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CHAPTER 12. Kinetics of Particles: Newton’s Second Law. 12.1 INTRODUCTION. Reading Assignment. 12.2 NEWTON’S SECOND LAW OF MOTION.
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CHAPTER 12 Kinetics of Particles: Newton’s Second Law
12.1 INTRODUCTION Reading Assignment
12.2 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 the resultant and in the direction of this resultant force. More accurately = Effect Cause
Inertial frame or Newtonian frame of reference – one in which Newton’s second law equation holds. Wikipedia definition.
Free Body Diagrams (FBD) This is a diagram showing some object and the forces applied to it. It contains only forces and coordinate information, nothing else. There are only two kinds of forces to be considered in mechanics: Force of gravity Contact forces
y x Example FBD A car of mass m rests on a 300 incline. FBD N F This completes the FBD. q q mg
y x Example FBD A car of mass m rests on a 300 incline. FBD Just for grins, let’s do a vector addition. N F q q mg
Newton’s Second LawNSL A car of mass m rests on a 300 incline. FBD NSL y N x F q q mg What if friction is smaller?
y x Newton’s Second LawNSL A car of mass m rests on a 300 incline. NSL N F q q oops mg
12.3 LINEAR MOMENTUM OF A PARTICLE. RATE OF CHANGE OF LINEAR MOMENTUM
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.
12.4 SYSTEMS OF UNITS Reading Assignment
12.5 EQUATIONS OF MOTION Rectangular Components or
O y x z For Projectile Motion In the x-y plane
y O x z Tangential and Normal Components m And as a reminder
12.6 DYNAMIC EQUILIBRIUM Take Newton’s second law, This has the appearance of being in static equilibrium and is actually referred to as dynamic equilibrium. Don’t ever use this method in my course … H. Downing
y x O z 12.7 ANGULAR MOMENTUM OF A PARTICLE. RATE OF CHANGE OF ANGULAR MOMENTUM
Angular Momentum of a Particle moment of momentum or the angular momentum of the particle about O. It is perpendicular to the plane containing the position vector and the velocity vector.
For motion in x-y plane y In Polar Coordinates x O
12.8 EQUATIONS OF MOTION IN TERMS OF RADIAL AND TRANSVERSE COMPONENTS Consider particle at r and q, in polar coordinates, y This latter result may also be derived from angular momentum. x O
This latter result may also be derived from angular momentum. y x O
y m x O 12.9 MOTION UNDER A CENTRAL FORCE. CONSERVATION OF ANGULAR MOMENTUM When the only force acting on particle is directed 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, Position vector and motion of particle are in a plane perpendicular to
y m x O Position vector and motion of particle are in a plane perpendicular to Since the angular momentum is constant, its magnitude can be written as Remember
y x O Conservation of Angular Momentum • Radius vector OP sweeps infinitesimal area • Areal velocity • Recall, for a body moving under a central force, • When a particle moves under a central force, its areal velocity is constant.
12.10 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. m • Newton’s law of universal gravitation - two particles of mass M and m attract each other with equal and opposite forces directed along the line connecting the particles, M • For particle of mass m on the earth’s surface,
12.11 TRAJECTORY OF A PARTICLE UNDER A CENTRAL FORCE For particle moving under central force directed towards force center, Second expression is equivalent to
Remember that Let Remember that
This can be solved, sometimes. 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.
12.12 APPLICATION TO SPACE MECHANICS Consider earth satellites subjected to only gravitational pull of the earth. r q O A
There are two solutions: General Solution Particular Solution r q O A From the figure choose polar axis so that The above equation for u is a conic section, that is it is the equation for ellipses (and circles), parabolas, and hyperbolas.
Conic Sections Ellipse Parabola Hyperbola Circle Eccentricity
Origin, located at earth’s center, is a focus of the conic section. r q O A Trajectory may be ellipse, parabola, or hyperbola depending on value of eccentricity.
Hyperbola, e>1 or C > GM/h2. The radius vector becomes infinite for q1 r q O A q1
Parabola, e = 1 or C = GM/h2. The radius vector becomes infinite for q2 r q O A
Ellipse, e < 1 or C < GM/h2. The radius vector is finite for all q, and is constant for a circle, for e = 0. r q O A
and Integration constant Cis determined by conditions at beginning of free flight, q =0, r = r0. Burnout O A Powered Flight Launching
Escape Velocity Remember that For Burnout O A Powered Flight Launching
If the initial velocity is less than the escape velocity, the satellite will move in elliptical orbits. If e = 0, then r q O A
Recall that for a particle moving under a central force, the areal velocity is constant, i.e., B a Periodic time or time required for a satellite to complete an orbit is equal to the area within the orbit divided by areal velocity, b A’ O’ C O A Where r0 r1
12.13 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 by astronomical observations by Johann Kepler (1571-1630) before Newton had developed his fundamental theory. • Each planet describes an ellipse, with the sun located at one of its foci. • The radius vector drawn from the sun to a planet sweeps equal areas in equal times. • The squares of the periodic times of the planets are proportional to the cubes of the semimajor axes of their orbits.
