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Curvilinear Motions in Newtonian Mechanics and Quantum Spin


Centripetal and Centrifugal Forces
Orbital Motions
Rigid Body Dynamics
Differential Rotation
Quantum Mechanical Spins

Centripetal and Centrifugal Forces

Curvilinear Motions According to Newton's First Law, force-free object always moves in a straight line. In other word, some kind of force would be involved when the object's motion deviates from a straight line, i.e., in a curvilinear motion. The pulling is called centripetal force such as the gravitational attrcation Fp = -GmM/r2. The centrifugal force is in a direction pushing outward, this is the force resisting the change and in the form Ff = mv2/r (Figure 01a). These two forces cancel out exactly in circular motion as depicted in most of the popular images. In this special case the mathematics is especially simple with :
radial velocity vr = dr/dt = 0, r = constant, angular velocity v = d/dt = v/r = constant, and the angular momentum L = mvr = mvA/ = constant, where A = r2 is the area of the circle.

Figure 01a Curvilinear Motion
[view large image]

The Kepler's three laws are very simple in this special case : 1. the circular trajectory is a special case of an ellipse, 2. Since both r and /t are constants, the area A sweeps out in interval t is always A = r2/2.
3. By equating Fp = Ff , it can be shown readily that the square of the period T2 = (42/GM) r3.


Orbital Motions

While the angular momentum L = mvr remains to be a constant for orbital motion in general, the equation of motion in the radial direction becomes more complicated since the centripetal and centrifugal forces do not balance as shown below.
Kepler's Law

Figure 01b Kepler's Three Law
[view large image]

This is the Kepler's 1st Law. The 2nd Law is expressed by dA/dt = L/2m = constant, where dA = (r2/2)d. While the 3rd Law is covered by T = ab/(dA/dt) giving T2 = (42/GM)a3.


Figure 02 Parabolic Motions [view large image]

Although there is no solution for three body problem in closed form, it turns out that there are points in a three body configuration with two large and a much smaller ones on which the forces are balanced. They can rotate together as a whole. The five such locations are called Lagrangian points L1, ... L5. Following is the mathematics to derive the location of the smaller object r at L1 with respect to M2; R is the distance between M1 and M2 (see Figure 03a). According to a theorem in two-body problem, the two body motion can be reduced to a one body equivalence with center of mass M = M1 + M2 and distance Rm = M1R/M (with origin at M2). Assuming circular motion, the equation for the balance of contripetal and contrifugal forces can be written as :
Lagrangian Points

Figure 03a Lagrangian Points [view large image]

This is exactly the distance to L1 - the parking lot for many of the artificial satellites. However, it is known to be unstable, thruster burns manoeuvre is required to keep the object in place.
The equation for L2 is similar with (R-r) replaced by (R+r) and changing sign for the 2nd and 4th terms. The approximate distance r is the same as the one for L1. BTW, see "Cubic Formula" for cubic root calculation. The approximate formula for r is obtained by neglecting the r5, r4 powers and taking only the leading term in the cubic formula.

See "Synchronous Rotation" for another application of Newtonian mechanics to celestial bodies.

Lorentz Force One more example of circular motion involves charged particle moving in magnetic field. It is part of the
"Lorentz Forces" :     F = e (v x B),
where v is the velocity and B the magnetic field. The motion is particularly simple when v is perpendicular to a constant B field giving F = Bev, which is balanced by the centrifugal force : Bev = mv2/r. Thus, the motion is circular with r = mv/eB = constant (Figure 03b). This is the principle in its application to cyclotron and synchrotron. The motion becomes a helix in case the velocity has a component along the z direction (see insert).

Figure 03b Lorentz Force
[view large image]

Actually, at relativistic speed the circular motion of a charged particle turns into a spiral circling inward as it losses energy via synchrotron radiation unless to keep replenished such as in the case of synchrotron.


Rigid Body Dynamics

The Newtonian's equations of motion cannot be solved in closed form for three particles and beyond. It seems that the case for many particles would be much worse. However, such system can be described with mathematical precision if the separation between the particles is constant, that is, it is a rigid body, which is a good approximation to many solid objects. It will be shown momentarily below that the dynamics can be reduced to two equations - one for the translational motion of the center of mass, and the other for the rotation around the center of mass (an additional vibrational mode has to be considered for microscopic oscillation of the individual constituents, see specific heat). As shown in Figure 04a, if the reference point for the particles in the solid is shifted to the center of mass such that i mir'i = 0, r'i = constant, and with the further restriction on the motion to be parallel to a plane, i.e., in 2-dimensional motion, then the equations of motion can be written down in terms of the force F and torque T summing over all the particles in the system :
Rigid Body Dynamics Moment of Inertia

Figure 04a Rigid Body Dynamics

Figure 04b Moment of Inertia

These formulas are applicable to 3-dimensional motion as well, and the rotational term can be neglected if the size of the solid is much smaller than the distance scale.
That's why a galaxy of size ~ 1022 cm can be treated as a point in the cosmological model at the scale of ~ 1028 cm. Another example is the planets with size no bigger than 1010 cm in orbit of about 1013 cm or further from the Sun. Actually, these examples are just one of the reductionist's views of the world, which maintains that details of lower levels can be neglacted in higher level (see "Effective Theories").

Static in Rigid Body OInternal Shear Force When a force is exerted at one point on a beam as shown in Figure 05a,b it will be transferred internally all along the beam. This is a special property of rigid body and is very useful in our daily life. The situation is drastically different when the beam comes to rest on two pivots as shown Figure 05,d. Since the beam is not moving, the total force and total moment (torque) must be zero. Thus, we have

Figure 05 Statics in Rigid Body [view large image]

Figure 06a Internal Shear Force [view large image]

P = F1 + F2, and F1d = F2(L-d) giving
F1 = [(L-d)/L]P, and F2 = (d/L)P.

The internal shear force Fs inside the beam is computed by the formula :

Fs = F1 + (all forces between A and x) Fi

with the convention that positive force points up, negative force down, and the force F2 at the other end doesn't count. Thus
Fs = F1 from point A up to the load point, and on to point B with Fs = -F2, see

If we flip the diagram in Figure 05,d, i.e., F1 and F2 become the loads and P is the reaction force at the pivot (also see the wood gatherer in Figure 06c). The mathematical formulas remain unchanged but with the reverse direction of all the forces (including the shear force).

Another example with continuous and constant load along the beam is shown in Figure Figure 06a where f is the load density in unit of N/m. The summation in the shear force computation is now replaced by the integral

The internal shear force is canceled out by the reaction of the material within the rigid body resulting in static equilibrium with no deformation.

Modulus The above illustrations assume that the beam is perfectly rigid. In the real world, all materials exhibit some degree of elastic property defined by the shear modulus S = Shear Stress / Strain = (P/A) / (L/L). The higher the number corresponds to stiffer material until it becomes infinity for a perfectly rigid body. Figure 05,e shows the deformation of a beam under a point load. The maximum amount of deflection is expressed by the formula max = (PL3) / (48EI), where E is the Young's modulus defined similarly to the shear modulus S except that the force is now applied perpendicular to the area; and I is the second moment of area in unit of (length)4. The material is said to be linear if it can return to its original shape (after running through damped harmonic oscillation, see also footnote) after the

Figure 06b Modulus [view large image]

removal of the load. All materials have this characteristic for very small displacement, but the modulus would vary with displacement at large displacement for non-linear materials. All materials will break with very large applying force or deformed to certain amount. For example, the steel rod will break at about 0.25 BPa (109N/m2).

Mechanical Advantage Figure 06b lists the values of three different kinds of modulus for various materials. The Bulk modulus measures the substance's resistance to uniform compression. The other modulus are for linear deformation.

Figure 06c shows some of the indispensable working tools that make modern life much easier. The principle of lever comes from the balance of moments (torques) Faa = Fbb, or Fb = (a/b) Fa (see the illustrations in Figure 06c). Once such configuration has been arranged, then it takes very little effort to move the load at the other end. The factor (a/b) is the "mechanical advantage", which expresses the fact that the lower this number corresponds to lesser applying force (effort) to lift the object (resistance).

Figure 06c Mechanical Advantage

Footnote : For a simply supported beam (as shown in Figure 05,e) released from linear compression, it will undergo damped harmonic motion with force P = kx. This force is derived from the elastic property of the material; in term of the deflection P = (48EI/L3)max. Upon identifing x = max, we obtain k = (48EI/L3). For a rectangular beam with thickness t, width b, I = (t3b)/12 (see illustration below ), then the fundamental frequency = (k/m)1/2 = (2t/L2)(E/)1/2, where is the density of the beam, and m = tbL.

BTW, there are three types of supports as shown


Differential Rotation

Another way to describe the many particles motion is to consider them collectively in the form of fluid motion - a velocity field depending on space and time, e.g., u(x,y,z,t). The mathematics is embodied in the Navier-Stokes Equations. For irrotational flow with , they assume the forms :
Differential Rotation

Figure 06d Differential Rotation

where u = velocity vector field, = thermodynamic internal energy, p = pressure, T = temperature, = density, = viscosity, KH = heat conduction coefficient, F = external force per unit mass = acceleration, , and . Further details can be found in "Fluid Dynamics and the Navier-Stokes Equations"

The differential rotation (Figure 06d) is one of the examples for its application. According to the mathematics of Spiral Flow in polar coordinates, the trajectory is in the form of an Archimedean spiral r = a + b (see insert in Figure 06d), while the solution of the velocity fields is :

ur2 = GM / r + 2G0r0 - r2u2,
r2u2 = r (GM + 2G 0r0 r) / (b2 + r2),
where M is the mass of the central black hole, while the mass on the disk within r is given by M' = 2 r dr, where = 0 (r0/r) is the surface density in unit of gm/cm2, 0 is its cutoff value at the edge of the galaxy r = r0, and the third term is related to the centrifugal force. The formula for (r u) shows a similar profile taking from the Milkway disk (Figure 06d). The observed curve takes into account of the dark matter in the halo. This kind of analysis is also applicable to the hurricane, and the drain in the sink although on a much smaller scale.

See "Formation of Spherical Body" for a hydrostatics application.


Quantum Mechanical Spins