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- Applied Mechanics - AM
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- Introduction to dynamics - ( 1 - 9 )
- Curvilinear motion of particles - ( 10 - 23 )
- Kinetics of particles newtons second law - ( 24 - 38 )
- Kinetics of particle: energy and momentum method - ( 39 - 51 )
- System of particles - ( 52 - 62 )
- Kinematics of rigid bodies - ( 63 - 71 )
- Plane motion of rigid bodies - ( 72 - 86 )
- Mechanical vibration - ( 87 - 92 )

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CHAPTER- 1: INTRODUCTION TO DYNAMICS Mechanics as the origin of Dynamics: Mechanics is defined as that science which describe and predicts the conditions of rest or motion of bodies under the action of forces. It is the foundation of most engineering sciences. It can be divided and subdivided as below: (i) Newtonian Mechanics (Engineering Mechancis) (ii) Relativistic Mechanics (It deals with the conditions involving seed of bodies close to the speed of light ) (iii) Quantum Mechanics (It deals with the conditions involving extremely small mass and size ie atomic distance) (a) Mechancis of rigid bodies (b) Mechanics of deformable bodies (c) Mechanics of fluids Mechanics of Compressible fluids Dynamics Statics Kinematics Mechanics of Incompressible fluids Kinetics Dynamics: It is which of Newtonian Mechanics which deals with the forces and their effects, while acting upon the bodies in motion. When we talk about the motion of the planets in our solar system, motion of a space craft, the acceleration of an automobile, the motion of a charged particle in an electric field, swinging of a pendulum, we are talking about Dynamics. Kinematics: It is that branch of Dynamics which deals with the displacement of a particles or rigid body over time with out reference to the forces that cause or change the motion. It is concerned with the position, velocity and acceleration of moving bodies as functions of time. Kinetics: It is that branch of Dynamics which deals with the motion of a particle or rigid body, with the reference to the forces and other factor that cause or influence the motion. For the study of motion Newton’s Second Law is widely used. Downloaded from www.jayaram.com.np/ -1

Downloaded from www.jayaram.com.np Chapter:- 2 Determination of motion of particles: • In general motion of particles (position, velocity and acceleration ) is expressed in terms of function as, X = f(x) , [ x = 6t2 +t3 ] * But in practice the relation of motion may be defined by any other equation with function of x, v,& t . a = f(t) a = f(x) a = f(v) etc. so these given relation are integrated to get the general relation of motion x = f(t) . Case-I: When acceleration is given as function of time [i.e a = f(t) ] [ a = 6t2+t3] We know, a = dv/dt → dv = adt or, dv = f(t) dt Now integrating both sides taking limit as time varies from 0 to t and velocity varies form vo to v. ∫ v t vo dv = ∫ f (t )dt 0 t v − v o = ∫ f (t )dt o t v = v o + ∫ f (t )dt .......(i ) o Again, velocity is given by, V = dx/dt → dx = vdt Again integration both sides of equation similarly form time 0 to t and position xo to x. We get, ∫ x xo t dx = ∫ vdt 0 ⎡v + t f (t )dt ⎤ dt Putting value of V form equation (i) ∫o ⎢⎣ 0 ∫0 ⎥⎦ t t x = xo + ∫ ⎡vo + ∫ f (t )dt ⎤ dt .........(ii ) ⎥⎦ 0⎢ 0 ⎣ Thus position is obtained from equation of a = f(t) x – xo = t # Find the velocity and position of a particles after its 5 sec from Rest, which moves with equation of a = 6t2-4t. Solution: Given equation a = f(t) → a = 6t2 – 4t xo = 0 , vo = 0 and t = 5. We know, V0 = [ ∫ t o f (t )dt = ∫ t o v = 2t 3 − 2t 2 Again, ] 5 0 ⎡ 6t 3 4t 2 ⎤ − f (t )dt = ∫ (6t − 4t )dt = ⎢ ⎥ 0 2 ⎦ ⎣ 3 5 2 5 0 = 200 m / s -By Er. Biraj Singh Thapa (Lecturer, Eastern College of Engineering, Biratnagar)/ -2

t ∫ ⎡⎢⎣v + ∫ X = xo + o 0 0 =0+ t f (t )dt ⎤ dt ⎥⎦ ∫ [0 + 200]dt = ∫ t 5 0 0 200dt = [200t ]0 5 = 100 m Therefore, x = 100m and v = 200m/s after 5 second of motion. Case-II When the acceleration is a given function of position [ i.e a = f(x) eg. x2+4x] We know, a = dv/dt = dv/dx . dx/dt = v.dv/dx or, vdv = adx or, vdv = f(x)dx [ ∵ a = f(x)] Now , Integrating both sides of above equation , taking limit as velocity varies from Vo to v as position p varies form xo to x. t vdv = ∫ x ⎡v2 ⎤ f ( x)dx ⇒ ⎢ ⎥ ⎣2⎦ i.e ∫ or, x v 2 v 02 − = ∫ f ( x)dx x 0 2 2 vo xo v x v0 = ∫ f ( x)dx xo 1 x 2 ∵ v = ⎡v02 + 2∫ f ( x)dx ⎤ .......(1) ⎢⎣ ⎥⎦ x0 Again We know, V = dx/dt ⇒ dx = vdt. Integrating both sides with limits as time varies from 0 to t and position from xo to x . i.e ∫ x xo t dx = ∫ vdt ........(1) o Putting value of v varies from equation (1) we get, x – xo = ∫ 1 2 ⎡v + 2 f ( x)dx ⎤ dt ∫xo ⎥⎦ ⎢⎣ 2 0 x 1 x 2 ∴ x = x0 + ∫ ⎡vo2 + 2∫ f ( x)dx ⎤ dt ⎥⎦ x0 0⎢ ⎣ t Case III : When acceleration is a given function of velocity (i.e a =f(v) eg. a = v2+v) We know, a = v dv/dx ⇒ f(v) = v dv/dx Or , dx = v dv/f(v) Integrating both sides taking limit as velocity varies form vo to v and position varies from xo to x x v v dv dv dx v x x = ⇒ − = 0 ∫x0 ∫v0 f (v) ∫v0 v f (v) dv v0 f (v ) e.g The acceleration of a particle is defined as a = -0.0125v2, the particle is given as initial velocity v0, find the distance traveled before its velocity drops to half. Solution: Given, a = -0.0125v2 i.e a = f(v) , Initial velocity vo, final velocity vo/2 For motion a = f(v) xo = 0, x = ? v ∴ x = x0 + ∫ v Downloaded from www.jayaram.com.np/ -3

Downloaded from www.jayaram.com.np v x = xo + ∫ v v0 dv f (v ) v0 2 v0 v 0 v −1 1 2 x= ∫ = dv dv 2 ∫ v 0.0125 0 v − 0.0125v v0 ⎡ ⎤ 1 [ln v]v20 = − 1 ln ⎢ v0 ⎥ =− 0.0125 0.0125 ⎣ 2v0 ⎦ Or, x = 24.08 m ans. 2.2 Uniform Rectilinear motion: * Uniform motion means covering equal distance over equal intervals of time. ie velocity = constant. We have, V = dx/dt = v [ v = constant velocity of body] ∫ ∴ dx = udt ⇒ x x0 t dx = ∫ vdt 0 [ Integrating both sides under limits as position varies from xo to x and time 0 to t] ∴ x – xo = vt x = xo +vt ∴ Change in position (or displacement) is equal to uniform velocity x change in time [ i.e s = vt] 2.3 Uniform Accelerated Rectilinear motion: If constant acceleration be ‘a’ then, dv/dt = a = constant ⇒ dv = adt Integrating both sides with limit v0 to v and 0 to t . We get, ∫ v v0 t dv = a ∫ dt ⇒ v − v0 = at 0 v = v 0 + at ........(1) Again for position , we have v = dx/dt …..(2) from 1 and 2. dx = (vo +at) dt , Integrating both sides over the limits ∫ x x0 dx = ∫ (v 0 + dt )at ⇒ x − x0 = v0 t + t 0 x = x0 + v0 t + Also, a = v 1 2 at 2 1 2 at 2 dv dx Or, vdv = adx Integrating both sides under limits ∫ v v0 x vdv = a ∫ dx x0 1 2 (v − v02 ) = a(x − x0 ) 2 v 2 = v 02 + 2a( x − x 0 ) xo = Initial position x = Final position v0 = Initial velocity. v = Final velocity 0 = Initial time t = Final time 2.4 Motion of several particles: Two or more particles moving in straight line. Equations of motion may be written for each particles as: -By Er. Biraj Singh Thapa (Lecturer, Eastern College of Engineering, Biratnagar)/ -4

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