1. Chapter 12 KINETICS OF PARTICLES: NEWTON’S SECOND LAW
Denoting by m the mass of a particle, by S 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
S F = ma
Introducing the linear momentum of a particle, L = mv, Newton’s second law can also be written as .
S 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.
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2. S Fx = max S Fy = may S Fz = maz
To solve a problem involving the motion of a particle, S F = ma should be replaced by equations containing scalar quantities. Using rectangular components of F and a, we have P ax az
S Fx = max S Fy = may S Fz = maz x z
Using tangential and normal components, y an dv dt
S Ft = mat = m at P v2 r
S Fn = man = m O x
Using radial and transverse components, aq . .. ar
S Fr = mar= m(r - rq2) r P .. . q .
S Fq = maq = m(rq + 2rq) x O

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

4. i j k x y z HO = mvx mvy mvz HO = Hz = m(xvy - yvx) . . S MO = HO y mvf P r O
In the case of a particle moving
in the xy plane, we have z = vz = 0. The angular momentum is perpendicular to the xy plane and is completely defined by its magnitude x z
HO = Hz = m(xvy - yvx) .
Computing the rate of change HO of the angular momentum HO , and applying Newton’s second law, we write .
S MO = HO
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.

5. 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 S MO = 0 at any given instant, it follows that HO = 0 for all values of t, and f P
mv0 r O f0 r0 P0 .
HO = 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 HO .

6. rmv sin f = romvo sin fo . r2q = h . . mv
Recalling that HO = rmv sin f, we have, for points PO and P f P
mv0 r
rmv sin f = romvo sin fo O f0 r0 P0
for the motion of any particle under a central force. . .
Using polar coordinates and recalling that vq = rq and HO = mr2q, we have .
r2q = h
where h is a constant representing the angular momentum per unit mass Ho/m, of the particle.

7. mv . f P
r2q = h
mv0 r O f0 r0 P0
The infinitesimal area dA swept by the radius vector OP as it rotates through dq is equal to r2dq/2 and, thus, r2q represents twice the areal velocity dA/dt of the particle. The areal velocity of a particle moving under a central force is constant. .
r dq dA P dq F O q

8. 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 r m F -F M
with equal and opposite forces F and -F directed along the line joining the particles. The magnitude F of the two forces is Mm r2
F = G
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 = gR2
where g = 9.81 m/s2 = 32.2 ft/s2 and R is the radius of the earth.

9. F d 2u + u = dq 2 mh 2u 2 1 GM r = u = + C cos q h2A q O r
A particle moving under a central force describes a trajectory defined by the differential equation
d 2u
dq 2 F
mh 2u 2
+ u =
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/r2 into this equation. Measuring q from the axis OA joining the focus O to the point A of the trajectory closest to O, we find 1 r GM h2
= u = C cos q

10. 1 r GM h2 = u = + C cos q 2GM r0 GM r0 vesc = vcirc = r qA q O r 1 r GM h2
= u = C cos q
This is the equation of a conic of eccentricity e = Ch2/GM. The conic is an ellipse if e < 1, a
parabola if e =1, and a hyperbola
if e > 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 v0 perpendicular to OA, we have h = r0v0.
The values of the initial velocity corresponding, respectively, to a parabolic and circular trajectory are
2GM r0 GM r0
vesc =
vcirc =

11. 2GM GM vesc= vcirc= r0 r0 2pab t = h r qO r
2GM r0 GM r0
vesc=
vcirc=
vesc is the escape velocity, which is the smallest value of v0 for which the particle will not return to its starting point.
The periodic time t of a planet or satellite is defined as the time required by that body to describe its orbit,
2pab h
t =
where h = r0v0 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.