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 derivativeU'
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 =
ai + bj + ck ,
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'. k ) i +
( k'. i ) j +
( i'. j ) 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 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 -
W^{2}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 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, rr* is a square matrix, while
r* r is merely the scalar
r^{2 }).
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!
(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:
E_{R} = ½ W . L
= ½ W* J W
If the rotation vector W
has coordinates
(W_{1 },
W_{2 },
W_{3 }) along the principal axes
or inertia, then the above is also equal to:
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 r^{2 }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
( x^{2} + y^{2} + z^{2} ) dm
Moment of inertia about the axis Oz :
ò
( x^{2} + y^{2} ) dm
Moment of inertia about the plane xOy :
ò
z^{2} 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 MR^{2}.
So, the moment of inertia of such a uniform shell about an axis going through its
center is:
J = ^{2}/_{3} M R^{2}
(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) r^{2} dr
and
J = ^{2}/_{3}
ò 4 p r(r) r^{4} dr
For an homogeneous sphere of radius R,
r is constant and we obtain:
J = ^{2}/_{5} M R^{2}
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).
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} M c^{2}
J_{yOz} = ^{1}/_{5} M a^{2}
J_{zOx} = ^{1}/_{5} M b^{2}
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 }
(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 = J_{2}/ 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 (J_{2 }) defined by:
J_{2} = [ C - ½ (A+B) ] / M a^{ 2}
Under its alternate name of dynamic form factor, J_{2}
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
J_{2} = 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:
(2007-09-13) Perpendicular Axis Theorem
Momenta of inertia of a lamina about an axis perpendicular to it.
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:
J_{z} =
J_{x} + J_{y}
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
( x^{2} + y^{2} ) dm
=
ò
y^{2} dm
+
ò
x^{2} 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 R^{2}
The theorem gives the moment of that thin disk
about an horizontal axis:
J_{x} = J_{y} = ¼ M R^{2}.
(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 + h^{2}/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 + h^{2}/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 h^{2} / 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^{2} times its mass.
Therefore, the moment of inertia of the cylinder about any
horizontal axis is:
J_{x} =
J_{y} = M ( R^{2} / 4
+ h^{2} / 12 )
The thin-disk formula holds for the vertical axis:
J_{z} = ½ M R^{2} Isotropic inertia
( J_{x} = J_{y} = J_{z })
is achieved when h = R Ö3.
(2011-04-19) Inertia of a right cone
(or conical frustum)
Moment of inertia of a cone about its axis of symmetry.
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:
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 R^{2}
= 5.898 g cm^{2}
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 R^{2}
= 7.074 g cm^{2}
All told, the moment of inertia of a ping-pong ball increased 20% in 2000.
(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 + ML^{2}.
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 + ML^{2 }) q'.
(J + ML^{2 }) 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 + 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.
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.
(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).
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.