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UNIT-IV Kinetics of Particles: Newton’s Second Law Contents Introduction Newton’s Second Law of Motion Linear Momentum of a Particle Systems of Units Equations of Motion Dynamic Equilibrium Sample Problem 12.1 Sample Problem 12.3 Sample Problem 12.4 Sample Problem 12.5 Sample Problem 12.6 Angular Momentum of a Particle Equations of Motion in Radial & Transverse Components Conservation of Angular Momentum Newton’s Law of Gravitation Sample Problem 12.7 Sample Problem 12.8 Trajectory of a Particle Under a Central Force Application to Space Mechanics Sample Problem 12.9 Kepler’s Laws of Planetary Motion 12 - 2 Kinetics of Particles We must analyze all of the forces acting on the wheelchair in order to design a good ramp High swing velocities can result in large forces on a swing chain or rope, causing it to break. 2-3 Introduction F ma • Newton’s Second Law of Motion • If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of resultant and in the direction of the resultant. • Must be expressed with respect to a Newtonian (or inertial) frame of reference, i.e., one that is not accelerating or rotating. • This form of the equation is for a constant mass system 12 - 4 Linear Momentum of a Particle • Replacing the acceleration by the derivative of the velocity yields dv F m dt d dL m v dt dt L linear momentum of the particle • 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. 12 - 5 Systems of Units • Of the units for the four primary dimensions (force, mass, length, and time), three may be chosen arbitrarily. The fourth must be compatible with Newton’s 2nd Law. • International System of Units (SI Units): base units are the units of length (m), mass (kg), and time (second). The unit of force is derived, kg m m 1 N 1 kg 1 2 1 2 s s • U.S. Customary Units: base units are the units of force (lb), length (m), and time (second). The unit of mass is derived, 1lb 1lb lb s 2 1lbm 1slug 1 2 2 ft 32.2 ft s 1ft s 12 - 6 Equations of Motion • Newton’s second law F ma • Can use scalar component equations, e.g., for rectangular components, Fx i Fy j Fz k ma x i a y j a z k Fx max Fy ma y Fz maz Fx mx Fy my Fz mz 12 - 7 Dynamic Equilibrium • Alternate expression of Newton’s second law, F m a 0 ma inertial vector • With the inclusion of the inertial vector, the system of forces acting on the particle is equivalent to zero. The particle is in dynamic equilibrium. • Methods developed for particles in static equilibrium may be applied, e.g., coplanar forces may be represented with a closed vector polygon. • Inertia vectors are often called inertial forces as they measure the resistance that particles offer to changes in motion, i.e., changes in speed or direction. • Inertial forces may be conceptually useful but are not like the contact and gravitational forces found in statics. 12 - 8 Free Body Diagrams and Kinetic The free body diagram isDiagrams the same as you have done in statics; we will add the kinetic diagram in our dynamic analysis. 1. Isolate the body of interest (free body) 2. Draw your axis system (e.g., Cartesian, polar, path) 3. Add in applied forces (e.g., weight, 225 lb pulling force) 4. Replace supports with forces (e.g., normal force) 5. Draw appropriate dimensions (usually angles for particles) y x 225 N 25o N Ff mg 12 - 9 Free Body Diagrams and Kinetic Diagrams Put the inertial terms for the body of interest on the kinetic diagram. 1. Isolate the body of interest (free body) 2. Draw in the mass times acceleration of the particle; if unknown, do this in the positive direction according to your chosen axes y x 225 N may 25o N max Ff mg F ma 12 - 10 Free Body Diagrams and Kinetic Draw the FBD andDiagrams KD for block A (note that the massless, frictionless pulleys are attached to block A and should be included in the system). 2 - 11 Free Body Diagrams and Kinetic 1. Isolate body Diagrams 2. Axes 3. 4. 5. 6. T T T Applied forces Replace supports with forces Dimensions (already drawn) Kinetic diagram y NB T x Ff-B T mg N1 Ff-1 = may = 0 max 2 - 12 Free Body Diagrams and Kinetic Diagrams Draw the FBD and KD for the collar B. Assume there is friction acting between the rod and collar, motion is in the vertical plane, and q is increasing 2 - 13 Free Body Diagrams and Kinetic 1. Isolate body Diagrams 2. Axes 3. 4. 5. 6. Applied forces Replace supports with forces Dimensions Kinetic diagram eq er maq mar q Ff = q mg N 2 - 14 Sample Problem 12.1 SOLUTION: • 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. A 200-lb block rests on a horizontal plane. Find the magnitude of the force P required to give the block an acceleration of 10 ft/s2 to the right. The coefficient of kinetic friction between the block and plane is mk 0.25. 12 - 15 Sample Problem 12.1 SOLUTION: • Resolve the equation of motion for the block into two rectangular component equations. Fx ma : y O P cos 30 0.25N 6.21lb s 2 ft 10 ft s 2 62.1lb x W 200 lb m g 32.2 ft s 2 lb s 2 6.21 ft F mk N 0.25N Fy 0 : N P sin 30 200 lb 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 0.25P sin 30 200 lb 62.1lb P 151lb 12 - 16 Sample Problem 12.3 SOLUTION: • Write the kinematic relationships for the dependent motions and accelerations of the blocks. • Write the equations of motion for the blocks and pulley. • Combine the kinematic relationships with the equations of motion to solve for the accelerations and cord tension. The two blocks shown start from rest. The horizontal plane and the pulley are frictionless, and the pulley is assumed to be of negligible mass. Determine the acceleration of each block and the tension in the cord. 12 - 17 Sample Problem 12.3 SOLUTION: O x y • Write the kinematic relationships for the dependent motions and accelerations of the blocks. y B 12 x A a B 12 a A • Write equations of motion for blocks and pulley. Fx m Aa A : T1 100 kg a A Fy mB aB : mB g T2 mB a B 300 kg 9.81m s 2 T2 300 kg a B T2 2940N - 300 kg a B Fy mC aC 0 : T2 2T1 0 12 - 18 Sample Problem 12.3 • Combine kinematic relationships with equations of motion to solve for accelerations and cord tension. O x y y B 12 x A a B 12 a A T1 100 kg a A T2 2940N - 300 kg a B 2940N - 300 kg 12 a A T2 2T1 0 2940 N 150 kg a A 2100 kg a A 0 a A 8.40 m s 2 a B 12 a A 4.20 m s 2 T1 100 kg a A 840 N T2 2T1 1680 N 12 - 19 Sample Problem 12.4 SOLUTION: • The block is constrained to slide down the wedge. Therefore, their motions are dependent. Express the acceleration of block as the acceleration of wedge plus the acceleration of the block relative to the wedge. The 12-lb block B starts from rest and slides on the 30-lb wedge A, which is supported by a horizontal surface. • Write the equations of motion for the wedge and block. • Solve for the accelerations. Neglecting friction, determine (a) the acceleration of the wedge, and (b) the acceleration of the block relative to the wedge. 12 - 20 SampleSOLUTION: Problem 12.4 • The block is constrained to slide down the wedge. Therefore, their motions are dependent. aB a A aB A • Write equations of motion for wedge and block. Fx m Aa A : N1 sin 30 m A a A y 0.5 N1 W A g a A x Fx mB a x mB a A cos 30 aB A : WB sin 30 WB g a A cos 30 a B a B A a A cos 30 g sin 30 Fy mB a y mB a A sin 30 : N1 WB cos 30 WB g a A sin 30 12 - 21 A Sample• Solve Problem 12.4 for the accelerations. 0.5 N1 W A g a A N1 WB cos 30 WB g a A sin 30 2W A g a A WB cos 30 WB g a A sin 30 aA gWB cos 30 2W A WB sin 30 32.2 ft s 2 12 lb cos 30 aA 230 lb 12 lb sin 30 a A 5.07 ft s 2 a B A a A cos 30 g sin 30 a B A 5.07 ft s 2 cos 30 32.2 ft s 2 sin 30 a B A 20.5 ft s 2 12 - 22 Group Problem Solving SOLUTION: • Write the kinematic relationships for the dependent motions and accelerations of the blocks. • Draw the FBD and KD for each block • Write the equations of motion for the blocks and pulley. • Combine the kinematic relationships with the equations of motion to solve for the accelerations and cord tension. The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the horizontal surface, determine (a) the acceleration of each block, (b) the tension in the cable. 2 - 23 Group Problem Solving SOLUTION: • Write the kinematic relationships for the dependent motions and accelerations of the blocks. xA yB This is the same problem worked last chapter- write the constraint equation x A 3 yB constants L Differentiate this twice to get the acceleration relationship. v A 3vB 0 a A 3aB 0 a A 3aB (1) 2 - 24 Group Problem Solving • Draw the FBD and KD for each block 2T T B mAg +y A T = maBy mBg Fx m A a A : Fy mB aB 9.81 m/s 2 aB 0.83136 m/s2 mA 30 kg 1 9 1 9 25 kg mB g = +x NA • Write the equation of motion for each block WB 3T mB aB (2) • Solve the three equations, 3 unknowns (3) (2) mB g 3(3mA aB ) mB aB maAx T m A aB From Eq (1) T 3m A aB (3) T 3 30 kg 0.83136 m/s2 T 74.8 N a A 2.49 2.49 m/s2 2 - 25 Concept Quiz (1) (2) (3) The three systems are released from rest. Rank the accelerations, from highest to lowest. a) (1) > (2) > (3) b) (1) = (2) > (3) c) (2) > (1) > (3) d) (1) = (2) = (3) e) (1) = (2) < (3) 2 - 26 Kinetics: Normal and Tangential Coordinates Aircraft and roller coasters can both experience large normal forces during turns. 2 - 27 Equations of Motion • Newton’s second law F ma • For tangential and normal components, F n man F ma F t t t m dv dt F n m v2 12 - 28 Sample Problem 12.5 SOLUTION: • Resolve the equation of motion for the bob into tangential and normal components. • Solve the component equations for the normal and tangential accelerations. The bob of a 2-m pendulum describes an arc of a circle in a vertical plane. If the tension in the cord is 2.5 times the weight of the bob for the position shown, find the velocity and acceleration of the bob in that position. • Solve for the velocity in terms of the normal acceleration. 12 - 29 Sample Problem 12.5 SOLUTION: • Resolve the equation of motion for the bob into tangential and normal components. • Solve the component equations for the normal and tangential accelerations. mg sin 30 mat Ft mat : at g sin 30 Fn man : at 4.9 m s 2 2.5mg mg cos 30 man an g 2.5 cos 30 an 16.03 m s 2 • Solve for velocity in terms of normal acceleration. an v2 v an 2 m 16.03 m s 2 v 5.66 m s 12 - 30 Sample Problem 12.6 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. Determine the rated speed of a highway curve of radius = 400 ft banked through an angle q = 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. • Resolve the equation of motion for the car into vertical and normal components. • Solve for the vehicle speed. 12 - 31 Sample Problem 12.6 • Resolve the equation of motion for the car into vertical and normal components. R cosq W 0 Fy 0 : W R cosq Fn man : R sin q 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. W an g W W v2 sin q cosq g • Solve for the vehicle speed. v 2 g tan q 32.2 ft s 2 400 ft tan 18 v 64.7 ft s 44.1 mi h 12 - 32 Group Problem Solving v SOLUTION: • Draw the FBD and KD for the collar. • Write the equations of motion for the collar. • Determine kinematics of the collar. The 3-kg collar B rests on the frictionless arm AA. The collar is held in place by the rope attached to drum D and rotates about O in a horizontal plane. The linear velocity of the collar B is increasing according to v= 0.2 t2 where v is in m/s and t is in sec. Find the tension in the rope and the force of the bar on the collar after 5 seconds if r = 0.4 m. • Combine the equations of motion with kinematic relationships and solve. 2 - 33 Group Problem Solving SOLUTION: • Given: v= 0.2 t2, r = 0.4 m • Find: T and N at t = 5 sec Draw the FBD and KD of the collar mat et = en T man N Write the equations of motion Fn man N m v2 Ft mat dv T m dt 2 - 34 Group Problem Solving e t Kinematics : find vt, an, at vt 0.2t 2 0.2(52 ) =5 m/s v 2 mat en = 2 5 an 62.5 (m/s2 ) 0.4 T q N man dv at 0.4t 0.4(5) 2 m/s 2 dt Substitute into equations of motion Fn man Ft mat N 3.0(62.5) T 3.0(2) N 187.5 N T 6.0 N 2 - 35 Group Problem Solving e t How would the problem change if motion was in the vertical plane? mat en = q N T You would add an mg term and would also need to calculate q man mg When is the tangential force greater than the normal force? Only at the very beginning, when starting to accelerate. In most applications, an >> at 2 - 36 Concept Question B C A A car is driving from A to D on the curved path shown. The driver is doing the following at each point: A – going at a constant speed C – stepping on the brake D B – stepping on the accelerator D – stepping on the accelerator Draw the approximate direction of the car’s acceleration at each point. 2 - 37 Kinetics: Radial and Transverse Coordinates Hydraulic actuators and extending robotic arms are often analyzed using radial and transverse coordinates. 2 - 38 Eqs of Motion in Radial & Transverse Components • Consider particle at r and q, in polar coordinates, 2 F ma m r r q r r Fq maq mrq 2rq 12 - 39 Sample Problem 12.7 SOLUTION: • Write the radial and transverse equations of motion for the block. • Integrate the radial equation to find an expression for the radial velocity. A block B of mass m can slide freely on a frictionless arm OA which rotates in a horizontal plane at a constant rate q0 . • Substitute known information into the transverse equation to find an expression for the force on the block. Knowing that B is released at a distance r0 from O, express as a function of r a) the component vr of the velocity of B along OA, and b) the magnitude of the horizontal force exerted on B by the arm OA. 12 - 40 Sample Problem 12.7 • Integrate the radial equation to find an SOLUTION: • Write the radial and transverse equations of motion for the block. Fr m ar : 0 m r rq F q m aq : F mrq 2rq 2 expression for the radial velocity. dv dv dr dv vr r r vr r r dt dr dt dr dv dv dr dv vr r r vr r r dt dr dt dr vr dvr rq 2 dr rq02 dr vr r 2 vr dvr q0 r dr 0 r0 vr2 q 02 r 2 r02 • Substitute known information into the transverse equation to find an expression for the force on the block. F 2mq 02 r 2 2 12 r0 12 - 41 Group Problem Solving SOLUTION: • Draw the FBD and KD for the collar. • Write the equations of motion for the collar. • Determine kinematics of the collar. • Combine the equations of motion with kinematic relationships and solve. The 3-kg collar B slides on the frictionless arm AA. The arm is attached to drum D and rotates about O in a horizontal plane at the rate q 0.75t where q and t are expressed in rad/s and seconds, respectively. As the arm-drum assembly rotates, a mechanism within the drum releases the cord so that the collar moves outward from O with a constant speed of 0.5 m/s. Knowing that at t = 0, r = 0, determine the time at which the tension in the cord is equal to the magnitude of the horizontal force exerted on B by arm AA. 2 - 42 Group Problem Solving SOLUTION: • Given: q 0.75t r (0) 0 r 5 m/s • Find: time when T = N Draw the FBD and KD of the collar maq eq er T = mar N Write the equations of motion Fr mar T m(r rq 2 ) Fq mB aq N m(rq 2rq ) 2 - 43 Group Problem Solving r 5 m/s Kinematics : find expressions for r and q q (0.75t ) rad/s r 0 dr t 0.5 dt 0 r (0.5t ) m q 0.75 rad/s2 r 0 Substitute values into ar , aq ar r rq 2 0 [(0.5t ) m][(0.75t ) rad/s]2 (0.28125t 3 ) m/s 2 aq rq 2rq [(0.5t ) m][0.75 rad/s2 ] 2(0.5 m/s)[(0.75t ) rad/s] (1.125t ) m/s2 Substitute into equation of motion Set T = N Fr mar : T (3 kg)(0.28125t ) m/s 3 Fq mB aq : N (3 kg)(1.125t ) m/s 2 2 (0.84375t 3 ) (3.375t ) t 2 4.000 t 2.00 s 2 - 44 Concept Quiz e2 Top View e1 v w The girl starts walking towards the outside of the spinning platform, as shown in the figure. She is walking at a constant rate with respect to the platform, and the platform rotates at a constant rate. In which direction(s) will the forces act on her? a) +e1 b) - e1 c) +e2 d) - e2 e) The forces are zero in the e1 and e2 directions2 - 45 Angular Momentum of a Particle Satellite orbits are analyzed using conservation of angular momentum. 2 - 46 Eqs of Motion in Radial & Transverse Components • Consider particle at r and q, in polar coordinates, 2 F ma m r r q r r Fq maq mrq 2rq • This result may also be derived from conservation of angular momentum, H O mr 2q d mr 2q dt m r 2q 2rrq r Fq Fq mrq 2rq 12 - 47 Angular Momentum of a Particle • H r mV moment of momentum or the angular O momentum of the particle about O. • H O is perpendicular to plane containing r and mV H O rmV sin i j k rm vq HO x y z mv x mv y mvz mr 2q • Derivative of angular momentum with respect to time, H O r mV r mV V mV r ma rF MO • It follows from Newton’s second law that the sum of the moments about O of the forces acting on the particle is equal to the rate of change of the angular momentum of the particle about O. 12 - 48 Conservation of Angular • When only force acting on particle is directed Momentum toward or away from a fixed point O, the particle is said to be moving under a central force. • Since the line of action of the central force passes through O, M O H O 0 and r mV H O constant • Position vector and motion of particle are in a plane perpendicular to H O . • Magnitude of angular momentum, H O rmV sin constant r0 mV0 sin 0 or H O mr 2q constant HO angular momentum r 2q h m unit mass 12 - 49 Conservation of Angular Momentum • Radius vector OP sweeps infinitesimal area dA 12 r 2 dq • Define dA 1 2 dq 1 2 2r 2 r q areal velocity dt dt • Recall, for a body moving under a central force, h r 2q constant • When a particle moves under a central force, its areal velocity is constant. 12 - 50 Newton’s Law of Gravitation • Gravitational force exerted by the sun on a planet or by the earth on a satellite is an important example of gravitational force. • Newton’s law of universal gravitation - two particles of mass M and m attract each other with equal and opposite force directed along the line connecting the particles, Mm F G 2 r G constant of gravitatio n 66.73 10 12 m3 kg s 2 34.4 10 9 ft 4 lb s 4 • For particle of mass m on the earth’s surface, MG m ft W m 2 mg g 9.81 2 32.2 2 R s s 12 - 51 Sample Problem 12.8 SOLUTION: • Since the satellite is moving under a central force, its angular momentum is constant. Equate the angular momentum at A and B and solve for the velocity at B. A satellite is launched in a direction parallel to the surface of the earth with a velocity of 18820 mi/h from an altitude of 240 mi. Determine the velocity of the satellite as it reaches it maximum altitude of 2340 mi. The radius of the earth is 3960 mi. 12 - 52 Sample Problem 12.8 SOLUTION: • Since the satellite is moving under a central force, its angular momentum is constant. Equate the angular momentum at A and B and solve for the velocity at B. rm v sin H O constant rA m v A rB m v B vB v A rA rB 18820mi h 3960 240mi 3960 2340mi v B 12550 mi h 12 - 53 Trajectory of a Particle Under a • For particle movingCentral under central force directed towards force center, Force m r rq 2 Fr F mrq 2rq Fq 0 • Second expression is equivalent to r 2q h constant , from which, q h r2 h2 d 2 1 and r 2 2 r r dq • After substituting into the radial equation of motion and simplifying, d 2u F u dq 2 mh 2u 2 where u 1 r • If F is a known function of r or u, then particle trajectory may be found by integrating for u = f(q), with constants of integration determined from initial conditions. 12 - 54 Application to Space Mechanics • Consider earth satellites subjected to only gravitational pull of the earth, d 2u F u dq 2 mh 2u 2 where u 1 r F GMm r2 GMmu 2 d 2u GM u constant 2 2 dq h • Solution is equation of conic section, 1 GM u 2 1 cosq r h Ch2 eccentricity GM • Origin, located at earth’s center, is a focus of the conic section. • Trajectory may be ellipse, parabola, or hyperbola depending on value of eccentricity. 12 - 55 Application to Space Mechanics • Trajectory of earth satellite is defined by 1 GM 2 1 cosq r h Ch2 eccentricity GM • hyperbola, > 1 or C > GM/h2. The radius vector becomes infinite for 1 1 1 GM 1 cosq1 0 q1 cos cos C h2 • parabola, = 1 or C = GM/h2. The radius vector becomes infinite for 1 cosq 2 0 q 2 180 • ellipse, < 1 or C < GM/h2. The radius vector is finite for q and is constant, i.e., a circle, for < 0. 12 - 56 Application to Space Mechanics • Integration constant C is determined by conditions at beginning of free flight, q =0, r = r0 , 1 GM Ch 2 2 1 cos 0 r0 h GM 1 GM 1 GM C 2 r0 h r0 r0 v0 2 • Satellite escapes earth orbit for 1 or C GM h 2 GM r0 v0 2 vesc v0 2GM r0 • Trajectory is elliptic for v0 < vesc and becomes circular for = 0 or C = 0, GM vcirc r0 12 - 57 Application• Recall to Space Mechanics that for a particle moving under a central force, the areal velocity is constant, i.e., dA 1 2 1 2 r q 2 h constant dt • Periodic time or time required for a satellite to complete an orbit is equal to area within the orbit divided by areal velocity, ab 2 ab h2 h where a 12 r0 r1 b r0 r1 12 - 58 Sample Problem 12.9 SOLUTION: • Trajectory of the satellite is described by 1 GM 2 C cosq r h Evaluate C using the initial conditions at q = 0. A satellite is launched in a direction parallel to the surface of the earth with a velocity of 36,900 km/h at an altitude of 500 km. Determine: a) the maximum altitude reached by the satellite, and b) the periodic time of the satellite. • Determine the maximum altitude by finding r at q = 180o. • With the altitudes at the perigee and apogee known, the periodic time can be evaluated. 12 - 59 Sample Problem 12.9 SOLUTION: • Trajectory of the satellite is described by 1 GM 2 C cosq r h Evaluate C using the initial conditions at q = 0. 1 GM C 2 r0 h r0 6370 500 km 6.87 106 m km 1000 m/km v 0 36900 h 3600 s/h 10.25 103 m s h r0v0 6.87 106 m 10.25 103 m s 70.4 109 m 2 s GM gR 2 9.81m s 2 6.37 106 m 1 6.87 10 m 6 398 1012 m3 s 2 70.4 m s 2 2 65.3 109 m-1 2 398 1012 m3 s 2 12 - 60 Sample •Problem 12.9 Determine the maximum altitude by finding r at q = 1 180o. 1 GM 398 1012 m3 s 2 9 1 2 C 65 . 3 10 2 2 r1 h m 70.4 m s r1 66.7 106 m 66700 km max altitude 66700 - 6370km 60300 km • With the altitudes at the perigee and apogee known, the periodic time can be evaluated. a 12 r0 r1 12 6.87 66.7 106 m 36.8 106 m b r0 r1 6.87 66.7 106 m 21.4 106 m 2 ab 2 36.8 106 m 21.4 106 m h 70.4 109 m 2 s 70.3 103 s 19 h 31min 12 - 61 Kepler’s Laws of Planetary Motion • Results obtained for trajectories of satellites around earth may also be applied to trajectories of planets around the sun. • Properties of planetary orbits around the sun were determined astronomical observations by Johann Kepler (1571-1630) before Newton had developed his fundamental theory. 1) Each planet describes an ellipse, with the sun located at one of its foci. 2) The radius vector drawn from the sun to a planet sweeps equal areas in equal times. 3) The squares of the periodic times of the planets are proportional to the cubes of the semimajor axes of their orbits. 12 - 62