A 3-dimensional vector U  of coordinates  (x,y,z) is a linear combinationof the three pairwise perpendicular unit vectors of the coordinate system:

U   =   xi  +  yj  +  zk

If the base vectors were constant,then thederivative U' would simply bethe vector of coordinates  ( x',y',z' ) because the three derivativesi',j',k'would vanish...  Otherwise, we have to use the general expression:

U'   =  x'i  +  y'j  +  z'k + xi'  +  yj'  +  zk'

The length of each base vector remains constant and they remain orthogonal. So, the derivatives of all their pairwise dot products is zero. For example, i . i' = 0  and i . j' + j . i' = 0. Thus, there are 3 numbersa,b,c such that:

Introducing    = ai +bj +ck ,  this tells us that:

i'   =  (i . i' ) i  + (j . i' ) j  + (k . i' ) k   =  i
Likewise, j'   =  j    and    k'   =  k

Those values of i',j' andk' turn the above expression for U'  into:

U'   =  x'i  +  y'j  +  z'k + U
 
Where    =  (j'. k ) i  + (k'. i ) j  + (i'. j ) k
  is calledthe rotation vector  of (i,j,k)

We may apply the same rule twice to obtain the second derivative of U and find that U''  is the sum of  four terms,corresponding to as many distinct typesof acceleration  which have been given the following names:

• Relative acceleration:  x''i  +  y''j  +  z''k
• Centripetal acceleration:  (U)  =  (.U)²U
• Coriolis  acceleration:  2 ( x'i  +  y'j  +  z'k )
• Euler acceleration: 'U

For rotational rigid motion, momentum, velocity and inertia are angular  quantities, whose definitions depend on the point  O chosen as origin for positions.  They are respectively called:

• Angular momentum. L =rp (mass of momentump located at r).
• Angular velocity: Another name for theabove rotation vector .
• Moment of inertia or, rather,tensor of inertia : J (a square matrix).

If  O  is the center of mass, then the following relation holds (where J  is given by an expression which we shallestablish next).

L   =  J

 Like L, J  is additive, which is to say that the value for an extended body is obtainedby adding up the contributions of all its mass elements.

L   =    rv  dm       where  v   =  v_o + r
Therefore,  L   =  (  r dm )v_o +  r (r ) dm

The first term vanishes if  O  is the center of mass, whereasthe second term is a linear function of  . As such, it can be expressed in the form J . 

An explicit expression for the tensor J  is best obtained by switchingfrom vectorial notations  (cross-products and/or dot-products) to matrix notations, whereby r*  is the adjoint  of r (i.e., r  is a column, r*  isa row,  r r*  is a square matrix, while r* r  is merely the scalar r).

r (r )  =  r ²   (r.)r  =  r ²   r r*

J  is properly a tensor  (i.e., a square matrix) but, when the axis of rotation is fixed, it's convenient to introducea scalar  J, called themoment of inertia about that axis, which may be defined as:

J   =  u*J u

In this, u is a unit  [column ]  vectoralong the given axis of rotation and u* is its transposed (u*  is arow of components). This yields a scalar  relation:

||L ||   =  L   =   J

Note that the diagonal elements in the matrix J are the moments of inertia about the axes  Ox, Oy and Oz,respectively.  (The opposites of the off-diagonal elements areknown as products of inertia.)

Since the tensor of inertia is clearly symmetrical, there is a particularchoice of the coordinate system in which the corresponding matrix is diagonal  (the products of inertia vanish). The axes of such a system are the body's so-called principal axes of inertia. Considerations of symmetry are commonly encountered whichdetermine the directions of those axes with little or no computation.

Bodies which are endowed with identical  principal moments of inertia are said to have isotropic inertia. Such a body has the same moment of inertia  J  about any axis going through its center of mass,  irrespective of its "slant" (without resorting to the "tensor of inertia" concept,this simple result would be very tedious  toprove by direct integration, even for a homogeneous cube).

When there is isotropic inertia,the angular momentum is always collinear with the rotationvector  . Otherwise, this need not be the case!

The rotational energy is:

E_R   =   ½ . L  =   ½*J

If the rotation vector  has coordinates (₁,₂,₃)  along the principal axesor inertia, then the above is also equal to:

(J₁₁²  + J₂₂²  + J₃₃² )

The total kinetic energy of a solid is equal to the sum of the above and what wouldbe the kinetic energy of its entire mass located at its center of gravity:

E   =   ½ Mv ² +  ½*J

The moment of inertia of an element of mass  m can generally be defined as  r² m where  r  is the distance to some reference object.

That reference object  is usually an axis about which rotationis considered.

However, we may also consider moments about a point or about aplane.  Such moments of inertia do not have any direct physical applicationto rotational motion but they can be convenient stepping stones  inthe computation of the physical moment of inertia of a rigid body about an axis...

In a Cartesian coordinate system, a rigid body is composed of infinitesimalmass elements  dm =  dx dy dz   where is the mass density at point (x,y,z). Typical (generalized) moments of inertia have the following expressions:

A few statements can be made which result immediately from such definitions.

For example, the sum of the moments of inertia about two perpendicular planes  is equal to the moment of inertia about the axis where they intersect.  (For a thin massive plate, this translates into a proof ofthe so-called perpendicular axis theorem presentedbelow.)

It's also readily observed thatthe sum of the (ordinary) moments of inertia about 3 mutually perpendicular axesmeeting at point  O  is equal to twice the moment of inertia about the point  O. Thus, if those 3 axial moments of inertia are known to be identical,each must be equal to  2/3  of the moment about the center. This makes it easy to compute the moment of inertia about an axis whichgoes through the center of a spherical distribution of mass,as discussed in thenext article.

For a thin hollow spherical shell  (like aping-pong ball) the whole mass M is at a distance R from the center and theinertia about the center is thus  MR. So, the moment of inertia of such a uniform shell about an axis going through itscenter is:

J   =  ²/₃ M R²

 Aspreviously noted,the moments of inertia about two perpendicular planes  add up to the moment of inertiaabout the axis  where they intersect. The conclusion follows from the remark that the momentof inertia of a planar distribution about an axis contained in its planeis the same as the moment of inertia about the plane perpendicular to the plate which intersects it along that axis.

More directly, the result can be construed as a consequence of the equality:

 ( x² + y² ) dm  =   y² dm +  x² dm     

Example :

The moment of inertiaabout a central vertical axis of an horizontal homogeneous disk of mass  M and radius  R  is easy to find directly... (:  M and J are proportional to  r dr and  r ³ dr .)

J_z   =   ½ M R²

The theorem gives the moment of that thin diskabout an horizontal  axis:

J_x   =   J_y   =   ¼ M R².

If  J  is the moment of inertia about an axis going through the centerof mass of a body of mass  M,  then the moment of inertia J'  about another parallel  axis ata distance  d  is given by the following relation:

J'   =   J  +  M d ²

Let's consider a thick horizontal plate and an horizontal axis through itscenter of mass  (at altitude z=0). Any horizontal cross-section at altitude  z  is a "thin plate" ofinfinitesimal height  dz  whose mass is  (M/h) dz. Theparallel axis theorem gives the moment of inertia of such a thin plate about our centralaxis and the moment of inertia of the entire thick  plate isobtained as a simple integral:

J   =

h/2   -h/2

( k + z 2) (M/h) dz    =     M ( k + h2/12 )

For a cross-section of negligible extent (k=0) that formula gives the moment ofinertia of a straight rod of length h, about a central axis perpendicular to it.

J   =   M h² / 12

Less trivially, we may consider a vertical cylinder of height  h whose cross-section is an homogeneous disk of radius  R. As shown above,the moment of such a disk about an horizontal  axis of symmetry is k = ¼ R  times its mass. Therefore, the moment of inertia of the cylinder about any horizontal  axis is:

J_x   =  J_y   =   M  ( R² / 4 +  h² / 12 )

Thethin-disk formula holds for the vertical axis: J_z = ½ M R²
Isotropic inertia ( J_x = J_y = J_z ) is achieved when  h = R 3.

J_z   =  ³/₁₀ M (R⁵ - r⁵)/(R³ - r³)

Note how this becomes J_z = ½ M R² when  R = r  (byl'Hospital's rule).

  of height  h,inner radius  r, outer radius  R :
[ For a solid cylinder, r = 0.  For a thin tube, r = R. ]

J_x   =  J_y   =   M  [ ( R² + r²) / 4 +  h² / 12 ]
J_z   =   M  ( R² + r²) / 2

  (ring)  ofinner radius  r  and outer radius  R :

J_x   =  J_y   =   M  [ ( R² + r²) / 4 +  (R-r)² / 32 ]
J_z   =   M  [ ( R² + r²) / 2   (R-r)² / 16 ]

  of radius  R :

J_x   =  J_y   =  J_z   =  ²/₅ M R²

 Spherical shell  of inner radius  r and outer radius  R :

J   =  ²/₅ M ( R⁵ - r⁵ ) /( R³ - r³ )

When such a spherical shell is fairly thin  (e.g., ping-pong ball) we may introduce the relative thickness  x = (R-r)/R. We thus have  r = R (1-x)  and obtain:

J   =  ²/₃ M  R ² [ 1 - x + 2 x ²/3 - x ⁴/45 - x ⁵/45- 2 x ⁶/135 - x ⁷/135 - ... ]

Old-fashioned ping-pong balls have a nominalradius of  19 mm  and a thickness of roughly  0.38 mm. So, x is very nearly  0.02  and we have:

J   =   0.6535  M R²  =   5.898 g cm²

Such "38 mm" balls have a nominal mass of  2.5 g. On October 1, 2000, those were replaced by "40 mm" balls (with a nominal mass of  2.7 g)  for official competitions. Assuming the material remained the same, this made the thickness decreaseby  7.4%.  and reduced the parameter x  by  12%,  down to  0.0176.

J   =   0.655  M R²  =   7.074 g cm²

All told, the moment of inertia of a ping-pong ball increased  20%  in 2000.

Pictured at right is the situation in the vertical plane containing the body'scenter of gravity  (C).  The axis of rotation goes through point  O.   is the angle from the vertical to the line  OC.

Let  M  be the mass of the body and  J  be its momentof inertia around an axis through the center of gravity. Let  L  be the distance from  O  to  C. By theparallel axis theorem, themoment around the actual axis of rotation is J + ML.

With respect to  O the only torque  is that ofthe weight  M g applied at point  C.  It equals thederivative of the angular momentum (J + ML) '.

(J + ML² ) ''  +  M g L sin   =   0

For small oscillations  (sin )  thisis an harmonic motion of period  T.

T   =   2			L + J/ML
			g

A simple pendulum  is the special case  J = 0; a point mass suspended by a massless string, which yields small oscillationsof period T = 2 (L/g).

The above has the same period as a simple pendulum of length L + J/ML. The point  S  beyond  C  at that distance from  O is called the center of oscillation  or center of percussion since a perpendicular impact at that particular point (affectionately called sweet spot) will not be felt at the pivot point  O.

In the compound pendulum  discussedabove,thegeometrical mean of thedistances  OC  and  CS  is always equal to the radius of gyration  R.

OC . CS   =   R ²   =   J/ M

In1673,Christiaan Huygenspointed out that this gives the pivot point  O  and the centerof percussion  S  ("sweet spot") interchangeable rôles.

Precision pendulums,called reversible pendulums, have been constructed which can be swingedfrom either point  with the same period of oscillation as a simple pendulumwhose length would be equal to the distance  OS.

The two (parallel) pivoting axes on a reversible pendulum are normally located at different  distances from the center of gravity  C (except in the very special case where they are both located at a distanceequal to the aforementionned radius of gyration  R). A symmetric pendulum with two pivots is rarely  a reversible one.

In the field of gravimetry, a reversible pendulum is often called an absolute  pendulum becauseknowledge of the distance between its two axes allows a direct conversionof its measured period into the absolute local value of the gravitationalfield  (other types of pendulums give only relative measurements andmust be calibrated to provide an absolute value).

For a rigid object with spherical symmetry, the moment of inertia about an axisthrough the center is  2/3  of the inertia about the center itself (as shownabove). So, if  (r)  is the density at a distance r  from the center, then the total mass  M and the moment of inertia  J  about any axis going through thecenter are:

M   =  4 p r(r) r² dr          and          J   =  ²/₃4 p r(r) r⁴ dr

For an homogeneous  sphere of radius R,  is constant and we obtain:

J   =  ²/₅ M R²

Homogeneous Ellipsoid :

For an homogeneous ellipsoid of equation  (x/a)² + (y/b)² +(z/c)²   < 1  we apply the same argument successively to 3 orthogonal stretching, starting with a sphere. This gives the ellipsoid's moments of inertia about the 3 coordinate planes.  Namely:

J_xOy   =  ¹/₅ Mc²       J_yOz   =  ¹/₅ Ma²       J_zOx   =  ¹/₅ Mb²

The pairwise sums of those are the moments of inertia about the coordinate axes.  So, we obtain the following expression for the matrix of inertia:

J   =    1/5 M

b 2 +c 2  0 0

0  a 2 +c 2  0

0 0  a 2 +b 2

The coefficient  0.330695  =  J₂/ H  (see below) is known much more accurately  (from astronomical observations)  thanthe actual density distribution of matter within the Earth. The former is thus viewed as an experimental constraint for any model of the latter.

The so-called dynamical flattening  (H)  of the Earth is thefollowing function of the three principal momenta of inertia of the Earth A < B < C.

H   =   [ C ½ (A+B) ]/ C

H can be derived from the observed period of precession of equinoxes. It is related to the second zonal Stokes parameter  (J₂)  defined by:

   J₂   =   [ C ½ (A+B) ]/ Ma ²

This appears in the multipole expansion of the gravitational field. Under its alternate name of second dynamic form factor,  J₂ is a primary parameter in the definition of thereference ellipsoid  (IUGG 1980)which also inposes the value of the equatorial radius a.

a  =  6378137 m           J₂  =  0.00108263

At face value, this gives  H = 0.00108263 / 0.330695 = 0.0032738
Also, the moment of inertia of the Earth about an equatorial axis is:

A   =   B   =  C    J₂ Ma ²  =   0.329612 Ma ²

The following dicussion is usually only applied to oblate spheroids,  butit's valid for any rigid mass distribution,  regardless of symmetry. If we use as origin of our coordinates the center of mass  O, then the dipolar component vanishes and the dominant correctionto the spherical field is quadrupolar.

Let  M  be the total mass of the solid and let's call a its largest radius  (from the centerof mass  O  to the farthest massive point). For an oblate spheroid,  that radius would be the equatorial radius.

By analogy withelectrostatics, the multipolar Newtonian potential is:

V(r)   =   V(ru)   =    -G     Ma n Jn(u) r r n

By definition,  the dimensionless scalar  J_n(u)  is the n^thform-factor in the direction of the unit vector u. 

The two bodies  (A and B) exert opposite torques on each other via some sort of muscle  (e.g., battery-powered electromagnets or massless springs ).

If the whole thing is spinless, the following expression of the angularmomentum integrated over time remains constant:

( J_A + M_Aa²)_A  +  ( J_B + M_Bb²)_B

In this, the two centers of gravity  A  and  B are assumed to be in a plane perpendicular to the axis of rotation, which intersects itat point  O,  as depicted here. The  M  and  J  symbols respectively denote the massesand the proper moments of inertia (around the direction of the common axis of rotation). The quantities a andb are the distances from O to A and B,respectively.  The angles   are the inclinations with respect to a fixed direction  Ox.

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