The entire mass of a physical object is considered to be concentrated at a single point. This leads to the idealization of a physical object as a particle with no size or shape, but finited mass.
Forces and moments responsible for the motion, even though they exist, are not taken into consideration.
Rectilinear motion
Rectilinear motion is a straight-line motion.
vavg=ΔtΔx,vinst=limΔt→0ΔtΔx=dtdxaavg=ΔtΔv,ainst=limΔt→0ΔtΔv=dtdv
Rectilinear motion
Uniform linear motion
Non-uniform linear motion
Motion with constant acceleration
Motion with non-constant acceleration
Uniform linear motion
v &= v_c = \frac{dx}{dt} \\
\int_{x_0}^x dx &= \int_0^t v_c dt \\
x - x_0 &= v_ct
\end{align}$$
Finally,
$$x = x_0 +v_ct$$
when the object is moving along a straight line path with constan velocity.
#### Non-uniform linear motion
Linear motion but with non-zero acceleration
##### Motion with constant acceleration
1. Galileo's first equation of motion ($a = \frac{dv}{dt}$)
$$a = \frac{dv}{dt} \rightarrow \int_{v_0}^v dv = \int_0^t a_cdt \Rightarrow v = v_0 + a_ct$$
2. Galileo's second equation of motion ($v = \frac{dx}{dt}$)
$$v = \frac{dx}{dt} \rightarrow \frac{dx}{dt} = v_0 + a_ct \rightarrow \int_{x_0}^x dx = \int_0^t (v_0 + a_ct) dt \Rightarrow x = x_0 + v_0t + \frac{1}{2} a_c t^2$$
3. Galileo's thrid equation of motion (chain rule, $a = v \frac{dv}{dx}$)
$$a_c = v \frac{dv}{dx} \rightarrow \int_{v_0}^v vdv = \int_{x_0}^x a_c dx \Rightarrow v^2 = v^2_0 + 2a_c(x - x_0)$$
##### Rectilinear motion with Non-constant acceleration
When performing integration, you can only have ==two variables== in the equation.
1. $a = a(t)$
$$a(t) = \frac{dv}{dt} \Rightarrow \int_{v_0}^v dv = \int_0^t a(t)dt$$
2. $a = a(v)$
$$a(v) = \frac{dv}{dt} \Rightarrow \int_{v_0}^v \frac{dv}{a(v)} = \int_0^t dt$$
3. $a = a(s)$
$$a(s) = \frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} = v \frac{dv}{ds} \Rightarrow \int_{v_0}^v vdv = \int_{s_0}^s a(s)ds$$
### Motion of a particle along curved path
A particle is moving along a curved path.
$$\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t}, \quad \vec{v}_{inst} = \frac{d\vec{r}}{dt}$$
$$\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t}, \quad \vec{a}_{inst} = \frac{d \vec{v}}{dt}$$
#### Direction of velocity
The direction of the instantaneous velocty is always ==tangential== to the path and points along the direction of the movement.
$$\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta \vec{r}}{\Delta t}$$
When $\Delta t$ approaches zero, the velocity vector is tangential to the path at the point.
#### Direction of acceleration
$$\vec{a} = \frac{d \vec{v}}{dt}$$
The acceleration would be non-zero, if the magnitude or direction may be changing.
The acceleration is pointing towards the ==concave== portion of the path. $\vec{v}'-\vec{v}$
### Cartesian system
The position vector of the particle
$$\vec{r} = x(t) \hat{i} + y(t) \hat{j}$$
$$\begin{align}
\vec{v} &= \frac{d \vec{r}}{dt} = \frac{d}{dt} (x\hat{i} + y\hat{j}) \\
&= \left( \frac{dx}{dt} \hat{i} + x \frac{d \hat{i}}{dt}\right) + \left( \frac{dy}{dt} \hat{j} + y \frac{d \hat{j}}{dt}\right) \\
&= \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j}
\end{align}$$
$$\begin{align}
\vec{v} &= \dot{x} \hat{i} + \dot{y} \hat{j} \\
\vec{a} &= \ddot{x} \hat{i} + \ddot{y} \hat{j}
\end{align}$$
### Tangent-Normal Coordinate system
The unit vectors in this system are directed in the tangent and normal direction at any given point of the path.
1. $\hat{e_t}$ is a unit vector along the ==tangential== direction at a point on a curved path.
2. $\hat{e_n}$ is a unit vector along the ==normal== direction which always point towards the ==concave== direction of a curve or towards the center of the osculating circle.
3. $\hat{e_t}$ and $\hat{e_n}$ are always perpendicular to each other and $\hat{e_t}$ points in the direction of movement.
The velocity of a particle along a curved path is always in the tangential direction.
$$\vec{v} = v \hat{e}_t$$
The acceleration,
$$\begin{align}
\vec{a} &= \frac{d \vec{v}}{dt} = \frac{d}{dt} (v \hat{e}_t) \\
&= \hat{e}_t \frac{dv}{dt} + v \frac{d \hat{e}_t}{dt} \\
&= \dot{v} \hat{e}_t + v \frac{d \hat{e}_t}{dt}
\end{align}$$
$\frac{d \hat{e_t}}{dt}$ 는 단위벡터 미분이지만, 방향이 계속 변하기 때문에 0은 아니다.
$$\begin{align}
\hat{e}_t &= \cos \theta \hat{i} + \sin \theta \hat{j} \\
\frac{d \hat{e}_t}{dt} &= \dot{\theta} (- \sin \theta \hat{i} + \cos \theta \hat{j}), \quad (\hat{k} \times \hat{i} = \hat{j} \ \text{and} \ \hat{k} \times \hat{j} = -\hat{i}) \\
\frac{d \hat{e}_t}{dt} &= \dot{\theta} \hat{k} \times (\cos \theta \hat{i} + \sin \theta \hat{j}) \\
\frac{d \hat{e}_t}{dt} &= (\dot{\theta} \hat{k}) \times \hat{e}_t, \quad (\hat{k} \times \hat{e}_t = \hat{e}_n)
\end{align}$$
This shows that the derivative of a unit vector is the cross product of two vectors.
1. a vector pointing towards the positive $\hat{k}$ direction with magnitude $\dot{\theta}$
2. the unit vector itself $\hat{e_t}$
$$\frac{d \theta}{dt} = \frac{d \theta}{ds} \frac{ds}{dt} = \frac{1}{\rho}v \Rightarrow \dot{\theta} = \frac{v}{\rho}$$
Finally, the acceleration is
$$\vec{a} = \dot{v} \hat{e}_t + \frac{v^2}{\rho} \hat{e}_n$$
급한 커브 $\iff$ $\rho$ 가 작다. 반지름이 작을수록 더 휜 것. 직선은 $\rho = \infty$
### Polar Coordinate system
Two variables are used to descriv a motion of a particle along a two dimensional path.
1. $\hat{e}_r$ is the unit vector that points radially outward in the direction of position vector.
2. $\hat{e}_{\theta}$ is the unit vector perpendicular to $\hat{e_r}$ and always points towards the direction of increasing $\theta$.
3. $\hat{e}_r, \hat{e}_{\theta} \ \text{and} \ \hat{k}$ form a righ-handed coordinate system and therefore, $\hat{e}_r \times \hat{e}_{\theta} = \hat{k}, \hat{k} \times \hat{e}_r = \hat{e}_{\theta} \ \text{and} \ \hat{k} \times \hat{e}_{\theta} = -\hat{e}_r$
The position vector
$$\vec{r} = r \hat{e}_r$$
The velocity vector
$$\begin{align}
\vec{v} &= \frac{d \vec{r}}{dt} = \frac{d}{dt} (r \hat{e}_r) \\
&= r \frac{d \hat{e}_r}{dt} + \hat{e}_r \frac{dr}{dt}, \quad \left(\frac{d \hat{e}_r}{dt} = (\dot{\theta} \hat{k}) \times \dot{\theta} \hat{e}_{\theta} \right) \\
\vec{v} &= \dot{r} \hat{e}_r + r \dot{\theta} \hat{e}_{\theta}
\end{align}$$
The acceleration vector
$$\begin{align}
\vec{a} &= \frac{d \vec{v}}{dt} = \frac{d}{dt} (\dot{r} \hat{e}_r + r \dot{\theta} \hat{e}_{\theta}) \\
&= \left[ \ddot{r} \hat{e}_r + \dot{r} \frac{d \hat{e}_r}{dt} \right] + \left[ r \dot{\theta} \frac{d \hat{e}_{\theta}}{dt} + r \hat{e}_{\theta} \frac{d \dot{\theta}}{dt} + \dot{\theta} \hat{e}_{\theta} \frac{dr}{dt}\right] \\
&= \ddot{r} \hat{e}_r + \dot{r} \dot{\theta} \hat{e}_{\theta} - r \dot{\theta}^2 \hat{e}_r + r \ddot{\theta} \hat{e}_{\theta} + \dot{r} \dot{\theta} \hat{e}_{\theta} \\
&= \ddot{r} \hat{e}_r + 2 \dot{r} \dot{\theta} \hat{e}_{\theta} - r \dot{\theta}^2 \hat{e}_r + r \ddot{\theta} \hat{e}_{\theta} \\
\vec{a}&= (\ddot{r} - r \dot{\theta}^2) \hat{e}_r + (2 \dot{r} \dot{\theta} + r \ddot{\theta}) \hat{e}_{\theta}
\end{align}$$
Converting polar coordinates to cartesian coordinates
$$\begin{align}
\hat{e}_r &= \cos \theta \hat{i} + \sin \theta \hat{j} \\
\hat{e}_{\theta} &= -\sin \theta \hat{i} + \cos \theta \hat{j}
\end{align}$$
### Relative motion analysis
$$\vec{r}_A = \vec{r}_B + \vec{r}_{A/B}$$
$\vec{r}_{A/B}$ is the position vector of $A$ with respect to $B$.
$A/B$ = $B$ 에 대한 $A$ = $B$ 가 관측자. 순서 바뀌면 부호 뒤집힘.
$$\begin{align}
\vec{v}_{A/B} &= \vec{v}_A - \vec{v}_B \\
\vec{a}_{A/B} &= \vec{a}_A - \vec{a}_B
\end{align}$$
The translating frame moves along with B as $B$ translates, but it does not change it's orientation.
측정 frame 이 고정 혹은 평행이동, 회전하면 위의 식들이 성립하지 않는다.