Relation between angular acceleration and linear acceleration

Equation for the relation between linear acceleration and angular acceleration:

Deduce the equation from the General Method:

We know that angular acceleration is

$\alpha=\frac{\Delta \omega}{\Delta t} \qquad (1)$

$\alpha=\frac{\Delta (v/r)}{\Delta t} \qquad \left( \because \omega=\frac{v}{r} \right)$

$\alpha=\frac{1}{r} \frac{\Delta v}{\Delta t}$

If $\Delta t \rightarrow 0$, then the above equation can be written as

$\alpha=\frac{1}{r} \: \underset{\Delta t \rightarrow 0}{Lim} \frac{\Delta v}{\Delta t}$

Where
$\underset{\Delta t \rightarrow 0}{Lim} \: \frac{\Delta v}{\Delta t} \rightarrow$  Instantaneous Acceleration $(a)$

$\alpha=\frac{a}{r}$

Deduce the equation from the Differential Method

We know that angular acceleration is

$\alpha=\frac{d \omega}{dt} \qquad (1)$

$\alpha=\frac{d (v/r)}{dt} \qquad \left( \because \omega=\frac{v}{r} \right)$

$\alpha=\frac{1}{r} \frac{dv}{dt} \qquad (2)$

Where
$\frac{dv}{dt}$ = Instantaneous Acceleration $(a)$

Now equation $(2)$ can be written as

$\alpha=\frac{a}{r}$

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Relation between angular velocity and linear velocity

Mathematical Relation between angular velocity $(\omega)$ and linear velocity$(v)$:

We know that the angular displacement of the particle is

$\Delta \theta= \frac{\Delta s}{r} \qquad(1)$

Where $r$ = The radius of a circle.

Now divide by $\Delta t$ on both side of equation $(1)$

$\frac{\Delta \theta}{\Delta t}=\frac{1}{r} \frac{\Delta s}{\Delta t} $

If $\Delta t \rightarrow 0$, then the above equation can be written as

$\underset{\Delta t \rightarrow 0}{Lim}\: \frac{\Delta \theta}{\Delta t}=\frac{1}{r}\: \underset{\Delta t \rightarrow 0}{Lim} \: \frac{\Delta s}{\Delta t} \qquad(2)$

Where
$\underset{\Delta t \rightarrow 0}{Lim}\: \frac{\Delta \theta}{\Delta t}$ = Instantaneous Angular Velocity $(\omega)$

$\underset{\Delta t \rightarrow 0}{Lim} \: \frac{\Delta s}{\Delta t}$= Instantaneous Linear Velocity $(v)$

Now equation $(2)$ can be written as

$\omega=\frac{1}{r}v$

$v=r\omega$

The above equation shows that linear or tangential velocity depends on the angular velocity and radius of the circular path. This means that if the radius of the circular path increases with the same angular velocity, then the linear velocity will also increase.

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Conservation's Law of linear momentum and its Derivation

Derivation of Conservation's law of Linear momentum from Newton's Second Law and Statement: Let us consider, A particle that has mass $m$ and is moving with velocity $v$ then According to Newton's second law the applied force on a particle is

Derivation From General Form	Derivation from Differential Form
$F=ma$ $F=m \frac{\Delta v}{\Delta t}$ $F= \frac{\Delta (mv)}{\Delta t}$ $F= \frac{\Delta P}{\Delta t} \qquad \left( \because P=mv \right)$ If the applied force on a body is zero then $\frac{\Delta P}{\Delta t}=0$ $\Delta P=0$ Here $\Delta P$ is change in momentum of the particle .i.e $P_{2}-P_{1}=0$ $P_{2}=P_{1}$ $P=Constant$	$F=ma$ $F=m \frac{dv}{dt}$ $F= \frac{dmv}{dt}$ $F= \frac{dP}{dt} \qquad \left( \because P=mv \right)$ If the applied force on a body is zero then $\frac{dP}{dt}=0$ $dP=0$ On integrating the above equation $P=constant$

From the above equation, we conclude the statement i.e.

When no external force is applied to a particle, the total momentum of a particle is always conserved.

Derivation of Conservation's law of Linear momentum from Newton's Third Law and Statement: Let us consider, The two-particle that have mass $m_{1}$ and $m_{2}$ are moving towards each other. If

The velocity of the particles $m_{1}$ before collision is = $u_{1}$

The velocity of the particles $m_{2}$ before collision is = $u_{2}$

The velocity of the particles $m_{1}$ after collision is = $v_{1}$

The velocity of the particles $m_{2}$ after the collision is = $v_{2}$

When these particles collide to each other than according to Newton's third law

$F_{12}=-F_{21}$

$m_{1}a_{1}=m_{2}a_{2}$

$m_{1}\frac{\Delta V_{1}}{\Delta t}=m_{2} \frac{\Delta V_{2}}{\Delta t} \quad (1) \qquad \left(\because a=\frac{\Delta V} {\Delta t} \right)$

Where
$\Delta V_{1}$= Change in Velocity of mass $m_{1}$ i.e $\left( v_{1}-u_{1} \right)$
$\Delta V_{2}$= Change in Velocity of mass $m_{2}$ i.e $\left( v_{2}-u_{2} \right)$

So from equation $(1)$

$m_{1}\frac{\left( v_{1}-u_{1} \right)}{\Delta t}=m_{2} \frac{\left( v_{2}-u_{2} \right)}{\Delta t}$

$m_{1}\left( v_{1}-u_{1} \right)=m_{2} \left( v_{2}-u_{2} \right)$

$m_{1} v_{1}- m_{1} u_{1}=m_{2}v_{2}- m_{2} u_{2}$

$m_{1} u_{1}+ m_{2} u_{2}=m_{1}v_{1} + m_{2} v_{2}$

From the above equation, we conclude the statement i.e.

When two particle collide two each other. If the total momentum of the particles before the collision and total momentum of the particles after is collision is same then it is known as conservation law of linear momentum.

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Alternating Current Circuit containing Capacitance only (C-Circuit)

Alternating Current Circuit Containing Capacitance Only (C-Circuit): Let us consider, A circuit containing a capacitor of capacitance $C$ only which is connected with an alternating EMF i.e electromotive force source i.e.

Let us consider, A circuit containing a coil of inductance $L$ only which is connected with an alternating EMF i.e electromotive force source i.e.

$E=E_{\circ}sin\omega t\qquad(1)$

When alternating emf is applied across the capacitor plates then the charge on capacitor plates varies continuously and correspondingly current flows in the connecting leads. Let the charge on the capacitor plates is $q$ and the current in the circuit at any instant is $i$. Since there is no resistance in the circuit then the instantaneous potential difference is $\frac{q}{C}$ across the capacitor plates must be equal to the applied emf i.e.

$\frac{q}{C} = E_{\circ} sin \omega t$

$q = CE_{\circ} sin \omega t \qquad(2)$

The instantaneous current $i$ in the circuit is, therefore

$i=\frac{dq}{dt} \qquad(3)$

Now substitute the value of $q$ from equation $(2)$ to equation $(3)$

$i=\frac{d}{dt} CE_{\circ} sin \omega t $

$i=CE_{\circ} \omega cos\omega t$

$i=\omega CE_{\circ} cos\omega t$

$i=\frac{E_{\circ}}{\frac{1}{\omega C}} cos\omega t$

$i=\frac{E_{\circ}}{X_{C}} cos\omega t$

Where $X_{C}= \frac{1}{\omega C}$ is known as capacitive reactance.

$i=i_{\circ} cos\omega t \qquad(4)$

Where the $i_{\circ}=\frac{E_{\circ}}{X_{C}}$ is the maximum current in the circuit. Now apply Ohm's law in this equation and we find that the term $X_{C}=\frac{1}{\omega C}$ has the dimension of resistance. It represents the 'effective opposition' of the capacitor to the flow of alternating current. It is known as the 'reactance of the capacitor' or 'capacitive reactance' and is denoted by $X_{C}$. The capacitive reactance is infinite for DC for which $f=0$.

$i=i_{\circ} sin \left( \omega t + \frac{\pi}{2} \right) \qquad(4)$

Now compare equation $(1)$ and equation $(4)$ which shows that an alternating circuit containing a capacitor only, the current leads the emf by a phase angle of $\frac{\pi}{2}$ or $90^\circ$ (or the emf lags behind the current by a phase angle of $\frac{\pi}{2}$). The phase diagram between EMF and the current of a capacitor is shown below-

The phasor diagram between the EMF and current of a capacitor is also shown in the given figure below-

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Alternating Current Circuit containing Inductance only (L-Circuit)

Alternating Current Circuit Containing Inductance only (L-Circuit): Let us consider, An alternating current circuit containing a coil of inductance $L$ only. This inductor is connected with an alternating EMF i.e electromotive force source i.e.

$E=E_{\circ}sin\omega t\qquad(1)$

The current $i$ in coil varies continuously then an opposite emf is induced in the coil whose magnitude is $L\frac{di}{dt}$ So the net instantaneous of the circuit:

$E_{\circ}sin\omega t -L\frac{di}{dt}=0$

$E_{\circ}sin\omega t =L\frac{di}{dt}$

$di=\frac{E_{\circ}}{L}sin\omega dt$

Now integrate the above equation then the above equation can be written as

$\int di=\int \frac{E_{\circ}}{L}sin\omega dt$

$\int di= \frac{E_{\circ}}{L} \int sin\omega dt$

$i= \frac{E_{\circ}}{L} \frac{-cos\omega t}{\omega}$

$i= -\frac{E_{\circ}}{\omega L} cos\omega t$

$i= -\frac{E_{\circ}}{X_{L}} cos\omega t$

Where $X_{L}= \omega L$ is known as inductive reactance.

$i= -i_{\circ} cos\omega t$

Where $i_{\circ}=\frac{E_{\circ}}{X_{L}}$ is known as the maximum value of current in the circuit. Now compare this equation to Ohm's law then we find that the term $X_{L}=\omega L$ has the dimensions of resistance. It defines the 'effective opposition' of the coil to the flow of alternating current. it is known as the 'reactance of the coil' or 'inductive reactance' and it is denoted by $X_{L}$. The inductive reactance $X_{L}$ is zero for DC at which frequency is zero.

$i= -i_{\circ} sin \left(\frac{\pi}{2}- \omega t \right)$

$i= i_{\circ} sin \left(\omega t - \frac{\pi}{2} \right) \qquad(2)$

Now compare equation $(1)$ and equation $(2)$ which shows that an alternating circuit containing an inductor only, the current lags behind the emf by a phase angle of $\frac{\pi}{2}$ or $90^\circ$ (or the emf leads the current by a phase angle of $\frac{\pi}{2}$). The phase diagram between EMF and the current of an inductor is shown below-

The phasor diagram between the EMF and current of an inductor is also shown in the given figure below-

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Electric field Intensity (Definition) and Electric field Intensity due to point charge

Definition of Electric Field Intensity:

The force acting on the per unit test charge in electric field is called the Electric field intensity. It is represented by $'E'$.

Let us consider that a test-charged particle of $q_{0}$ Coulomb is placed at a point in the electric field and a force $F$ acting on them so the electric field intensity at that point

$ \overrightarrow{E}=\frac{\overrightarrow{F}}{q_{0}}$

SI Unit:$\quad Newton/Coulomb$ $ (N/C)$ $\quad Kg-m^{2}/sec^{3} A$

Dimension: $\left [ML^{2}T^{-3}A^{-1} \right ]$

Physical Significance of Electric Field:

The force experienced by a charge is different at different points in space. So electric field intensity also varies from point to point. In general, Electric field intensity is not a single vector quantity but it is a set of infinite vector and each point in space have a unique electric field intensity. So electric field is an example of the vector field.

Electric Field Intensity due to a Point Charge:

Let us consider that an isolated point charge of $+q$ Coulomb is placed at a point $O$ in a medium whose dielectric constant is $K$. If a test charge particle of $q_{0}$ Coulomb is placed at a point $P$ in the electric field at a distance $r$ from point $O$. So the electric field intensity (magnitude form) at point $P$

Electric field intensity due to a point charge

$E=\frac{F}{q_{0}}\qquad (1)$

According to Coulomb's Law:

$F=\frac{1}{4\pi \varepsilon _{0}K}\frac{qq_{0}}{r^{2}}\qquad (2)$

From equation $(1)$ and equation $(2)$, we can write

$E=\frac{1}{4\pi \varepsilon _{0}K}\frac{q}{r^{2}} \qquad (3)$

For air or vacuum $K=1$, Then from equation $(3)$

$E=\frac{1}{4\pi \varepsilon _{0}}\frac{q}{r^{2}} \qquad (4)$

Where

$\frac{1}{4\pi \varepsilon _{0}}=9\times10^{9} N-m^{2}/C^{2}$

From equation $(4)$

$E=9\times 10^{9} \frac{q}{r^{2}}\qquad (5)$

In Vector Notation:

$\overrightarrow{E}=\frac{1}{4\pi \varepsilon _{0}K}\frac{q}{r^{2}}r\hat{}$

If a system containing $'n'$ point charge $q_{1},q_{2},q_{3},..........q_{n}$ then electric field intensity due to the system of charge particle will be equal to the vector sum of the intensities i.e.

$\overrightarrow{E}=\overrightarrow{E_{1}}+\overrightarrow{E_{2}}+\overrightarrow{E_{3}}+....+\overrightarrow{E_{n}}$

$\overrightarrow{E}=\frac{1}{4\pi \varepsilon _{0}K}\sum_{i=0}^{n}\frac{q_{i}}{r_{i}^{2}}r_{i}\hat{}$

Where $r_{i}$ is the distance from point $'P'$ to charge $q_{i}$.

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