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(2007-07-25)   Rotation of a Frame of Reference  (Rigid Kinematics)
True variation of a vector when the frame of reference moves.

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

U   =   x i  +  y j  +  z k

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

U'   =   x' i  +  y' j  +  z' k  +  x i'  +  y j'  +  z k'

The length of the three base vectors 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 numbers a, b, c such that:

i . i'	=	0		i . j'	=	-c		i . k'	=	b
j . i'	=	c		j . j'	=	0		j . k'	=	-a
k . i'	=	-b		k . j'	=	a		k . k'	=	0

Introducing   W  =  a i + b j + c k ,   this tells us that:

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

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

U'   =   x' i  +  y' j  +  z' k  +  W ´ U
 
Where  W   =   ( k . j' )  i  +  ( i . k' )  j  +  ( j . i' )  k
W  is called the  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  4 terms.  When  U  is the position of a point, these are 4 distinct types of  acceleration  which have been given the following names:

• Relative acceleration:   x'' i  +  y'' j  +  z'' k
• Centripetal acceleration:   W ´ (W ´ U)   =   (W.U) W - W² U
• Coriolis acceleration:   2 W ´ ( x' i  +  y' j  +  z' k )
• Euler acceleration:  W' ´ U

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(2007-09-12)   Dynamics of the Rotation of a Rigid Body
How momentum does remain the product of inertia by velocity...

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 = r ´ p  (for a mass of momentum p at position r).
• Angular velocity:  Another name for the above rotation vector  W.
• 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 shall establish next).

L   =   J W

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

L   =   ò  r ´ v  dm         where   v   =   v_o  +  W ´ r
Therefore,   L   =   ( ò  r dm ) ´ v_o  +  ò  r ´ ( W ´ r ) dm

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

An explicit expression for the tensor  J  is best obtained by switching from 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*  is a row,   r r*  is a square matrix, while   r* r  is merely the scalar  r² ).

r ´ ( W ´ r )   =   r ² W  -  (r.W) r   =   r ² W  -  r r* W

So,   J   =   ò  ( r 2 Î - r r* ) dm     =

ò

y 2 + z 2  - y x - z x

- x y  x 2 + z 2  - z y

- x z - y z  x 2 + y 2

dm

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

J   =   u* J u

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

|| L ||   =   L   =   J w

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 are known as  products of inertia.)

Since the tensor of inertia is clearly symmetrical, there is a particular choice 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 which determine the directions of those axes with little or no computation.

For example, the principal directions of inertia of an homogeneous  brick are parallel to the brick's edges.  In the special case of a cube, the principal moments of inertia are identical.  Therefore, the inertia tensor of a cube is a scalar multiple of the identity matrix  Î.

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  to prove by direct integration, even for a homogeneous cube).

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

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(2007-09-13)   Moments of Inertia about a Point or a Plane
Mathematical fictions that are useful to compute the inertia about an axis.

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

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

However, we may also consider moments about a point or about a plane.  Such moments of inertia do not have any direct physical application to rotational motion but they can be convenient  stepping stones  in the 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 infinitesimal mass elements   dm = r dx dy dz   where r is the mass density at point (x,y,z).  Typical (generalized) moments of inertia have the following expressions:

Moment of inertia about the  point  O :		ò  ( x2 + y2 + z2 ) dm
Moment of inertia about the  axis  Oz :		ò  ( x2 + y2 ) dm
Moment of inertia about the  plane  xOy :		ò  z2  dm

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 of the so-called  perpendicular axis theorem presented below.)

It's also readily observed that the sum of the (ordinary) moments of inertia about 3 mutually perpendicular axes meeting 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 which goes through the center of a spherical distribution of mass, as discussed in the next article.

For a thin hollow spherical shell  (like a ping-pong ball)  the whole mass M is at a distance R from the center and the inertia about the center is thus  MR².  So, the moment of inertia of such a uniform shell about an axis going through its center is:

J   =   ²/₃ M R²

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(2007-09-13)   Moment of Inertia of a Spherical Distribution
An easy computation even for a  nonhomogeneous  sphere.

For a rigid object with  spherical symmetry, the moment of inertia about an axis through the center is  2/3  of the inertia about the center itself  (as shown above).  So, if  r(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 the center are:

M   =   ò 4 p r(r) r² dr           and           J   =   ²/₃ ò 4 p r(r) r⁴ dr

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

J   =   ²/₅ M R²

Homogeneous Ellipsoid :

A uniform solid remains uniform when it's streched by a factor (c/R) along the direction of  Oz.  If such a stretch leaves the total mass M unchanged, then the moments of inertia about planes parallel to the direction of stretch are unchanged.  However, the moment of inertia about a perpendicular plane is multiplied by the  square  of the stretching factor (c/R).

Applying the same argument successively to 3 orthogonal stretching direction, we see that the expressions valid for an homogeneous sphere yield the moments of inertia about the 3 coordinate planes of an homogeneous ellipsoid of equation   (x/a)² + (y/b)² + (z/c)²   < 1 .  Namely:

J_xOy   =   ¹/₅ M c²         J_yOz   =   ¹/₅ M a²         J_zOx   =   ¹/₅ M b²

The pairwise sums of those expressions 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

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(2007-09-13)   Perpendicular Axis Theorem
Momenta of inertia of a thin plate about an axis perpendicular to it.

The moment of inertia of a thin plate (a planar mass distribution) about an axis  Oz  perpendicular to its plane is equal to the sum of the moments of inertia about two orthogonal axes  Ox  and  Oy  within  the plane  (intersecting at the point  O  where the perpendicular axis crosses the plate).  In a nutshell: Jz   =   Jx + Jy

Proof :   As previously noted, the moments of inertia about two perpendicular  planes  add up to the moment of inertia about the  axis  where they intersect.  The conclusion follows from the remark that the moment of inertia of a planar distribution about an axis contained in its plane is 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:

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

Example :

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

J_z   =   ½ M R²

The above theorem gives the moment of that thin disk about an  horizontal  axis:

J_x   =   J_y   =   ¼ M R².

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(2007-09-13)   Parallel Axis Theorem
Koenig's theorem  applied to the moment of inertia about an axis.

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

J'   =   J  +  M d ²

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(2007-09-16)   Moment of Inertia of a Thick Plate
If a thin plate of mass M has inertia  k M  about a central axis of its plane, then a similar plate of thickness  h  has inertia  J = M (k + h²/12).

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

J   =

ò

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 of inertia 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 R²  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 )

Isotropic inertia  ( J_x = J_y = J_z )  is achieved when  h = R Ö3.

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(2007-09-13)   Inertia of  homogeneous  solids of mass M
Moments of inertia about principal axes through the center of mass.

Tube  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

Torus  (ring)  of inner 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 ]

Sphere  of radius  R :

J_x   =   J_y   =   J_z   =   ²/₅ M R²

Ellipsoid  of  principal semiaxes  a, b, c :	Jx  =  1/5 M (b 2 + c 2 ) Jy  =  1/5 M (a 2 + c 2 ) Jz  =  1/5 M (a 2 + b 2 )
Brick  of  edges  a, b, c :   ( for a cube,  J  =  1/6 M a 2 )	Jx  =  1/12 M (b 2 + c 2 ) Jy  =  1/12 M (a 2 + c 2 ) Jz  =  1/12 M (a 2 + b 2 )

Hollow ball.   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 nominal radius 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 decrease by  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 by  20%  in 2000.

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(2008-02-15)   Rigid Pendulum
Rigid body moving about a fixed horizontal axis.

Pictured at right is the situation in a plane containing the body's center of gravity  C  and perpendicular to the axis of rotation  (which goes through O).  q  is the angle from the vertical to  the line  OC.

Let  M  be the mass of the body and  J  be its moment of inertia around an axis through the center of gravity.  Let  L  be the distance from  O  to  C.  By the parallel axis theorem, the moment around the actual axis of rotation is  J + ML².

With respect to  O  the only  torque  is that of the weight  M g  applied at point  C.  That's equal to the derivative of the angular momentum  (J + ML² ) q'.  So:

(J + ML² ) q''  +  M g L  sin q   =   0

For small oscillations  (sin q » q)  this is an harmonic motion of period  T.

T   =   2 p	Ö		L + J/ML
			g

A  simple pendulum  is the special case  J = 0;  a point mass suspended by a massless string, which yields small oscillations of period  T = 2p Ö(L/g).  The above has the same period as a simple pendulum of length  L (1+J/ML² ).

See also :   Conical Pendulum