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 dW=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 $(\omega)$ 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.