IV. RESULTADOS
4.3 DISCUSIÓN DE RESULTADOS
where we again use the identities ˙1= ∇ 1· v +∂ 1
∂t = 0, ˙2= ∇ψ 2· v +∂ 2
∂t = 0.
For the Lagrange prescription, Ff = 0, and we can now see that, if the curve that the particle is moving on is fixed, i.e., 1= 1(r) and 2= 2(r), then Fcdoes no work. Otherwise, this constraint force is expected to do work because its normal components force part of the velocity vector of the particle to be vc. As in the case of a single constraint, for the Coulomb prescription, except when vc= 0, it is not possible to predict if work is done on the particle by this force.
We now consider some examples. Recall that the planar pendulum consists of a particle of mass m that is attached by a rod of length L0 to a fixed point O. The particle is also constrained to move on a vertical plane. In short, 1 = r · er− L0= 0 and 2= r · E3= 0. With a little work, we find that ∇ 1= er and∇ 2= E3. For this mechanical system, Lagrange’s prescription is appropriate:
Fc= λ1er+ λ2E3.
For this system, λ1ercan be interpreted as the tension force in the rod and λ2E3can be interpreted as the normal force exerted by the plane on the particle. If we let L0= L0(t), the prescription for the constraint force will not change.
A system that is related to the planar pendulum can be imagined as a particle moving on a rough circle whose radius L0= L0(t). The particle is subject to the same constraints as it is in the planar pendulum; however, Lagrange’s prescription is not valid. Instead, we now have
Fc= λ1er+ λ2E3− µd
λ21+ λ22 ˙θ
˙θeθ, where we used the fact that v− vc= L0˙θeθ.
Three Constraints
The reader may have noticed that our expressions for the constraint force when we employed Coulomb’s prescription were not valid when the particle was stationary relative to the surface or curve that it was constrained to move on. This is because we view this case as corresponding to the motion of the particle subject to three constraints: i(r, t)= 0, i = 1, 2, 3. As mentioned earlier, when a particle is subject to three constraints, the three equations i(r, t)= 0 can in principle be solved to determine the motion r(t) of the particle. We denote the resulting solution by f(t), i.e., r(t)= f(t). In other words, the motion is completely prescribed. In this case, the sole purpose of F= ma is to determine the constraint force Fc.
For the case in which the particle is subject to three constraints, the Lagrange prescription and a prescription based on static Coulomb friction are equivalent.
This equivalence holds in spite of the distinct physical situations these prescriptions pertain to.
To examine the equivalence, let us first use Lagrange’s prescription:
Fc= λ1∇ 1+ λ2∇ 2+ λ3∇ 3.
Here, λ1, λ2, and λ3are functions of time. Because the three constraints are tacitly assumed to be independent,{∇ 1,∇ 2,∇ 3} forms a basis for E3. Consequently, Lagrange’s prescription provides a vector Fcwith three independent components.
Coulomb’s static friction prescription for a particle that is not moving relative to the curve or surface on which it lies is
Fc= N + Ff,
where the magnitude of Ff is restricted by the static friction criterion:
Ff ≤ µs||N|| ,
where µs is the coefficient of static friction. Again, the Coulomb prescription pro-vides a vector Fcwith three independent components. In other words, both prescrip-tions state that Fcconsists of three independent unknown functions of time.
If we now assume that the resultant force F has the decomposition F= Fc+ Fa, where Fa are the nonconstraint forces, then we see how Fcis determined from F= ma:
Fc= −Fa+ ma = −Fa+ m¨f.
This solution Fcwill be the same regardless of whether one uses Lagrange’s pre-scription or Coulomb’s prepre-scription.
Nonintegrable Constraints
Our discussion of constraint forces has focused entirely on the case of integrable constraints. If a nonintegrable constraint,
f· v + e = 0,
is imposed on the particle, we need to discuss a prescription for the associated con-straint force. To this end, we adopt a conservative approach and use the prescription
Fc= λf.
The main reason for adopting this prescription is as follows: In the event that the nonintegrable constraint turns out to be integrable, then the prescription we employ will agree with Lagrange’s prescription we discussed earlier.
As a further example, suppose the motion of the particle is subject to two con-straints, one of which is integrable:
(r, t)= 0, f· v + e = 0.
Using Lagrange’s prescription, we find that the constraint force acting on the parti-cle is
Fc= λ1∇ + λ2f.
2.7 Conservations 45
Suppose that the applied force acting on the particle is Fa; then the equations gov-erning the motion of the particle are
(r, t)= 0, f· v + e = 0,
˙r= v, ˙v= 1
m(Fa+ λ1∇ + λ2f).
This set of equations constitutes eight equations for the eight unknowns: λ1, λ2, r, and v.
2.7 Conservations
For a given particle and system of forces acting on the particle, a kinematical quan-tity is said to be conserved if it is constant during the motion of the particle. The conserved quantities are often known as integrals of motion. The solutions of many problems in particle mechanics are based on the observation that either a mo-mentum or an energy (or both) is conserved. At this stage in the development of the field, most of these conservations are obvious and are deduced by inspection.
However, for future purposes it is useful to understand the conditions for such conservations. We shall consider numerous examples of these conservations later on.
Conservation of Linear Momentum
The linear momentum G of a particle is defined as G= mv. Recalling the integral form of the balance of linear momentum,
G(t)− G (t0)=
t
t0
F(τ)dτ,
we see that G(t) is conserved during an interval of time (t0,t) ift
t0F(τ)dτ= 0. The simplest case of this conservation arises when F(τ)= 0.
Another form of this conservation pertains to a component of G in the di-rection of a given vector b(t) being conserved. That is, dtd(G· b) = 0. For this to happen,
G˙· b = ˙G · b + G · ˙b = F · b + G · ˙b = 0.
In words, if F· b + G · ˙b = 0, then G · b is conserved.
Examples of conservation of linear momentum arise in many problems. For example, consider a particle under the influence of a gravitational force F= −mgE3. For this problem, the E1and E2components of G are conserved. Another example is to consider a particle impacting a smooth vertical wall. Then the components of G in the two tangential directions are conserved. For these two examples, the vector b is constant.
Conservation of Angular Momentum
The angular momentum of a particle relative to a fixed point O is HO= r × G. To establish how HOchanges during the motion of a particle, a simple calculation is needed:
H˙O= v × G + r × ˙G = v × mv + r × F = r × F.
It is important to note that we used F= ma during this calculation. The final result is known as the angular momentum theorem for a particle:
H˙O= r × F.
In words, the rate of change of angular momentum is equal to the moment of the resultant force.
Conservation of angular momentum usually arises in two forms. First, the en-tire vector is conserved, and, second, a component, say c(t), is conserved. For the first case, we see from the angular momentum theorem that HOis conserved if F is parallel to r. Problems in which this arises are known as central force problems. Dating to Newton, they occupy an important place in the history of dynamics. When the angular momentum theorem is used, it is easy to see that the second form of conservation, HO· c is constant, arises when r × F · c + HO· ˙c
= 0.
Conservation of Energy
As a prelude to discussing the conservation of energy, we first need to discuss the work–energy theorem. This theorem is a result that is established by use of F= ma and relates the time rate of change of kinetic energy to the mechanical power of F:
T˙ = F · v.
This theorem is the basis for establishing conservation of energy results for a single particle.
The proof of the work–energy theorem is very straightforward. First, recall that T= 12mv· v. Differentiating T, we find
T˙= d dt
1 2mv· v
= 1
2(m ˙v· v + mv · ˙v) = m˙v · v.
However, we know that m ˙v= F, and so substituting for m˙v, we find that ˙T = F · v, as required.
To examine situations in which the total energy of a particle is conserved, we first divide the forces acting on the particle into the sum of a resultant conservative force P= −∂U∂r and a nonconservative force Pncon: F= P + Pncon. Here, U is the sum