## Bidimensional generic motion[edit | edit source]

Considering a curvilinear trajectory on which we choose an origin $O$ and a way of movement:

**Definition ** (Curvilinear abscissa (l))

We define **curvilinear abscissa** the portion of bend which connects $x_{1}$ with $O$. If $x_{1}$ is among the positive $x$, according to the chosen way of movement, the curvilinear abscissa $Ox_{1}$ is positive; otherwise if $x_{1}$ is among the negative $x$, according to the chosen way of movement, the curvilinear abscissa is negative. The velocity of the point defines the concordance of symbol between the chosen way of movement and the way of movement of the point: positive velocity makes the point move according to the chosen way of movement, negative velocity makes the point move in the opposite way.

**Average velocity**

$V_{m}(t_{1},t_{2})={\frac {x(t_{2})-x(t_{1})}{t_{2}-t_{1}}}$

is represented by the vector with the same track of $x_{1}x_{2}$ and the same direction of motion. But we notice how this definition of average velocity gives very general information.

We use now the definition of derivative to calculate the **instantaneous velocity** (or just **velocity**):

$V=\lim _{\Delta t\rightarrow 0}V_{m}(t,t+\Delta t)=\lim _{\Delta t\rightarrow 0}{\frac {\Delta x}{\Delta t}}={\frac {dx}{dt}}$

whose vector is tangential to the trajectory in

$x_{1}$ where we find the point at the instant

$t$ considered.

If we apply a second time the derivative we find the **instantaneous acceleration** (or just **acceleration**):

$a=\lim _{\Delta t\rightarrow 0}{\frac {\Delta v}{\Delta t}}={\frac {dv}{dt}}={\frac {d}{dt}}\cdot {\frac {dx}{dt}}={\frac {d^{2}x}{dt^{2}}}$

whose vector is parallel to the banding ray in

$x_{1}$, and perpendicular to the vector

$v(t)$.

${\hat {i}},{\hat {j}},{\hat {k}}$ are versor and they are constant all over the space.
${\vec {x}}(t)$ is the position versor. Breaking it up on the axes x, y e z we find:

${\vec {x}}(t)=x(t)\cdot {\hat {i}}+y(t)\cdot {\hat {j}}+z(t)\cdot {\hat {k}}\ .$

From the decomposition of ${\vec {x}}(t)$ we break up on the three axis ${\vec {v}}(t)$:

${\vec {v}}(t)={\frac {d{\vec {x}}}{dt}}={\frac {dx}{dt}}{\hat {i}}+{\frac {dy}{dt}}{\hat {j}}+{\frac {dz}{dt}}{\hat {k}}\ .$

We also know that:

${\vec {v}}(t)=v_{x}{\hat {i}}+v_{y}{\hat {j}}+v_{z}{\hat {k}}\ ,$

and we find the equivalence:

$v_{x}={\frac {dx}{dt}};\qquad v_{y}={\frac {dy}{dt}};\qquad v_{z}={\frac {dz}{dt}}\ .$

We can also break up the acceleration vector ${\vec {a}}(t)$:

${\vec {a}}(t)={\frac {d{\vec {v}}}{dt}}={\frac {dv_{x}}{dt}}{\hat {i}}+{\frac {dv_{y}}{dt}}{\hat {j}}+{\frac {dv_{z}}{dt}}{\hat {k}}\ .$

We also know that:

${\vec {a}}(t)=a_{x}{\hat {i}}+a_{y}{\hat {j}}+a_{z}{\hat {k}}$

as we have done for the velocity, we find that:

$a_{x}={\frac {dv_{x}}{dt}}={\frac {d^{2}x}{dt}};\qquad a_{y}={\frac {dv_{y}}{dy}}={\frac {d^{2}y}{dt}};\qquad a_{z}={\frac {dv_{z}}{dt}}={\frac {d^{2}z}{dt}}\ .$

**Definition ** (Circular motion)

We define **circular motion** the specific curvilinear motion in which the trajectory is a circumference of centre $O$ and ray $R$

**Definition ** (Curvilinear coordinate)

We define *curvilinear coordinate* (indicated with $s$) the oriented length of the arch

The equation of motion is of the form

$\theta =\theta (t)$

If the circular motion is uniform, the speed is constant: $\omega =\omega _{0}$ and

${\frac {d\theta }{dt}}=\omega _{0};\quad d\theta =\omega _{0}dt;$

$\int _{\theta _{0}}^{\theta (t)}d\theta =\int _{t_{0}}^{t}\omega _{0}dt;$

$[\theta ]_{\theta _{0}}^{\theta (t)}=\omega _{0}[t]_{t_{0}}^{t};\quad \theta (t)-\theta _{0}=\omega _{0}t-\omega _{0}t_{0};$

As

$t_{0}=0s,\quad \omega _{0}t_{0}=0$, we find:

$\theta (t)=\theta _{0}+\omega _{0}t$

we now impose $\theta _{0}=0_{\text{rad}}$,

$\theta =\omega t$

We now have to apply what said on the decomposition of $v(t)$ and $a(t)$ in the general curvilinear motion, and we find:

$\left\{{\begin{aligned}&x(t)=r\cos \theta =r\cos(\omega t)\\&y(t)=r\sin(\omega t)\\\end{aligned}}\right.$

$\left\{{\begin{aligned}&V_{x}(t)={\frac {dx}{dt}}=-\omega r\sin(\omega t)\\&V_{y}(t)=\omega r\cos(\omega t)\\\end{aligned}}\right.$

$\left\{{\begin{aligned}&a_{x}(t)={\frac {dV_{x}}{dt}}=-\omega ^{2}r\cos(\omega t)\\&a_{y}(t)={\frac {dV_{y}}{dt}}=-\omega ^{2}r\sin(\omega t)\\\end{aligned}}\right.$

where $a$ is called **centripetal acceleration**

**Definition ** (Centripetal acceleration)

We define centripetal acceleration, the acceleration which causes the bending of trajectory without modifying the absolute value of the speed ω

→ This is why we talk about of *uniform motion* even though there is an acceleration

$\theta ={\frac {S}{R}}$

$\omega ={\frac {d\theta }{dt}}={\frac {V}{R}}$

As

$V={\frac {dS}{dt}}$ we find that

$\omega ={\frac {1}{R}}\cdot {\frac {dS}{dt}}$

.

$|{\bar {a}}|={\sqrt {a_{x}^{2}+a_{y}^{2}}}={\sqrt {(\omega ^{2}R\cos(\omega t))^{2}+(\omega ^{2}R\sin(\omega t))^{2}}}=\omega ^{2}R{\sqrt {\cos ^{2}(\omega t)+\sin ^{2}(\omega t)}}$

As

$\cos ^{2}\alpha +\sin ^{2}\alpha =1$ we find that

$\cos ^{2}(\omega t)+\sin ^{2}(\omega t)=1$ so

$|{\bar {a}}|=\omega ^{2}R\cdot 1=\omega ^{2}R$

We know that

$\omega ={\frac {V}{R}}$ and we can find the acceleration as a function of velocity

$a_{c}={\frac {V^{2}}{R}}$