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Motion of body in a vertical circle and its Practical Applications

Calculation of Motion of a body in a vertical Circle:   Let us consider, A body that has mass $m$ moving in a vertical circle of radius $r$. If at any instant the body is at position $P$ with angular displacement $\theta$ from the lower position $L$ of the circle. As shown in the figure below. The various forces acting on the body are: The weight $mg$ of the body, acting vertically downwards Tension $T$in the string acting along the $PO$. Now $mg$ can be resolved into two components: 1.) The horizontal component $mg cos\theta$ opposite to $T$ 2.) The vertical component $mg sin\theta$ act along tangent to the circle at $P$ So the net force on the body at position $P$ provides the necessary centripetal force required by the body $T-mg cos\theta = \frac{m v^{2}}{r}$ $T= \frac{m v^{2}}{r} + mg \: cos\theta \qquad(1)$ Tension and velocity at the highest position $H$ of the verticle circle: At the highest position $H$ the tension $T_{H}

Definition of work done and its essential condition

Work: When a force is applied on a body and a body is displaced along the line of action of force then this is called the work done by force. Work is a scalar quantity. Unit: $Joule$, $N-m$ Dimension: $[ML^{2}T^{-2}]$ Example: When a boy kicks a football to move it, the boy is said to have done some work. Let us consider, A body that has mass $m$. If a force $F$ applied on a body at an angle $\theta$ and body is displaced from position $A$ to position $B$ with distance $d$ then work done by force: $W= horizontal \: component \: of \: the \: force \times displacement $ $W=F\:cos\theta \times d$ $W=F d\:cos\theta \qquad(1)$ $W=\overrightarrow{F}. \overrightarrow{d}$ So, we can conclude from the above equation that "The work done by the force is the scalar product of the force and displacement ". Note: When the force is applied at an angle $\theta$ then the vertical component of the force is balanced by weight $mg$ and the

Principle and Proof of Law of Conservation of Energy

Law of Conservation of Energy: According to this principle The energy is neither created nor destroyed. The energy can be changed from one form to another form. i.e. when there is not any external influence is applied on the particle then the total energy of the particle is always conserved. Proof of Conservation's Law of Energy: Let us consider, A particle is freely falling from height $h$ under the gravitational acceleration $g$. So the total energy of the particle at different points : Calculation of Total Energy at Point $A$: The initial velocity of the particle at point $A$ is $(v_{A})=0$ The kinetic energy of the particle at point $A$ is $(K_{A})=\frac{1}{2}mv_{A}^{2}$ $K_{A}=0\qquad \left( \because v_{A}=0 \right)$ The potential energy of the particle at point $A$ is $(U_{A})=mgh$ The total energy of a particle at point $A$ is $(E_{A})=K_{A}+U_{A}$ Now substitute the value of $K_{A}$ and $U_{A}$ in the above equation. Now the above e

Definition and Practical Applications of Centripetal Force

Definition of Centripetal Force: When a particle moves in a circular path then a force act, toward the centre of the circle, on a particle. This type of force is called the centripetal force. This force is also known as a radial force because the direction of force is toward the centre of the circle. Let us consider, A particle of mass $m$ moving around a circular path of radius $r$ with linear velocity $v$. So the force on a particle: $F=ma$ Where $a$ is centripetal acceleration i.e. $a=\frac{v^{2}}{r}$. So the force on a particle can be written as $F=m\frac{v^{2}}{r}$ $F=\frac{mv^{2}}{r}$ $F=m\frac{(r \omega)^{2}}{r} \qquad \left( \because v=r \omega \right)$ $F=mr\omega^{2}$ $F=mr\left(\frac{2\pi}{T}\right)^{2} \qquad \left( \because \omega=\frac{2\pi}{T} \right)$ $F=mr\frac{4\pi^{2}}{T^{2}}$ $F= \frac{4 \pi^{2} m r}{T^{2}} $ $F=4 \pi^{2} m r n^{2} \qquad \left( \because n=\frac{1}{T} \right)$ Practical Applications of

Definition and Derivation of Centripetal Acceleration

Definition: When a particle moves in a circular path then acceleration act on the particle which has a direction toward the center of the circle. This acceleration is called centripetal acceleration. Derivation of Centripetal Acceleration: Let us consider, A particle that has mass $m$ moving with velocity $v$ in a circular path of radius $r$. If a particle is moving from point $P_{1}$ to point $P_{2}$ by covering distance $\Delta s$ on the circumference of the circle by making an angular displacement of $\theta$ at the center $O$ of the circle. The direction of velocity of the particle at point $P_{1}$ and $P_{2}$ is $\overrightarrow{v_{1}}$ and $v_{2}$. Now take the change in velocity from point $P_{1}$ to $P_{2}$ by vector subtraction method as shown in figure below: To find the expression for the centripetal acceleration, Now take two similar triangles $\Delta OP_{1}P_{2}$ and $\Delta ABC$ from the figure: $\frac{OP_{1}}{AB}=\frac{P_{1}P_{2}}{BC}$ Now su

Relation between angular acceleration and linear acceleration

Derivation of equation of the relation between linear acceleration and angular acceleration: Deduce the equation from the General form Deduce the equation from the Differential form 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 $\underset{\Delta t \rightarrow 0}{Lim} \: \frac{\Delta v}{\Delta t}$ = Instantaneous Acceleration $(a)$ $\alpha=\frac{a}{r}$ 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}

Relation between angular velocity and linear velocity

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 $\omega=\frac{1}{r}v$ $v=r\omega$ This is the relation between linear velocity and angular velocity.

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

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,

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} co

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.

Refraction of light and its Properties

Definition of Refraction of light→ When a light goes from one medium to another medium then light bends from the path. This bending of the light phenomenon is knowns as the refraction of light . When light goes from a rarer medium to a denser medium, light bends toward the normal as shown in the figure below: Propagation of light from rarer medium to denser medium When light goes from a denser medium to a rarer medium, the light goes away from normal as shown in the figure below. Propagation of light from denser medium to rarer medium Properties of refraction of light→ The incident ray, normal ray, and refracted ray lie on the same point. According to Snell's law - The ratio of the sine of the incident angle to the sine of the refractive angle is always constant. This constant value is known as the refractive index of the medium. $\frac{sin \: i}{sin \: r}=constant(_{1}n_{2}) $ $\frac{sin \: i}{sin \: r}= (_{1}n_{2})$

Alternating Current Circuit containing Resistance only (R-Circuit)

Alternating Current Circuit containing Resistance (R-Circuit): Let us consider, An alternating current circuit containing resistance $R$ only. This resistance $R$ is connected with an alternating EMF i.e electromotive force source i.e. $E=E_{\circ}sin\omega t\qquad(1)$ The potential difference across the circuit $E=iR$ Then from equation $(1)$ $iR=E_{\circ}sin\omega t $ $i=\frac{E_{\circ}}{R}sin\omega t $ $i=i_{\circ}sin\omega t \qquad(2) $ Where $i_{\circ}$ is the peak value or amplitude of the current in the circuit which has value $i_{\circ}=\frac{E_{\circ}}{R}$. Now compare the equation $(1)$ and equation $(2)$ which shows that if a circuit is containing a resistor only then the current is always in phase with the applied EMF i.e electromotive force. The phase diagram between EMF and the current of resistance is shown below- The phasor diagram between the EMF and current of resistance is also shown in the given figure below-

Derivation of Addition of Velocity in Special Relativity

Addition of Velocities: Let us consider two frames $S$ and $S'$, frame $S'$ is moving with constant velocity $v$ relative to frame $S$ along the positive direction of the X-axis. Let us express the velocity of the body in these frames. Suppose that body moves a distance $dx$ in time $dt$ in frame $S$ and through a distance $dx'$ in the time $dt'$ in the system $S'$ from point $P$ to $Q$. Then Addition of velocity $\frac{dx}{dt} = u\qquad \frac{dx'}{dt'}= u'\qquad (1)$ From Lorentz's inverse transformation $x=\alpha\left ( x'+vt' \right )\qquad(2)$ $t=\alpha '\left ( t'+\frac{v\cdot x'}{c^{2}} \right )\qquad(3)$ Differentiate the equation $(2)$ and equation $(3)$ with respect to $t'$ $\frac{dx}{dt'}= \alpha \left ( \frac{dx'}{dt'}+v \right )\qquad(4)$ $\frac{dt}{dt'}= \alpha ' \left ({1}+\frac{v}{c^{2}} \cdot \frac{

Work done by a rotating electric dipole in uniform electric field

Derivation of Work done by a rotating electric dipole in uniform electric field : Let us consider, An electric dipole AB, made up of two charges $+q$ and $-q$, is placed at a very small distance $2l$ in a uniform electric field $\overrightarrow{E}$. If dipole $AB$ rotates at angle $θ$ from its equilibrium position. If $A'$ and $B'$ are the new position of a dipole in the electric field. Then force on $+q$ charge particle due to electric field→ $ \overrightarrow{F_{+q}}=q\overrightarrow{E}\qquad (1)$ Then force on $-q$ charge particle due to electric field→ $ \overrightarrow{F_{-q}}=q\overrightarrow{E}\qquad(2)$ Work done by rotating an electric dipole   So work done by a force on $+q$ charge particle to bring from position $A$ to position $A'$→ $ \overrightarrow{W_{+q}}=\overrightarrow{F_{+q}}·\overrightarrow{AC}$ $ \overrightarrow{W_{+q}}=q\overrightarrow{E}· \overrightarrow{AC}\qquad(3)$ Similarly, work done by the forc

Electric field Intensity due a uniformly charged Spherical Shell

Electric field intensity at a different point in the field due to the uniformly charged spherical shell: Let us consider, a spherical shell of radius $R$ in which $+q$ charge is distributed uniformly on the surface of the sphere. Now find the electric field intensity at different points due to the spherical shell. These different points are: Electric field intensity at an external point of the spherical shell Electric field intensity on the surface of the spherical shell Electric field intensity at an internal point of the spherical shell 1. Electric field intensity at an external point of the spherical shell: If point $O$ is the center of the spherical shell, The electric field at the outside of the spherical shell can be determined by the following steps → First, take the point $P$ outside the sphere Draw a spherical surface of radius r which p