Kinetics of particle
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Kinetics of particle

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    Kinetics of particle Kinetics of particle Presentation Transcript

    • Presentation Made by : • Danyal Haider (01-133132-014) • Kamran Shah (01-133132-027)
    • • PRESENTATION TOPICS • Relation between force , mass and acceleration . • Applications of Newton second law of motion. • Equations of motions . • Introduction to kinetics of particles .
    • Relation between force , mass and acceleration : Relation between force , mass and acceleration is refered to Newton second law of motion . FORCE :• Force is the action of one body on another body . A force tends to move a body in the direction of its action . The action of the force is charaterized by its magnitude , by the direction of its action , and its point of application . OR
    • Force is a vector quantity having both magnitude and direction . The magnitude of the force can be measured using a spring scale. Unit of force is Newton .
    • MASS : Mass is a measure of Inertia of a body . Mass is a scalar quantity . Unit of mass is Kilogram (kg) . ACCELERATION : • The rate of change of velocity is called acceleration . • Acceleration is a vector quantity so it can be changed in two ways: 1. changing its magnitude 2. changing its direction • Acceleration can be positive or negative. If an object’s speed increases with time, it will have positive acceleration. If an object’s speed decreases with time, it will have negative acceleration that’s also known as deceleration. Si unit of acceleration is meters /second squared .
    • “NEWTON” Sir Isaac Newton (4 January 1643 – 31 March 1727) was an English physicist and mathematician. He is famous for his work on the laws ofmotion, optics, gravity, and calculus. In 1687, Newton published a book called the Principia Mathematica in which he presents his theory of universal gravitation and three laws of motion.
    • Newton’s second law: • A particle will have an acceleration proportional to the magnitude of the resultant force acting on it and in the direction of the resultant force . • The resultant of the forces acting on a particle is equal to the rate of change of linear momentum of the particle . • The sum of the moments about “O” of the forces acting on a particle is equal to the rate of change of angular momentum of the particle . • When a particle of mass m is acted upon by a force the acceleration of the particle must satisfy, F=ma
    • • If force acting on particle is zero, particle will not accelerate, i.e., it will remain stationary or continue on a straight line at constant velocity.
    • Linear Momentum Conservation Principle: • If the resultant force on a particle is zero, the linear momentum of the particle remains constant in both magnitude and direction . • Replacing the acceleration by the derivative of the velocity yields ΣF = m(dv/dt) = d/dt (mv) =dL/dt Where , L = linear momentumof the particle
    • SOLUTION: Q#1 • • • • • • A 200-lb block rests on a horizontal plane. Find the magnitude of the force P required to give the block an acceleration or 10 ft/s2 to the right. The coefficient of kinetic friction between the block and plane is u^k = 0.25 • Resolve the equation of motion for the block into two rectangular component equations. • Unknowns consist of the applied force P and the normal reaction N from the plane. The two equations may be solved for these unknowns.
    • SOLUTION: • Resolve the equation of motion for the block into two rectangular component equations. ΣFx = ma : Pcos30 - 0.25N = (6.21lb.s2/ft)(10ft/s2)= 6.21lb ΣFy = 0 : N − Psin 30° − 200lb = 0 • Unknowns consist of the applied force P and the normal reaction N from the plane. The two equations may be solved for these unknowns. N = P sin 30° + 200 lb P cos 30° - 0.25(P Sin 30° + 200 lb) = 62.1 lb P = 151 lb
    • • PRESENTATION TOPICS • Relation between force , mass and acceleration . • Applications of Newton second law of motion. • Equations of motions . • Introduction to kinetics of particles .
    • Applications of Newton second law of motion : • When a force is applied on a body of mass(m), there is a change in velocity w.r.t time , and change in velocity means acceleration is produced in a body .
    • We know that , Acceleration(a) is directly proportional to Force(F) • In the above figure, if the person applies lesser force on the vehicle . The less acceleration is produced in a body . • On the other hand , if the person applies greater force on the vehicle . The more acceleration is produced in a body .
    • • PRESENTATION TOPICS • Relation between force , mass and acceleration . • Applications of Newton second law of motion. • Equations of motions . • Introduction to kinetics of particles .
    • Equations of motion PROOF OF EQUATION OF MOTION. 1st Equation of motion. Vf = vi + at Proof. Consider a body moving with initial velocity vi .if a force F act on the body and produces an acceleration a in it, then after time t seconds. Its final velocity becomes vf. Then change in velocity of the body is vf - vi, and the rate of change of velocity of the body is equal to the acceleration acting on the body,i.e, a = vf – vi/t vf – vi = at vf = vi + at
    • 2nd Equation of Motion. S = vit + at2/2 Proof. According to the first equation of motion, we have Vf = vi + at eq no (1) If the body has constant acceleration, then its average velocity is given by Vav = vi + vf/2 eq no (2) Distance covered by the body in time t is giver by S = vav * t eq no (3) Putting the mauve of Vav from eq (2) in eq (3) we get S = t * (vi + vf)/2 Putting the value of vf from eq,(1) in eq,(4), we get S = t (vi + vi + a t)/2 = 2vit + at2/2 S = vi t + at2/2
    • Third equation of motion. 2as = vf2 – vi2 Proof. We know that S = vav . t eq(1) Vav = vi + vf/2 eq(2) From 1st equation of motion Vf = vi + at t= vf – vi/a this is eq (3) Putting the value of vav and t from eq,(2) and (3), in eq,(1), S = (vi + vf)/2* (vf – vi)/a 2as = (vf +vf). (vf – vi) 2as = vf2 – vi2
    • Some Rules to be used in the above three equation. • If the body starts its motion from rest then use vi=0 • If a moving body stops its motion then vf=0 • If the velocity of the body is incressing then its acceleration is positive and its directed is the same as that of velocity. • If the velocity of the body is decreasing then its acceleration is negative and its directed is opposite to that of the velocity .
    • Some Numerical Problems. Q # 2: What force is required to move a mass of 5kg through 150 meters in 10 sec. if the mass is originally at rest. Given data m=5kg s=150m t=10sec For this we have to find the acceleration acting on the body and according to the gives data we apply second equation of motion. F=? S=vit+1/2at2 150=0*10*a*100/2 a=3m/sec2 F= ma F=5*3=15N
    • Equation of motion for a freely falling body. If a body is moving freely upward of falling downward then its acceleration a will be the acceleration due to gravity. i.e g=9.8m/sec2 and its distance S will show the height h of the body. Then the equation of motion will take the form. Downward motion Vf=vi+at S=vit+1/2at2 2as=vf2-vi2 vf=vi+gt h=vit+1/2gt2 2gh=vf2-vi2 Upward motion Vf=vi+at vf=vi-gt S=vit+at2/2 s=vit-gt2/2 2as=vf2-vi2 -2gt=vf2-vi2
    • Q # 3:. A 50kg shell of conical shape is dropped from a height of 1000m. It falls freely under gravity and sinks in into the ground . Find (a) Its velocity of the moment it touches the ground. (b). the time it takes to reach the groun. (c). the deceleration of the shell as it penetrates into the ground. Given data is m=50kg vi=0 s=1m using 2nd equation of motion find the vf. 2gh=vf2-vi2 2*(9.8)*1000 = vf2-0 Vf = 140ms-1
    • Now find the t? Using equation of motion. From 1st equation on motion. Vf = vi+gt 140 = 0+9.8*t t = 140/9.8= 14.3sec t = 14.3sec (c) for penetrating into ground. For ground the velocity of 140ms-1 is intial velocity. In ground it comes to rest after covering a distance of 1m. Given data is Vi’ = 140ms-1 Vf ’ = 0 S = 1m a= ? Using equation motion From 3rd equation motion 2as = vf2-vi2 2*a*1 = 0(140)2 a= -19600/2 a = -9800ms-1
    • • PRESENTATION TOPICS • • • • Relation between force , mass and acceleration . Applications of Newton second law of motion. Equations of motions . Introduction to kinetics of particles .
    • Introduction to kinetics of particles Difference between kinematics and kinetics : Kinematics:- kinematics is the study of motion (not focusing upon forces , just the motion) . Kinetics :- kinetics is the study of motion in relation to forces . Introduction to kinetics of particles According to Newton second law, a particle will accelerate when it is subjected to unbalanced forces . kinetics is the study of the relations between unbalanced forces and resulting changes in motion . Methods of solution of problems of particle kinetics:There are three basic methods of solution of problems of particle kinetics . These three methods are summarized as follows :
    • (1)DIRECT APPLICATION OF NEWTON SECOND LAW:First we applied the Newton second law F=ma to determine the instantaneous relation between force and acceleration they produce . Then solve a problem using x-y coordinates for plane motion problems and x-y-z coordinates for space motion problems . (2)WORK-ENERGY EQUATIONS:Next, we apply F=ma with respect to displacement and derived the equations for work and energy . These equations enable us to relate the initial and final velocities to the work done during an interval of forces . (3)IMPULSE-MOMENTUM EQUATIONS:- Then we write Newton second law in terms of force equals time rate of change of linear momentum and moment equals time rate of change of angular momentum . Then solve these relations with respect to time and derived the impulse and momentum equations .
    • Q#4: Determine the rated speed of a highway curve of radius ρ = 400 ft banked through an angle θ = 18o. The rated speed of a banked highway curve is the speed at which a car should travel if no lateral friction force is to be exerted at its wheels. SOLUTION:  The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path . The forces acting on the car are its weight and a normal reaction from the road surface.  Resolve the equation of motion for the car into vertical and normal components.  Solve for the vehicle speed.
    • • Resolve the equation of motion for the car into vertical and normal components. ΣFy = 0 : R Cos θ – W = 0 R = W/Cos θ ΣFn = man : R sin θ = (W/g) *an (W/Cos θ)*Sin θ =(W/g)* (v ^2/ ρ ) SOLUTION: • The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path.The forces acting on the car are its weight and a normal reaction from the road surface. • Solve for the vehicle speed. v^2 = g ρ Tan θ =(32.2ft/s^2)(400 ft)Tan 18° v = 64.7 ft/s = 44.1mi/h
    • Thank you Any Question please !!!!!