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Final Answers
© 2000-2015   Gérard P. Michon, Ph.D.    

     Rigid Bodies

 Housemark (gmerki) of 
 Nicolaus Copernicus
The proper motion of a sphere
 is rotation in a circle
.
Nicolaus Copernicus  (1473-1543)

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Moment of Inertia by Eric W. Weisstein
 
Wikipedia :   Moment of Inertia   |   List of Moments of Inertia

Video :   MIT OpenCourseWare   Classical Mechanics by  Walter Lewin.
UC Berkeley  Physics for Future Presidents by  Richard A. Muller.

 
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Dynamics of a Rigid Body
in Classical  (Newtonian)  Mechanics


(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 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 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   =   ( j' . ki  +  ( k' . ij  +  ( i' . jk
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  four terms, corresponding to as many 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:    Leonhard Euler 
 (1707-1783)  Gaspard Coriolis 
 (1707-1783) 2 W ´ ( x' i  +  y' j  +  z' k )
  • Euler acceleration:  W' ´ U


(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 momentumL = r ´ p  (mass of momentum p located at r).
  • Angular velocity:  Another name for the above rotation vector  W.
  • Moment of inertia or, rather, tensor of inertiaJ  (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  LJ  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   =   vo  +  W ´ r
Therefore,   L   =   ( ò  r dm ) ´ vo  +  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  WQED

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   rr  is merely the scalar  r).

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
- y x
- z x
  - x y
 x 2 + z
- z y
  - x z
- y z
 x 2 + y
-
|
-
  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!


(2013-02-21)   Kinetic Energy of a Solid
Sum of its rotational energy and the kinetic energy of its center of mass.

The rotational energy is:

ER   =   ½ W . L   =   ½ W* J W

If the rotation vector  W  has coordinates  (W1 , W2 , W3 )  along the principal axes or inertia, then the above is also equal to:

½ ( J1 W12  +  J2 W22  +  J3 W32 )

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

E   =   ½ M v 2  +  ½ W* J W


(2007-09-13)   Moments of Inertia about a Point or a Plane
Mathematical fictions aimed at computing the inertia about an  axis.

The moment of inertia of an element of mass  dm  can generally be defined as  rdm  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


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

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 :

 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). 

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:

JxOy   =   1/5 M c2         JyOz   =   1/5 M a2         JzOx   =   1/5 M b2

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
0
0
  0
 a 2 + c
0
  0
0
 a 2 + b
-
|
-


(2010-12-13)   Moment of Inertia of the Earth, about its polar axis
C  =  0.330695 M a 2   [ where  a is the  equatorial  radius of the Earth ]

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

The seismologist and applied mathematician  Keith Edward Bullen (1906-1976; FRS 1949)  is credited with the remark that the above numerical coefficient could be derived from the observed motions of the Earth and the Moon.  From seismological data, Bullen established, in the 1930's, models of the Earth's density compatible with astronomical observations which remained standard for decades.

The so-called  dynamical flattening  (H)  of the Earth is the following 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) ] / M a 2

Under its alternate name of  dynamic form factor,  J2  is a primary parameter in the definition of the reference 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 M a 2   =   0.329612 M a 2

Steady Change in [the] Flattening of the Earth: The Precession Constant and its Long-Term Variation.
by Milan Bursa, Erwin Groten   Zdislav Sima.   The Astronomical Journal  135  (2008) 1021


(2007-09-13)   Perpendicular Axis Theorem
Momenta of inertia of a  lamina about an axis perpendicular to it.
 Lamina   A planar mass distribution is called a  lamina.  The moment of inertia of such a  thin plate  about an axis  Oz  perpendicular to its plane is the sum of the moments of inertia about two orthogonal axes  Ox  and  Oy  within  the plane  (the intersection  O  is 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      QED

Example :

The moment of inertia Thin Disk 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 .)

Jz   =   ½ M R2

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

Jx   =   Jy   =   ¼ M R2.


(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


(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 )     QED

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:  Solid Cylinder

Jx   =   Jy   =   M  ( R2 / 4  +  h2 / 12 )

The thin-disk formula holds for the vertical axis:  Jz = ½ M R2
Isotropic inertia  ( Jx = Jy = J)  is achieved when  h = R Ö3.


 Solid Conical Frustum (2011-04-19)   Inertia of a right cone  (or conical frustum)
Moment of inertia of a cone about its axis of symmetry.

Jz   =   3/10  M (R5 - r5 ) / (R3 - r3 )

Note how this becomes  Jz = ½ M R2  when  R = r   (by l'Hospital's rule).


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

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

Jx   =   Jy   =   M  [ ( R2 + r2 ) / 4  +  h2 / 12 ]
Jz   =   M  ( R2 + r2 ) / 2

 Torus

Torus  (ring)  of inner radius  r  and outer radius  R :

Jx   =   Jy   =   M  [ ( R2 + r2 ) / 4  +  (R-r)2 / 32 ]
Jz   =   M  [ ( R2 + r2 ) / 2  -  (R-r)2 / 16 ]

 Billiard ball

Sphere  of radius  R :

Jx   =   Jy   =   Jz   =   2/5 M R2

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

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  20%  in 2000.


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

Pictured at right is the situation in the vertical plane containing the body's center of gravity  (C).  The axis of rotation goes through point  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.  It equals the derivative of the angular momentum  (J + MLq'.

(J + MLq''  +  M g L  sin q   =   0

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

vinculum
T   =   2 p   Ö   L + J/ML
vinculum
g

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 + 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.  Hammer

That's a very desirable feature if the "pendulum" is in fact a hammer with your wrist at point  O.  It is most comfortable to hit a nail with the surface of the hammer at point  S.  Experienced carpenters who "know" their favorite hammers will do that pretty naturally.

See also :   Conical Pendulum


(2009-08-19)   Reversible Pendulum
Asymmetrical pendulum with two pivots yielding the  same  period.

In the  compound pendulum  discussed above, the geometrical mean of the distances  OC  and  CS  is always equal to the  radius of gyration  R.

OC . CS   =   R 2   =   J / M

In 1673, Christiaan Huygens pointed out that this gives the pivot point  O  and the center of percussion  S  ("sweet spot")  interchangeable  rôles.

Precision pendulums, called  reversible pendulums, have been constructed which can be swinged from either point  with the same period of oscillation as a simple pendulum whose 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 distance equal 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 because knowledge of the distance between its two axes allows a direct conversion of its measured period into the absolute local value of the gravitational field  (other types of pendulums give only relative measurements and must be calibrated to provide an absolute value).

See also :   Airy's coal pit


(2010-01-29)   Simple Joint  (with or without a total spin)
Two isolated interacting rigid bodies rotating around a common axis.

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 angular momentum integrated over time remains constant:

( JA + MA a2 ) qA   +   ( JB + MB b2 ) qB

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 it at point  O,  as depicted here.  The  M  and  J  symbols respectively denote the masses and the proper moments of inertia  (around the direction of the common axis of rotation).  The quantities a and b are the distances from O to A and B, respectively.  The angles  q  are the  inclinations  with respect to a fixed direction  Ox.

 Come back later, we're
 still working on this one...

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