Linear velocity: $\vec{v}=\vec{w} \times \vec{n}$

Transverse acceleration: $\vec{A}_t=\vec{\alpha} \times \vec{r}$

Radial acceleration: $\vec{A}_n=\vec{\omega} \times(\vec{\omega} \times \vec{n})$

Position vector: $\vec{r}=r \hat{e}_r,$

Differentiating with respect to time,

$\vec{v} =\dot{r} \hat{e}_r+(r \dot{\theta}) \hat{e}_\theta$

$=v_r \hat{e}_r+v_\theta \hat{e}_\theta$

For a particle fixed, $\dot{r}=0$, $v_0 = 0$ and $v_\theta=r \dot \theta$.

$\vec{\omega}=\omega \hat{e} _ z$, $\alpha_ 2 \vec{r}=r \hat{e}_r,$

$\therefore \vec{\omega} \times \vec{r}=\left(\omega \hat{e} _ z\right) \times\left(r \hat{e}_r\right)$

$\vec{A}=\frac{d \vec{v}}{d t}=\frac{d}{d t}\left(\dot{r}_n+r \dot{\theta} \hat{e} _{\theta}\right)$

$=\left(\ddot{r}- r\dot{\theta}^2\right) \hat{e_ r}+( r \ddot{\theta}+2 \dot{r} \dot{\theta}) \hat{e_{\theta}}$

For a fixed particles, $\dot{r}=0, \ddot{r}=0$

$\vec{A}_t =\vec{\alpha} \times \vec{r}$

$=\left(\ddot{\theta} \hat{e}_ z\right) \times\left(r \hat{e}_n\right) =r \ddot{\theta} \hat{e} _\theta$

$\vec{A}_ {\mu }=\vec{\omega} \times(\vec{\omega} \times \vec{r})=\dot{\theta} \hat{e}_ z \times\left(r\dot{\theta} \hat{e}_{\theta}\right)$

$=(-r \dot{\theta}^2) \hat{e}_ r+(r \ddot{\theta}) \hat{e}_{\theta}$.

$=A_r \hat{e} _ r + A _ t \hat{e} _ {\theta} =r \dot{\theta}^2 \hat{e} _r$

Force can cause acceleration.

Torque can cause angular acceleration.

Torque $=\vec{\tau} = \vec{r} \times \vec{F}$, given by right hand screw rule.

$\vec{\tau} = r F \operatorname{sin} \theta=r F_{\perp}$

$\vec{\tau} = F r \sin \theta=F r_{\perp}$

$[\vec \tau]$ has the dimensions of energy.

$\vec\tau=0$, if $\vec{F} =0$

$\vec{r} =0$, $\theta =0 \text { or } 180^{\circ}$

$\tau=r m g \sin \theta = r m g(\frac{x}{r})=x m g$

$\vec{\tau}=\vec{r} \times \vec{f}=$

$\left|\begin{array}{lll}\hat{\imath} & \hat{\jmath} & \hat{k} \\ x & -y & 0 \\ 0 & mg & 0 \end{array}\right|$

$=\hat{k}(xmg) = mgx\hat{k}$

let us consider a particle of mass $m$, and its position it is a with respect coordinate system it is at a particular position vector $\vec{r}$ and its momentum is $\vec{p}$ then its orbital angular momentum,

$\vec{l}=\vec{r} \times \vec{p}$

$=r p \sin \theta=r p_\perp$

$=p(r \sin \theta)=p r_\perp$

$\vec{l}=0, \vec{r}=0 \text { or, } \vec{p}=0 \quad\text { the } \theta=0 \text { or } 180^0$

$\vec{\tau}$ = Torque (or moment of a force ) is rotational analogue of a force.

$\vec{l}$ = Angular momentum is the rotational analogue of linear momentum.

$\vec{l} =\vec{r} \times \vec{p}$

$\frac{d}{d t}(\vec{l}) =\frac{d}{d t}(\vec{r} \times \vec{p})=\frac{d \vec{r}}{d t} \times \vec{p}+\vec{r} \times \frac{d \vec{p}}{d t}=\vec{r} \times \vec{F}=\vec{\tau}$

$\frac{d \vec{l}}{d t} =\vec{\tau}$

Compare,

$\frac{d \vec{p}}{d t}=\vec{F}$

$\vec{L} =\vec{l}_{+} \vec{l}_2+\cdots+\vec{l}_n$,

$\vec{l}_ i = \vec{r} _ i \times \vec{p} _ i$, $(\vec{p}_ i= \vec{m}_ i \vec{l}_i )$

$\vec{L}=\Sigma _ {i=1}^n \vec{l} _ i= \Sigma _ {i=1}^n(\vec{r} _ i \times \vec{p}_i)$

$\frac{d \vec{L}}{d t}=\frac{d}{d t}(\Sigma _ {i=1}^n \vec{l}_ i)=\Sigma_ {i=1}^n \frac{d \overrightarrow{l_ i}}{d t}=\Sigma_ i \vec{\tau}_ i = \vec\tau_i$

$\vec\tau_ i = \vec{r}_ i \times \vec{F}_ i = \vec{r} _ c \times(\vec{F _ i}^{E x t}+\vec{F_i} ^{Int})$

$\vec\tau = \vec\tau_ {Ext}$, $\frac{d\vec{L}}{d t}=\vec{\tau}_{E x t}$, and $\frac{d \vec{p}}{d t}=\vec{F} _{\text {Ext }}$

If $\vec{\tau}_{\text {ext }}=0$, $\frac{d \vec{L}}{d u}=0 \Rightarrow \vec{L}$ is a constant of motion.

$\frac{d \vec{p}}{d t}=\vec{F},$

If $\vec{f}=0$ , then ${\vec{p}}$ is a constant of the motion.

$P$ is conserved.

Let's consider a particle which is moving with a constant velocity.

$\vec{v}$ is a constant.

Momentum: $\vec{p}=m \vec{v}$

The direction of $\vec{p}$ into the page.

$\vec{l}= \vec{r} \times m \vec{v}=r m v \sin \hat{\theta}, \quad$

$r=|\vec{r}|, v=|\vec{v}|$

$\sin \theta=\frac{O M}{r}$

$l= (r \sin \theta)(m v) = (OM)(m v) \Rightarrow$

l is constant of motion.