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# 数学代写|数学建模代写Mathematical Modeling代考|Oscillations of a pendulum in a gravity field

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## 数学代写|数学建模代写Mathematical Modeling代考|Oscillations of a pendulum in a gravity field

Oscillations of a pendulum in a gravity field. Consider a slightly more complicated example of applying the Hamiltonian principle, along with the detailed consideration of the initial stage of constructing a model – the description of a mechanical system.

Let a pendulum be hung on a fixed (motionless) hinge. The pendulum is a mass $m$ attached on the edge of a rod of length $l$ (Fig.12). The hinge is considered to be ideally smooth, in the sense that there are no losses by friction. The motionless hinge means that no energy is passed from it to the system “rod-weight”, i.e. no work is performed by the rod. The hinge is considered weightless and absolutely rigid, i.e. its kinetic and potential energies are equal to zero, and the weight cannot move along it. The weight is small in comparison with the length of the rod (material point), the acceleration of gravity $g$ is constant, the air resistance is neglected, the oscillations occur in the fixed vertical plane obviously, the vector of the initial velocity of the weight lies in this plane.

It is clear that after these simplifying assumptions, the position of a pendulum is determined only by one of the generalized coordinates; we select the angle $\alpha(t)$ of the deviation of the rod from the vertical. A generalized velocity in this case is the angular velocity $d \alpha / d t$.

The kinetic energy of the system is given by the formula
$$E_k=\frac{1}{2} m v^2=\frac{1}{2} m\left(l \frac{d \alpha}{d t}\right)^2=\frac{1}{2} m l^2\left(\frac{d \alpha}{d t}\right)^2$$
and the potential energy is given by the expression
$$E_p=m g h=-m g(l \cos \alpha-1)$$
where $h$ is the deviation of the pendulum from the lowest position by the vertical. In further calculations we omit the quantity $m g l$ in $E_p$, in so far as the potential energy is determined to be within a constant.

Now it is not difficult to estimate the Lagrange function (1) and the action $(2)$ :
\begin{aligned} & L\left(\alpha, \frac{d \alpha}{d t}\right)=m l\left[\frac{1}{2} l\left(\frac{d \alpha}{d t}\right)^2+g \cos \alpha\right] \ & S[\alpha]=m l \int_{t_1}^{t_2}\left[\frac{1}{2} l\left(\frac{d \alpha}{d t}\right)^2+g \cos \alpha\right] d t \end{aligned}

## 数学代写|数学建模代写Mathematical Modeling代考|Conclusion

Conclusion. The examples of this use of the Hamiltonian principle for constructing models of mechanical systems enable one to draw up a rather precise program of actions, in a general form described in subsection 1. The universality of the the successive procedures not depending on the details of the concrete system, certainly, are the attractive features of variational principles. In the simple cases above mentioned models can easily be constructed in other ways. However for many other, more complicated objects, the variational principles appear to be actually the only method of constructing models. So, for example, the mechanical parts of the majority of robotic devices consist of a great number of various elements mutually connected in various ways. Their mathematical models include a large number of equations, obtained with the help of the variational principles. This approach is also successfully applied to other kinds of systems (physical, chemical, biological), for which the appropriate general statements about a character of their evolution (behavior) are formulated.

The circumstance that the Hamiltonian principle and other approaches give identical models is natural, in so far as they describe the same initial objects. Certainly, such a coincidence is guaranteed only when initial assumptions about the object are the same. If its idealization (as one of the initial phases of constructing a model) will be carried uniformly, the different modes of constructing models should give identical results. For example, in the system “ball-spring”, let an additional constant external force acts on the ball $F_0$. Then from Newton’s second law it is easy to derive the equation of motion of the ball
$$m \frac{d^2 r}{d t^2}=-k r+F_0$$

(compare with (5) of section 2). Applying the Hamiltonian principle to such a system, it is necessary to take into account the presence of this force. Obviously, the definitions of generalized coordinates, velocity and kinetic energy $E_k$ remain the same. At the same time, the expression for the potential energy is modified essentially (compare with subsection 2) – on the quantity equal to work performed by this force:
$$E_p=k \frac{r^2}{2}+\int_0^r F_1 d r=k \frac{r^2}{2}+F_0 r$$

## 数学代写|数学建模代写Mathematical Modeling代考|Oscillations of a pendulum in a gravity field

12) 。铰链被认为是理想的光滑，因为没有摩擦损失。静止铰链意味着没有能量从它传递到系统“杆

$$E_k=\frac{1}{2} m v^2=\frac{1}{2} m\left(l \frac{d \alpha}{d t}\right)^2=\frac{1}{2} m l^2\left(\frac{d \alpha}{d t}\right)^2$$

$$E_p=m g h=-m g(l \cos \alpha-1)$$

$$L\left(\alpha, \frac{d \alpha}{d t}\right)=m l\left[\frac{1}{2} l\left(\frac{d \alpha}{d t}\right)^2+g \cos \alpha\right] \quad S[\alpha]=m l \int_{t_1}^{t_2}\left[\frac{1}{2} l\left(\frac{d \alpha}{d t}\right)^2+g \cos \alpha\right] d t$$

## 数学代写|数学建模代写Mathematical Modeling代考|Conclusion

$$m \frac{d^2 r}{d t^2}=-k r+F_0$$
(与第 2 节 (5) 比较) 。将哈密顿原理应用于这样的系统，有必要考虑这种力的存在。显然，广义坐标、速度 和动能的定义 $E_k$ 保持不变。同时，对势能的表达式进行了实质性修改（与第 2 小节相比) — 在等于该力所做 功的量上:
$$E_p=k \frac{r^2}{2}+\int_0^r F_1 d r=k \frac{r^2}{2}+F_0 r$$

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