### 物理代写|理论力学作业代写Theoretical Mechanics代考|Mechanics of the Free Mass Point

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## 物理代写|理论力学作业代写Theoretical Mechanics代考|Velocity and Acceleration

The motion of a mass point is characterized by:
position vector: $\mathbf{r}(t)$,
velocity vector: $\quad \mathbf{v}(t)=\dot{\mathbf{r}}(t)$.
acceleration vector: $\mathbf{a}(t)=\ddot{\mathbf{r}}(t)$.
Higher time derivatives do not interest us in mechanics; very often they even fail to exist because in many realistic cases the acceleration is not a continuous function of time.

The typical task for mechanics consists of the calculation of the path line (trajectory) $\mathbf{r}(t)$ on the basis of a given acceleration $\mathbf{a}(t)=\ddot{\mathbf{r}}(t)$. For this purpose one has obviously to integrate $\mathbf{a}(t)$ twice with respect to time. After each integration an integration constant appears which remains undetermined unless we have two initial conditions at our disposal to fix these constants. In this connection let us assume that we know the velocity and the position of the mass point (particle) at a certain time $t_{0}$, i.e.
$$\mathbf{a}(t) \text { for all } t, \mathbf{v}\left(t_{0}\right) \text {, and } \mathbf{r}\left(t_{0}\right)$$
are given. Then the velocity of the particle is determined by
$$\mathbf{v}(t)=\mathbf{v}\left(t_{0}\right)+\int_{t_{0}}^{t} d t^{\prime} \mathbf{a}\left(t^{\prime}\right)$$
and the position vector by:
$$\mathbf{r}(t)=\mathbf{r}\left(t_{0}\right)+\mathbf{v}\left(t_{0}\right)\left(t-t_{0}\right)+\int_{t_{0}}^{t}\left[\int_{t_{0}}^{t^{\prime}} d t^{\prime \prime} \mathbf{a}\left(t^{\prime \prime}\right)\right] d t^{\prime}$$
Before we inspect these relations using simple examples we want to formulate the characteristic parameters of a mass point $\mathbf{r}(t), \mathbf{v}(t), \mathbf{a}(t)$ in different systems of coordinates.

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Simple Examples

(a) Mass Point on a Straight Line
We can describe the motion without referring to any special system of coordinates. If $\mathbf{c}$ is a vector in the direction of the motion and $\mathbf{b}$ a vector perpendicular to it then we can write for the position vector of the mass point (Fig. 2.4):
$$\mathbf{r}(t)=\mathbf{b}+\alpha(t) \mathbf{c}$$
From this, the respective time derivatives give us the velocity and acceleration:
$$\mathbf{v}(t)=\dot{\alpha}(t) \mathbf{c} ; \quad \mathbf{a}(t)=\ddot{\alpha}(t) \mathbf{c}$$
(b) Uniform Straight-Line Motion
Therewith it is meant the most simple form of motion, namely the one without any acceleration:
$$\mathbf{a}(t)=0 ; \quad \mathbf{v}(t)=\mathbf{v}_{0} \quad \text { for all } t$$

The third summand in $(2.2)$ then disappears:
$$\mathbf{r}(t)=\mathbf{r}\left(t_{0}\right)+\mathbf{v}{0}\left(t-t{0}\right) .$$
This formally agrees with (2.26). The motion is thus carried out rectilinearly in the direction of the constant velocity vector $\mathbf{v}{0}$. It is called ‘uniform’ since the same distances are covered in equal time intervals (Fig. 2.5). (c) Uniformly Accelerated Motion Now we assume a constant acceleration $$\mathbf{a}(t)=\mathbf{a}{0}$$
That means in (2.2):
\begin{aligned} \int_{t_{0}}^{t}\left[\int_{t_{0}}^{t^{\prime}} d t^{\prime \prime} \mathbf{a}\left(t^{\prime \prime}\right)\right] d t^{\prime} &=\int_{t_{0}}^{t}\left[\mathbf{a}{0}\left(t^{\prime}-t{0}\right)\right] d t^{\prime}=\ &=\mathbf{a}{0}\left(\frac{t^{2}}{2}-\frac{t{0}^{2}}{2}\right)-\mathbf{a}{0} t{0}\left(t-t_{0}\right)=\ &=\frac{1}{2} \mathbf{a}{0}\left(t-t{0}\right)^{2} . \end{aligned}
We therewith get as path line:
$$\mathbf{r}(t)=\mathbf{r}\left(t_{0}\right)+\mathbf{v}\left(t_{0}\right)\left(t-t_{0}\right)+\frac{1}{2} \mathbf{a}{0}\left(t-t{0}\right)^{2} .$$
The velocity of the mass point increases linearly with time (Fig. 2.6):
$$\mathbf{v}(t)=\mathbf{v}\left(t_{0}\right)+\mathbf{a}{0}\left(t-t{0}\right)$$

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Fundamental Laws of Dynamics

Up to now we have restricted ourselves to describe the motion of a mass point without investigating the primary cause of the motion. From now on, the latter will be the focus of our considerations. The goal is to develop procedures by which one can derive the explicit movement of the mass point from a known driving cause.

We start with a few very general remarks concerning the challenges and possibilities of every physical theory; here, however, with the special perspective on Classical Mechanics. Like any physical theory mechanics also is based on
definitions and theorems
The definitions are reasonably separated into basis definitions and following definitions:

By basis definitions we mean concepts like position, time, mass, …, which are no further commented on in the course of the theory. Following definitions are entities derived from the basis definitions such as velocity, acceleration, momentum, … Analogously we have also to dccompose the theorenīs:

Axioms are a matter of basic empirical facts which are mathematically not provable and will not be further justified within the theory. In the framework of Classical Mechanics these are ‘Newton’s axloms of motion’. By concluslons we understand the actual results of the physical theory. By use of the concept of the “mathematical proof’ they emerge out of the basis definitions and axioms which together are called the postulates of the theory.

The ‘ultimate judge’ of any physical theory is the experiment. The value of a theory is measured by the degree of agreement of its conclusions with the manifestations of nature. It is known today that Classical Mechanics is not able to correctly describe all movements and manifestations of the inanimate nature. In particular in atomic and subatomic regions modifications have become necessary. But one can regard Classical Mechanics as a self-consistent limiting case of a higher all-embracing theory, if it is finally found.

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Velocity and Acceleration

r(吨)=r(吨0)+在(吨0)(吨−吨0)+∫吨0吨[∫吨0吨′d吨′′一种(吨′′)]d吨′

## 物理代写|理论力学作业代写Theoretical Mechanics代考|Simple Examples

(a) 直线上

r(吨)=b+一种(吨)C

(b) 匀速直线运动

r(吨)=r(吨0)+在0(吨−吨0).

∫吨0吨[∫吨0吨′d吨′′一种(吨′′)]d吨′=∫吨0吨[一种0(吨′−吨0)]d吨′= =一种0(吨22−吨022)−一种0吨0(吨−吨0)= =12一种0(吨−吨0)2.

r(吨)=r(吨0)+在(吨0)(吨−吨0)+12一种0(吨−吨0)2.

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