1. Chapter 12 KINETICS OF PARTICLES: NEWTON’S SECOND LAW Denoting by m the mass of a particle, by  F the sum, or resultant, of the forces acting on the particle, and by a the acceleration of the particle relative to a newtonian frame of reference, we write  F = ma Introducing the linear momentum of a particle, L = mv, Newton’s second law can also be written as  F = L. which expresses that the resultant of the forces acting on a particle is equal to the rate of change of the linear momentum of the particle. Sharif University-Aerospace Dep. Fall 2004

2. To solve a problem involving the motion of a particle,  F = ma should be replaced by equations containing scalar quantities. Using rectangular components of F and a, we have  F x = ma x  F y = ma y  F z = ma z x y P anan O atat x y z axax ayay azaz P x P aa O arar Using tangential and normal components,  F t = ma t = m dv dt v2v2 Using radial and transverse components,.....  F r = ma r = m(r - r  2 )  r  F n = ma n = m  F  = ma  = m(r  + 2r  )

3. x y z P r  HOHO O The angular momentum H O of a particle about point O is defined as the moment about O of the linear momentum mv of that particle. H O = r x mv We note that H O is a vector perpendicular to the plane containing r and mv and of magnitude H O = rmv sin  Resolving the vectors r and mv into rectangular components, we express the angular momentum H O in determinant form as H O = i j k x y z mv x mv y mv z mvmv

4. x y z P r  HOHO O H O = i j k x y z mv x mv y mv z In the case of a particle moving in the xy plane, we have z = v z = 0. The angular momentum is perpendicular to the xy plane and is completely defined by its magnitude H O = H z = m(xv y - yv x ) Computing the rate of change H O of the angular momentum H O, and applying Newton’s second law, we write  M O = H O.. which states that the sum of the moments about O of the forces acting on a particle is equal to the rate of change of the angular momentum of the particle about O. mvmv

5. 00 mv 0 r0r0 O r P0P0 P  mv When the only force acting on a particle P is a force F directed toward or away from a fixed point O, the particle is said to be moving under a central force. Since  M O = 0 at any given instant, it follows that H O = 0 for all values of t, and H O = constant We conclude that the angular momentum of a particle moving under a central force is constant, both in magnitude and direction, and that the particle moves in a plane perpendicular to H O..

6. 00 mv 0 r0r0 O r P0P0 P  mv rmv sin  = r o mv o sin  o Using polar coordinates and recalling that v  = r  and H O = mr 2 , we have r 2  = h. where h is a constant representing the angular momentum per unit mass H o /m, of the particle. Recalling that H O = rmv sin , we have, for points P O and P.. for the motion of any particle under a central force.

7.  dd r d  dA P F O The infinitesimal area dA swept by the radius vector OP as it rotates through d  is equal to r 2 d  /2 and, thus, r 2  represents twice the areal velocity dA/dt of the particle. The areal velocity of a particle moving under a central force is constant. 00 mv 0 r0r0 O r P0P0 P  mv r 2  = h..

8. M m F -F r An important application of the motion under a central force is provided by the orbital motion of bodies under gravitational attraction. According to Newton’s law of universal gravitation, two particles at a distance r from each other and of masses M and m, respectively, attract each other with equal and opposite forces F and -F directed along the line joining the particles. The magnitude F of the two forces is F = G Mm r 2 where G is the constant of gravitation. In the case of a body of mass m subjected to the gravitational attraction of the earth, the product GM, where M is the mass of the earth, is expressed as GM = gR 2 where g = 9.81 m/s 2 = 32.2 ft/s 2 and R is the radius of the earth.

9. A particle moving under a central force describes a trajectory defined by the differential equation d 2 u d  2 + u = F mh 2 u 2 where F > 0 corresponds to an attractive force and u = 1/r. In the case of a particle moving under a force of gravitational attraction, we substitute F = GMm/r 2 into this equation. Measuring  from the axis OA joining the focus O to the point A of the trajectory closest to O, we find 1r1r = u = + C cos  GM h 2 A  O r

10. 1r1r = u = + C cos  GM h 2 This is the equation of a conic of eccentricity  = Ch 2 /GM. The conic is an ellipse if  < 1, a The values of the initial velocity corresponding, respectively, to a parabolic and circular trajectory are v esc = 2GM r 0 v circ = GM r 0 parabola if  =1, and a hyperbola if  > 1. The constants C and h can be determined from the initial conditions; if the particle is projected from point A with an initial velocity v 0 perpendicular to OA, we have h = r 0 v 0. A  O r

11. v esc = 2GM r 0 v circ = GM r 0 v esc is the escape velocity, which is the smallest value of v 0 for which the particle will not return to its starting point. The periodic time  of a planet or satellite is defined as the time required by that body to describe its orbit, 2  ab h  = where h = r 0 v 0 and where a and b represent the semimajor and semiminor axes of the orbit. These semiaxes are respectively equal to the arithmetic and geometric means of the maximum and minimum values of the radius vector r. A  O r