Let us say a P particle is moving along some path.

We are studying this with respect to some reference frame in which we have fixed coordinate axis.

$\overrightarrow{p p^{\prime}} =\Delta \vec{r}$ =$\vec{r}(t+\Delta t)-\vec{r}(t)$

$\overrightarrow{v_p}= \text{limit}_{\Delta t \rightarrow 0} \frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}$

$\vec{r}(t+\Delta t)= (t+\Delta x) \hat{\imath}+(y+\Delta y) \hat{\jmath}$

$\vec{r}(t)=x \hat{\imath}+y \hat{\jmath}$

$\Delta \vec{r}= \Delta x \hat{\imath}+\Delta y \hat{\jmath}$

$\vec{V}$ =$\frac{\Delta x}{\Delta t} \hat{\imath}$ + $\frac{\Delta y}{\Delta t} \hat{\jmath} =v_2 \hat{\imath}+v_t \hat{\jmath}$

$\vec{v}=v_x \hat{\imath}+v_y \hat{\jmath}$

$V=|\vec{v}|=\sqrt{v_x^2+v_y^2}$ or $v$.

$\operatorname{tan} \theta=\frac{v_y}{v_x}$

Acceleration $\vec{a}$, is given as the rate of change of velocity.

$\Delta \vec{v}=\vec{v}(t+\Delta t)-\vec{v}(t)$

$\vec{a}=\operatorname{lim}_{\Delta t \rightarrow 0} \frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}$

$\vec{a}=a_x \hat{\imath}+a_y \hat{\jmath}$

$a_x$ = x component of accelaration = $\frac{\Delta v_x}{\Delta t}$

$a_y$ = y component of accelaration = $\frac{\Delta v_y}{\Delta t}$

$a_x=\frac{d v_x}{d t}= \frac{d}{d t}\left (\frac{d x}{d t}\right)=\frac{d^2 x}{d t^2}$

$a_y=\frac{d v_y}{d t}=\frac{d}{d t}\left(\frac{d y}{d t}\right)=\frac{d^2 y}{d t^2}$

When a particle moves in a straight line, $\vec{v}, \vec{a}$ are along the same direction.

When a particle is moving in a curved path,

Velocity $\vec{v}$ at any instant is tangent to the path.

What about the acceleration $\vec{a}$ ?

Acceleration $=\frac{\Delta \vec{v}}{\Delta t}$.

The direction of acceleration has to be along $\Delta \vec{v}$.

Accelaration on a curved path is due to two components.

a) change in speed of the particle along $\hat{e}_t$

b) $2^{\text {nd }}$ component is perpendicular to the $\hat{e_t}$, because direction of $\vec{v}$ changes due to the curved nature of path.

$2^{\text {nd }}$ component perpendicular to the $\hat{e_t}$, points towards the centre of circle in which the particle is moving.

$a_n \propto (\text { speed })^2$

If locally, the particle is moving in a circle of radius $R$.

$a_n=\frac{v^2}{R}$

R = radius of curvature

Particle following a circular path.

Uniform $\Rightarrow$ constant speed $=v$

$\vec{a_p}=\frac{v^2}{R}$

$\Delta \theta \rightarrow$ angle covered by the particle with respect to the centre of circle,

$\Delta \theta \rightarrow$ angular displacement

$\Delta t \rightarrow$ time taken for particle to go from $P \rightarrow P^{\prime}$

Angular velocity =$\frac{\Delta \theta}{\Delta t}=\omega$

$\omega=\frac{\Delta \theta}{\Delta t}$

$v=\frac{PP^{\prime}}{\Delta t}$

speed = $\frac{\Delta S}{\Delta t}$=$\frac{\widehat{P P^{\prime}}}{\Delta t}$=$\frac{R \Delta \theta}{\Delta t}$

=R$\omega$

acceleration =$\frac{v^2}{R}$=$\frac{R^2 \omega^2}{R}$=R$\omega^2$

acceleration towards centre $\rightarrow$ centripetal acceleration.

Time taken for one revolution

=time period =T

Number of revolution made in 1sec = frequency =$\frac{1}{T}$ =$\nu$

$\omega$=$\frac{ \text {angular displacement} }{\text { time }}$=$\frac{2 \pi}{T}$

$\omega=\frac{2 \pi}{T}=2 \pi \nu$

$v=R \omega=2 \pi R \nu$

$a=R \omega^2=4 \pi^2 \nu^2 R$

If circular motion is not uniform speed, the speed of the particle changes.

$\Rightarrow \omega$ is not constant.

$\alpha \rightarrow$ angular acceleration =$\frac{d \omega}{d t}$

$\vec{v} =R \omega \hat{e_t}$

$\vec{a} =R \frac{d \omega}{d t} \hat{e_t}+R \omega^2 \hat{e}_n$

$\quad =R \alpha \hat{e}_t+R \omega^2 \hat{e}_n$

$\vec{a}=$ constant

$\vec{v}_0$ at $t=0, \vec{v}$ at time t

$\vec{r}_0$ at $t=0, \vec{r}$ at time t.

$\vec{a}=\frac{\vec{v}-\vec{v}_ 0}{(t-0)}=\frac{\vec{v}-\vec{v}_0}{t}$

$\vec{v}=\overrightarrow{v_ 0}+\overrightarrow{a_0 t}$

$v_ x=v_ {0 x} + a_x t$

$v_ y=v_ 0 y + a_y t$

$\vec{a}$ = $a_ x \hat{\imath}+a_y \hat{\jmath}$

$\vec{v}$ = $v_x \imath+v _ y \jmath$

$\overrightarrow{v_ 0}$ = $v_ {0x} \hat{\imath}$+ $v_{0y}\hat{\jmath}$

$\vec{r}$ (position vector) at time $t$

given $\vec{r}_0$ at $t=0$.

Average velocity =$\frac{\overrightarrow{v_0}+\vec{v}}{2}$

displacement =$\vec{r}-\overrightarrow{r_0}$= $(\frac{\overrightarrow{v_0}+\vec{v}} {2}) t$

=$\frac{\overrightarrow{v_0}+(\overrightarrow{v_0}+\vec{a} t) t}{2}$

$\vec{r}-\vec{r}_0=\vec{v} _0 t+\frac{1}{2} \vec{a} t^2$

$\vec{r}=\vec{r}_0+\vec{v} _0 t+\frac{1}{2} \vec{a} t^2$

$x=x_0+v _ {0 x} t+\frac{1}{2} a _ x t^2$

$y=y_0+v _ 0 y t+\frac{1}{2} a _y t^2$

Equation of the path: relate x to y.

Initial velocity $\vec{v_0}$

The speed $v_0$ and the angle $\theta_0$.

$v_{0 x}=v_0 \cos \theta_0$

$v_{0 y}=v_0 \sin \theta_0$

We wish to find coordinate of projectile (p) at a later time.

$\vec{a}=-g \hat{\jmath}, a_x=0, a_y=-g$

$\vec{v}_0=\left(v_0 \cos \theta_0\right) \hat{\imath}+\left(v_0 \sin \theta_0\right) \hat{\jmath}$

$\vec{v}_0$=(0,0) ${x_0}=0$, ${y_0}=0$

x=$(v_0 t)=(v_0 \cos \theta_0) t$

$y=(v_0 \sin \theta_0) t-\frac{1}{2} g t^2$

Eliminate t

$v_x=v_0 \cos \theta_0$

$v_y=v_0 \sin \theta_0-g t$

t=$\frac{x}{v_0 \cos \theta_0}$

y=$v_0 \sin \theta_0 \frac{x}{v_0 \cos \theta_0}$ -$\frac{1}{2}$ g $\frac{x^2} {v_0^2\cos^2 \theta_0}$

=$\tan \theta_0 x-\frac{1}{2} \frac{g}{v_0^2 \cos ^2 \theta_0} x^2$

=$a x+b x^2$

A particle which is thrown as a projectile has the path taken by that is of a parabola.

$v_y=0$

$v_y=v_{y_0}-g t$

$\Rightarrow t$=$\frac{v_{y_0}}{g}$=$\frac{v_0 \sin \theta_0}{g}$

maximum height y(t)

$(v_0 \sin \theta_0)$ $\frac{v_0 \sin \theta_0}{g}$-$\frac{1}{2} g \cdot \frac{v_0^2 \sin \theta_0^2}{g^2}$

=$\frac{v_0{ }^2 \sin ^2 \theta_0}{2 g}$

Range: Time taken for flight =$T_f$

time at which $y=0$.

0 = $v_0 \sin \theta_0 t$ - $\frac{1}{2} g t^2$

$T_f$ = $\frac{2 v_0 \sin \theta_0}{g}$

R = x = $(v_0 \cos \theta_0) T_f$

= $(v_0 \cos \theta_0)$ $\cdot$ $\frac{2 v_0 \sin \theta_0}{g}$

= $\frac{v_0^2 \sin (2 \theta_0)}{g}$

R= $\frac{v_0^2 \sin 2 \theta_0}{g}$

Maximum R for a given $v_0$ occurs at $\theta_0=45^0$

Neglected any other force which would act on the body other than gravity.

Air may exect force on the body

Drag

Lift