˙+˙cos+˙sin´++ = 0 (Λ)
˙˙sin+˙˙cos−sin− ³ 2˙+˙cos´+ = 0 (Λ) − ¡ 2˙¢− = 0 (Λ) −cos− (˙sin+˙) + = 0 (Λ) The generalized forces can be related to and using the definition
=F()·
r
where()is the vectorial sum of the forces acting at The components of vector= (+sin +cos)
and , and are in the directions defined in the figure which leads to the generalized forces = −cos+sin
= (−cos+sin) (−cos)−(sin+cos)sin=−
=
Solving the above 7equations gives that
¨sin+˙2−cos+ = 0
This last equation can be derived by Newtonian mechanics from consideration of the forces acting. The above equations of motion can be used to calculate the motion for the following conditions. a) Mass not slipping:
This occurs if = ≤ which also implies that 0 That is a situation where the system is holonomic with =˙ =˙ ˙ =˙ which can be solved using the generalized coordinate approach with only one independent coordinate which can be taken to be .
b) Mass slipping:
Here the no-slip constraint is violated and thus one has to explicitly include the generalized forces
and assume that sliding friction is given by = c) Reaction force is negative:
Here the mass is not subject to any constraints and it is in free fall.
The above example illustrates the flexibility provided by Lagrangian mechanics that allows simultane- ous use of Lagrange multipliers, generalized forces, and scalar potential to handle combinations of several holonomic and nonholonomic constraints for a complicated problem.