- An Introduction To The Use Of Generalized Coordinates In Mechanics And Physics
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Description Product Details Click on the cover image above to read some pages of this book! More Books in Mathematics See All. Math Art Truth, Beauty, and Equations. In Stock. Maths Skills for Success at University. Teaching Mathematics Foundations to Middle Years. Woo's Wonderful World of Maths. Antifragile Things That Gain from Disorder. The Art of Statistics Learning from Data. The Maths of Life and Death. There are rigid position constraints among some of the particles. We assume that all of these constraints are of the form.

The potential forces are derived as the negative gradient of the potential energy, and may depend on the positions of the other particles and the time. In general, the scalar constraint forces change as the system evolves. Formally, we can reproduce Newton's equations with the Lagrangian The velocity of F does not appear in the Lagrangian, and F itself appears only linearly. So the Lagrange equations associated with F are. The Lagrange equations for the coordinates of the particles are Newton's equations 1. Now that we have a suitable Lagrangian, we can use the fact that Lagrangians can be reexpressed in any generalized coordinates to find a simpler Lagrangian.

The strategy is to choose a new set of coordinates for which many of the coordinates are constants and the remaining coordinates are irredundant. Let q be a tuple of generalized coordinates that specify the degrees of freedom of the system without redundancy. Let c be a tuple of other generalized coordinates that specify the distances between particles for which constraints are specified.

## An Introduction To The Use Of Generalized Coordinates In Mechanics And Physics

The c coordinates will have constant values. The combination of q and c replaces the redundant rectangular coordinates x. Our new coordinates are the components of q , c , and F. Substituting these into Lagrangian 1. The Lagrange equations are derived by the usual procedure. Rather than write out all the gory details, let's think about how it will go.

We can use this result to simplify the Lagrange equations associated with q and c. The Lagrange equations associated with q are the same as if they were derived from the Lagrangian Notice that the constraint forces do not appear in the Lagrange equations for q because in the Lagrange equations they are multiplied by a term that is identically zero on the solution paths.

The Lagrange equations for c can be used to find the constraint forces. The Lagrange equations are a big mess so we will not show them explicitly, but in general they are equations in D 2 c , Dc , and c that will depend upon q , Dq , and F. If we are not interested in the constraint forces, we can abandon the full Lagrangian 1. In this exercise we will recapitulate the derivation of the Lagrangian for constrained systems for a particular simple system.

Consider two massive particles in the plane constrained by a massless rigid rod to remain a distance l apart, as in figure 1. There are apparently four degrees of freedom for two massive particles in the plane, but the rigid rod reduces this number to three. We can uniquely specify the configuration with the redundant coordinates of the particles, say x 0 t , y 0 t and x 1 t , y 1 t. Write Newton's equations for the balance of forces for the four rectangular coordinates of the two particles, given that the scalar tension in the rod is F.

Write the Lagrangian in these coordinates, and write the Lagrange equations. They should be the same equations as you derived for the same coordinates in part d.

Show that the Lagrangian 1. Be sure to examine the equations for the constraint forces as well as the position of the pendulum bob. Show that the Lagrange equations for Lagrangian 1. The derivation of a Lagrangian for a constrained system involves steps that are analogous to those in the derivation of a coordinate transformation. We can make a Lagrangian for the unconstrained system of particles in rectangular coordinates. In general there will be more coordinates than real degrees of freedom; the constraints will eliminate the redundancy. We then choose a convenient set of irredundant generalized coordinates that incorporate the constraints to describe our system.

We express the redundant rectangular coordinates and velocities in terms of the irredundant generalized coordinates and generalized velocities, and we use these transformations to reexpress the Lagrangian in the generalized coordinates. To carry out a coordinate transformation we specify how the configuration of a system expressed in one set of generalized coordinates can be reexpressed in terms of another set of generalized coordinates.

We then determine the transformation of generalized velocities implied by the transformation of generalized coordinates.

A Lagrangian that is expressed in terms of one of the sets of generalized coordinates can then be reexpressed in terms of the other set of generalized coordinates. These are really two applications of the same process, so we can make Lagrangians for constrained systems by composing a Lagrangian for unconstrained particles with a coordinate transformation that incorporates the constraint. The Lagrangian. This is a suitable Lagrangian for a set of unconstrained free particles with potential energy V. Let q be a tuple of irredundant generalized coordinates and v be the corresponding generalized velocity tuple.

Here we view this as a coordinate transformation. What is unusual about this as a coordinate transformation is that the dimension of x is not the same as the dimension of q. From this coordinate transformation we can find the local-tuple transformation function see section 1. A Lagrangian for the constrained system can be obtained from the Lagrangian for the unconstrained system by composing it with the local-tuple transformation function from constrained coordinates to unconstrained coordinates:.

To illustrate this we will find a Lagrangian for the driven pendulum introduced in section 1. A program that computes this Lagrangian is. It is interesting that the drive enters only through the specification of the constraints.

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A program implementing this coordinate transformation is. This is the same as the Lagrangian of equation 1. We have found a very interesting decomposition of the Lagrangian for constrained systems. One part consists of the difference of the kinetic and potential energy of the constituents. The other part describes the constraints that are specific to the configuration of a particular system. If the pendulum is released, at rest, with nonzero displacement from the local vertical, it will oscillate in an apparent plane.

However, the apparent plane of oscillation precesses as the Earth rotates. One way to specify the position of the bob is to erect a Foucault pendulum at the North Pole and rotate it to a point on the surface of the Earth at the appropriate colatitude and a fixed longitude. Because the Earth is rotating this is a time-varying transformation. There are two parts of this transformation. The rectangular coordinates of the bob for a pendulum at the North Pole are:. The transformation of coordinates is: Lagrangians are not in a one-to-one relationship with physical systems—many Lagrangians can be used to describe the same physical system.

The function F on the path at time t is. In general, the total time derivative of a local-tuple function F is that function D t F that when composed with a local-tuple path is the time derivative of the composition of the function F with the same local-tuple path:. The total time derivative D t F is not the derivative of the function F.

Nevertheless, the total time derivative shares many properties with the derivative. Demonstrate that D t has the following properties for local-tuple functions F and G , number c , and a function H with domain containing the range of G. The variational principle states that the action integral along a realizable trajectory is stationary with respect to variations of the trajectory that leave the configuration at the endpoints fixed.

So either Lagrangian can be used to distinguish the realizable paths. The addition of a total time derivative to a Lagrangian does not affect whether the action is stationary for a given path. So if we have two Lagrangians that differ by a total time derivative, the corresponding Lagrange equations are equivalent in that the same paths satisfy each. Moreover, the additional terms introduced into the action by the total time derivative appear only in the endpoint condition and thus do not affect the Lagrange equations derived from the variation of the action, so the Lagrange equations are the same.

The Lagrange equations are not changed by the addition of a total time derivative to a Lagrangian.

Let F t , q be a function of t and q only, with total time derivative. Show explicitly that the Lagrange equations for D t F are identically zero, and thus that the addition of D t F to a Lagrangian does not affect the Lagrange equations. The driven pendulum provides a nice illustration of adding total time derivatives to Lagrangians. The equation of motion for the driven pendulum see section 1. This intuitive interpretation was not apparent in the Lagrangian derived as the difference of the kinetic and potential energies in section 1.

However, we can write an alternate Lagrangian with the same equation of motion that is as easy to interpret as the equation of motion:. With this Lagrangian it is apparent that the effect of the accelerating pivot is to modify the acceleration of gravity. Note, however, that it is not the difference of the kinetic and potential energies. Let's compare the two Lagrangians for the driven pendulum. The addition of such terms to a Lagrangian does not affect the equations of motion. If the local-tuple function G , with arguments t , q , v , is the total time derivative of a function F , with arguments t , q , then G must have certain properties.

From equation 1. So if G is the total time derivative of F then. If there is more than one degree of freedom these partials are actually structures of partial derivatives with respect to each coordinate. The partial derivatives with respect to two different coordinates must be the same independent of the order of the differentiation. Note that we have not shown that these conditions are sufficient for determining that a function is a total time derivative, only that they are necessary. For each of the following functions, either show that it is not a total time derivative or produce a function from which it can be derived.

This form is Galilean invariant. Start with a Lagrangian for free particles, which is only the sum of their kinetic energies:. Carry out a coordinate transformation from old to new coordinates that consists of a shift and a uniform translation. Show that the additional terms are a total time derivative. Lagrange's equations are ordinary differential equations that the path must satisfy. They can be used to test if a proposed path is a realizable path of the system. However, we can also use them to develop a path, starting with initial conditions. The state of a system is defined to be the information that must be specified for the subsequent evolution to be determined.

Remember our juggler: he or she must throw the pin in a certain way for it to execute the desired motion. The juggler has control of the initial position and orientation of the pin, and the initial velocity and spin of the pin. Our experience with juggling and similar systems suggests that the initial configuration and the rate of change of the configuration are sufficient to determine the subsequent motion. Other systems may require higher derivatives of the configuration.

For Lagrangians that are written in terms of a set of generalized coordinates and velocities we have shown that Lagrange's equations are second-order ordinary differential equations. If the differential equations can be solved for the highest-order derivatives and if the differential equations satisfy the Lipschitz conditions, 71 then there is a unique solution to the initial-value problem: given values of the solution and the lower derivatives of the solution at a particular moment, there is a unique solution function.

Given irredundant coordinates the Lagrange equations satisfy these conditions. This is the information required to specify the dynamical state. A complete local description of a path consists of the path and all of its derivatives at a moment. The complete local description of a path can be reconstructed from an initial segment of the local tuple, given a prescription for computing higher-order derivatives of the path in terms of lower-order derivatives.

The state of the system is specified by that initial segment of the local tuple from which the rest of the complete local description can be deduced. The complete local description gives us the path near that moment. Actually, all we need is a rule for computing the next higher derivative; we can get all the rest from this.

Assume that the state of a system is given by the tuple t , q , v. Each of these functions depends on lower-derivative components of the local tuple. All we need to deduce the path from the state is a function that gives the next-higher derivative component of the local description from the state. We use the Lagrange equations to find this function. Solving this system for D 2 q , one obtains the generalized acceleration along a solution path q :.

That initial segment of the local tuple that specifies the state is called the local state tuple, or, more simply, the state tuple. We can express the function that gives the acceleration as a function of the state tuple as the following procedure. It takes a procedure that computes the Lagrangian, and returns a procedure that takes a state tuple as its argument and returns the acceleration. Once we have a way of computing the acceleration from the coordinates and the velocities, we can give a prescription for computing the derivative of the state as a function of the state.

We represent a state by an up tuple of the components of that initial segment of the local tuple that determine the state. The Lagrange equations are a second-order system of differential equations that constrain realizable paths q. We can use the state derivative to express the Lagrange equations as a first-order system of differential equations that constrain realizable coordinate paths q and velocity paths v :. For example, we can find the first-order form of the equations of motion of a two-dimensional harmonic oscillator:. The zero in the first element of the structure of the Lagrange equations residuals is just the tautology that time advances uniformly: the time function is just the identity, so its derivative is one and the residual is zero.

The equations in the second element constrain the velocity path to be the derivative of the coordinate path. The equations in the third element give the rate of change of the velocity in terms of the applied forces. A set of first-order ordinary differential equations that give the state derivative in terms of the state can be integrated to find the state path that emanates from a given initial state.

Numerical integrators find approximate solutions of such differential equations by a process illustrated in figure 1. The procedure state-advancer can be used to find the state of a system at a specified time, given an initial state, which includes the initial time, and a parametric state-derivative procedure. The arguments to state-advancer are a parametric state derivative, harmonic-state-derivative , and the state-derivative parameters mass 2 and spring constant 1. A procedure is returned that takes an initial state, up 1 up 1 2 up 3 4 ; a time increment, 10 ; and a relative error tolerance, 1.

The output is an approximation to the state at the specified final time. Consider the driven pendulum described in section 1. To examine the evolution of the driven pendulum we need a mechanism that evolves a system for some interval while monitoring aspects of the system as it evolves.

The procedure evolve provides this service, using state-advancer repeatedly to advance the state to the required moments. The procedure evolve takes a parametric state derivative and its parameters and returns a procedure that evolves the system from a specified initial state to a number of other times, monitoring some aspect of the state at those times. To generate a plot of the angle versus time we make a monitor procedure that generates the plot as the evolution proceeds: The initial segments of the two orbits are indistinguishable.

After about 75 seconds the two orbits diverge and become completely different. This extreme sensitivity to tiny changes in initial conditions is characteristic of what is called chaotic behavior. Write a program to evolve the motion of a particle subject to this Lagrangian and display the orbit in the plane. Observe that it describes an ellipse with its center at the origin, for a wide variety of initial conditions.

Observe that it describes an ellipse with a focus at the origin, for a wide variety of initial conditions. Observe that it describes a trefoil with its center at the origin. If a Foucault pendulum is erected at the North Pole, it will precess exactly once in a day. If it is erected at the Equator it will not precess at all. It is widely reported that the precession rate is proportional to the cosine of the colatitude.

### Where did they come from?

Evolve the Foucault pendulum, using the Lagrangian you constructed in exercise 1. How does the rate of precession compare to the predicted rate? You should expect to see an error caused by the fact that the local vertical, as defined by a plumb bob, is not directed to the center of the Earth. Is this perfect? A function of the state of the system that is constant along a solution path is called a conserved quantity or a constant of motion.

In this section, we will investigate systems with symmetry and find that symmetries are associated with conserved quantities. For instance, linear momentum is conserved in a system with translational symmetry, angular momentum is conserved if there is rotational symmetry, energy is conserved if the system does not depend on the origin of time. We first consider systems for which a coordinate system can be chosen that expresses the symmetry naturally, and later discuss systems for which no coordinate system can be chosen that simultaneously expresses all symmetries.

If a Lagrangian L t , q , v does not depend on some particular coordinate q i , then. The derivative of a component is equal to the component of the derivative, so this is the same as. The function P is called the momentum state function. The value of the momentum state function is the generalized momentum. We refer to the i th component of the generalized momentum as the momentum conjugate to the i th coordinate. The generalized momentum component conjugate to any cyclic coordinate is a constant of the motion.

Its value is constant along realizable paths; it may have different values on different paths. As we will see, momentum is an important quantity even when it is not conserved. Given the coordinate path q and the Lagrangian L , the momentum path p is. The momentum path is well defined for any path q. If the path is realizable and the Lagrangian does not depend on q i , then p i is a constant function.

The constant value of p i may be different for different trajectories. For the free particle the usual linear momentum is conserved for realizable paths. For a particle in a central force field section 1. The momentum state function is. It has two components. If the central-potential problem had been expressed in rectangular coordinates, then all of the coordinates would have appeared in the Lagrangian. In that case there would not be any obvious conserved quantities. Nevertheless, the motion of the system does not depend on the choice of coordinates, so the angular momentum is still conserved.

We see that there is great advantage in making a judicious choice for the coordinate system. If we can choose the coordinates so that a symmetry of the system is reflected in the Lagrangian by the absence of some coordinate component, then the existence of a corresponding conserved quantity will be evident. Momenta are conserved by the motion if the Lagrangian does not depend on the corresponding coordinate.

The energy function has a constant value along any realizable trajectory if the Lagrangian has no explicit time dependence; the energy E may have a different value for different trajectories. A system that has no explicit time dependence is called autonomous. In some cases the energy can be written as the sum of kinetic and potential energies. If T is a homogeneous function of degree 2 in the generalized velocities then. Notice that if V depends on time the energy is still the sum of the kinetic energy and potential energy, but the energy is not conserved.

Show that in this case the kinetic energy contains terms that are linear in the generalized velocities. An example in which terms that were linear in the velocity were removed from the Lagrangian by adding a total time derivative has already been given: the driven pendulum. A particle of mass m slides off a horizontal cylinder of radius R in a uniform gravitational field with acceleration g. If the particle starts close to the top of the cylinder with zero initial speed, with what angular velocity does it leave the cylinder?

Use the method of incorporating constraint forces that we introduced in section 1. One important physical system is the motion of a particle in a central field in three dimensions, with an arbitrary potential energy V r depending only on the radius. The kinetic energy has three terms:.

Let's first look at the generalized forces the derivatives of the Lagrangian with respect to the generalized coordinates. We compute these with a partial derivative with respect to the coordinate argument of the Lagrangian:. The corresponding momentum component is conserved. Compute the momenta:. We can show this by writing the z component of the angular momentum in spherical coordinates:. The choice of the z -axis is arbitrary, so the conservation of any component of the angular momentum implies the conservation of all components. Thus the total angular momentum is conserved.

We can choose the z -axis so all of the angular momentum is in the z component. Planar motion in a central-force field was discussed in section 1. A spherical pendulum is a massive bob, subject to uniform gravity, that may swing in three dimensions, but remains at a given distance from the pivot. Formulate a Lagrangian for a spherical pendulum driven by vertical motion of the pivot.

What symmetry ies can you find? Find coordinates that express the symmetry ies. What is conserved? Give analytic expression s for the conserved quantity ies. Consider the situation of two bodies of masses M 0 and M 1 in circular orbit about their common center of mass. What is the behavior of a third particle, gravitationally attracted to the other two, that must move in the plane of their circular orbit?

Assume that the third particle has such small mass that we can neglect its effect on the orbits of the two massive particles. The third particle, of mass m , moves in a field derived from a time-varying gravitational potential energy. We have:. Let a be the constant distance between the two bodies. If we put the center of mass at the origin of the coordinate system then the distances of the two particles from the origin are:. The radii of the circles are the distances given above.

Kepler's law gives the angular frequency of the orbit:. The gravitational potential energy function is:. It is convenient to examine the motion of the third particle in a rotating coordinate system where the massive particles are fixed. We can transform to the rotating rectangular coordinates as we did on page The resulting Lagrangian is the Lagrangian for the free particle with the addition of two gravitational potential energy terms:.

As a program we can write:. Notice that the Lagrangian in rotating coordinates is independent of time. So the energy state function defined by this Lagrangian is a conserved quantity. Let's compute it. If we separate this into a velocity-dependent part and a velocity-independent part we get. This constant of motion of the restricted three-body problem is called the Jacobi constant. Derive the Lagrange equations for the restricted three-body problem, given the Lagrangian 1. Identify the Coriolis and centrifugal force terms in your equations of motion. We have seen that if a system has a symmetry and a coordinate system can be chosen so that the Lagrangian does not depend on the coordinate associated with that symmetry, then there is a conserved quantity associated with the symmetry.

However, there are more general symmetries that no coordinate system can fully express. For example, motion in a central potential is spherically symmetric the dynamical system is invariant under rotations about any axis , but the expression of the Lagrangian for the system in spherical coordinates exhibits symmetry around only one axis. More generally, a Lagrangian has a symmetry if there is a coordinate transformation that leaves the Lagrangian unchanged. A continuous symmetry is a parametric family of symmetries. Noether proved that for any continuous symmetry there is a conserved quantity.

On a realizable path q we can use the Lagrange equations to rewrite the first term of equation 1. Thus the state function I ,. This conserved quantity is called Noether's integral. It is the product of the momentum and a vector associated with the symmetry. For example, consider the central-potential Lagrangian in rectangular coordinates:. Since the Lagrangian is preserved by any continuous rotational symmetry, all components of the vector angular momenta are conserved for the central-potential problem.

The procedure calls Rx angle-x q , Ry angle-y q , and Rz angle-z q rotate the rectangular tuple q about the indicated axis by the indicated angle. Formulate a Lagrangian for frictionless motion on this surface.

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## mamarihipa.tk: an-introduction-to-the-use-of-generalized-coordinates-in-mechanics-and-physics-b

Formulate a parametric transformation that represents this symmetry and show that the Lagrangian you formulated is invariant under this transformation. Compute the Noether integral associated with this symmetry. An essential step in the derivation of the local-tuple transformation function C from the coordinate transformation F was the deduction of the relationship between the velocities in the two coordinate systems. We did this by inserting coordinate paths into the coordinate transformation function F , differentiating, and then generalizing the results on the path to arbitrary velocities at a moment.

The last step is an example of a more general problem of abstracting a local-tuple function from a path function. Two paths that have the same local description up to the n th derivative are said to osculate with order n contact. For example, a path and the truncated power series representation of the path up to order n have order n contact; if fewer than n derivatives are needed by a local-tuple function, the path and the truncated power series representation are equivalent.

Let O be a function that generates an osculating path with the given local-tuple components. The number of components of the local tuple that are required is finite, but unspecified. One way of constructing O is through the truncated power series. We take the argument of f and construct an osculating path with this local description. The procedure osculating-path takes a number of local components and returns a path with these components; it is implemented as a power series. It then uses Gamma-bar to abstract f-bar to arbitrary local tuples in the primed coordinate system.

The calculation that led up to the state transformation 1. Given a procedure F implementing a local-tuple function and a path q , we construct a new procedure compose F Gamma q. The procedure DF-on-path implements the derivative of this function of time. We then abstract this off the path with Gamma-bar to give the total time derivative. Use the procedure Gamma-bar to construct a procedure that transforms velocities given a coordinate transformation. Given a Lagrangian, the Lagrange equations test paths to determine whether they are realizable paths of the system.

The Lagrange equations relate the path and its derivatives. The fact that the Lagrange equations must be satisfied at each moment suggests that we can abstract the Lagrange equations off the path and write them as relations among the local-tuple components of realizable paths. The operator E is called the Euler—Lagrange operator.

In terms of this operator the Lagrange equations are. The procedure Euler-Lagrange-operator implements E :. Notice that the components of the local tuple are individually specified. Using equation 1. Let F and G be two Lagrangian-like functions of a local tuple, C be a local-tuple transformation function, and c a constant. Demonstrate the following properties:. An advantage of the Lagrangian approach is that coordinates can often be chosen that exactly describe the freedom of the system, automatically incorporating any constraints. We may also use coordinates that have more freedom than the system actually has and consider explicit constraints among the coordinates.

For example, the planar pendulum has a one-dimensional configuration space. We have formulated this problem using the angle from the vertical as the configuration coordinate. Alternatively, we may choose to represent the pendulum as a body moving in the plane, constrained to be on the circle of the correct radius around the pivot. We would like to have valid descriptions for both choices and show they are equivalent.

In this section we develop tools to handle problems with explicit constraints. The constraints considered here are more general than those used in the demonstration that the Lagrangian for systems with rigid constraints can be written as the difference of kinetic and potential energies see section 1. How do we formulate the equations of motion? One approach would be to use the constraint equation to eliminate one of the coordinates in favor of the rest; then the evolution of the reduced set of generalized coordinates would be described by the usual Lagrange equations.

The equations governing the evolution of coordinates that are not fully independent should be equivalent. We can address the problem of formulating equations of motion for systems with redundant coordinates by returning to the action principle. Realizable paths are distinguished from other paths by having stationary action. Stationary refers to the fact that the action does not change with certain small variations of the path.

What variations should be considered? We have seen that velocity-independent rigid constraints can be used to eliminate redundant coordinates. In the irredundant coordinates we distinguished realizable paths by using variations that by construction satisfy the constraints. Thus in the case where constraints can be used to eliminate redundant coordinates we can restrict the variations in the path to those that are consistent with the constraints.

So how does the restriction of the possible variations affect the argument that led to Lagrange's equations refer to section 1. Actually most of the calculation is unaffected. The condition that the action is stationary still reduces to the conditions 1. So we have. Note that these are functions of time; the variation at a given time is tangent to the constraint at that time.

Together, equations 1. A vector that is orthogonal to all vectors orthogonal to a given vector is parallel to the given vector. Thus, the residual of Lagrange's equations is parallel to the normal to the constraint surface; the two must be proportional:.

The Lagrange equations associated with the coordinates q are just the modified Lagrange equations 1. So the Lagrange equations for this augmented Lagrangian fully encapsulate the modification to the Lagrange equations that is imposed by the addition of an explicit coordinate constraint, at the expense of introducing extra degrees of freedom.

Notice that this Lagrangian is of the same form as the Lagrangian equation 1. Consider the Lagrangian. This new Lagrangian has no extra degrees of freedom. Applying the Euler—Lagrange operator E see section 1. This allows us to deduce that D 2 q is also a state-dependent function composed with the path.

The evolution of the system is determined from the dynamical state. The pendulum can be formulated as the motion of a massive particle in a vertical plane subject to the constraint that the distance to the pivot is constant see figure 1. In this formulation, the kinetic and potential energies in the Lagrangian are those of an unconstrained particle in a uniform gravitational acceleration.

A Lagrangian for the unconstrained particle is. The constraint that the pendulum moves in a circle of radius l about the pivot is Forming the derivatives and substituting into the other two equations, we find. Due to the pressure far away, it is kept equal to , the "bubble" starts collapsing; find the collaps time of the "bubble".

Here we use the radius of the "bubble" as the generalized coordinate; there is no potential energy, but there is work done by the pressure,. What is left to do, is to express the kinetic energy of the fluid in terms of. So, the kinetic energy can be found as. This equation could be used to find the acceleration ; however, we need to know the collapse time; so we put , and express in terms of and :. Thus, we were able to obtain an answer, which contains a dimensionless integral: substituting allowed us to get rid of the dimensional quantities under the integral if possible, always use this technique to convert integrals into dimensionless numbers.

This result could be left as is, since finding an integral is a task for mathematicians. The mathematicians, however, have been up to the task: where denotes the gamma function. So, we can write. Finally, to close the topic of the generalized coordinates , it should be mentioned that this technique can be developed into generic theories — Lagrangian and Hamiltonian formalisms , which are typically taught as a main component of the course of theoretical mechanics.

In particular, the Hamiltonian formalism makes it possible to prove the conservation of adiabatic invariant , as well as the KAM Kolmogorov-Arnold-Mozer theorem, as well as to derive conservation laws from the symmetry properties of the Hamiltonian or Lagrangian using the Noether's theorem. The Hamiltonian approach differs from what is described here by using the generalized momentum , instead of the generalized velocity.

For the most typical cases when the kinetic energy is proportional to the square of the generalized velocity, one can just use the effective mass defined above :. However, for the practical application of problem solving, the simplified approach to the generalized coordinates provided above is just enough! Minimum or maximum? Fast or slow? Force diagrams or generalized coordinates? Are Trojans stable? Images or roulette? Upon taking time-derivative of this equality, we obtain , from where we can express the acceleration of the generalized coordinate: Note that most often, is constant, because the kinetic energy is proportional to , and plays the role of an effective mass.

Hence, we find that Upon taking time derivative of this equation and cancelling out , we obtain an expression for the wedge acceleration: As another example, let us consider an old IPhO problem 5 th IPhO in Sofia, , Problem No 1. However, if we make use of the conservation of the centre of mass there are no external horizontal forces , we can express the displacement of the bricks with respect to the wedge via the displacement of the wedge : What is left to do, is to write substitute by , take time derivative of the full energy, and express.

As a result, we obtain which gives us immedieately the circular frequency of small oscillations,. So, the energy conservation law is written as from where. Now, assuming that we have heavy wedge, and the system moves leftwards, the Newton II law for the wedge can be written as , and hence,. So, the kinetic energy can be found as So, the energy balance can be written as This equation could be used to find the acceleration ; however, we need to know the collapse time; so we put , and express in terms of and : Thus, we were able to obtain an answer, which contains a dimensionless integral: substituting allowed us to get rid of the dimensional quantities under the integral if possible, always use this technique to convert integrals into dimensionless numbers.