Chapter 18. Kinetics of Rigid Bodies in Three Dimensions
Introduction
Rigid Body Angular Momentum in Three Dimensions
Principle of Impulse and Momentum
Kinetic Energy
Motion of a Rigid Body in Three Dimensions
Euler’s Equations of Motion and D’Alembert’s Principle
Motion About a Fixed Point or a Fixed Axis
Motion of a Gyroscope. Eulerian Angles
Steady Precession of a Gyroscope
Motion of an Axisymmetrical Body Under No Force
Three dimensional analyses are needed to determine the forces and moments
on the gimbals of gyroscopes, the joints of robotic welders, and the supports
of radio telescopes.
Introduction
• The fundamental relations developed for the
plane motion of rigid bodies may also be applied
to the general motion of three dimensional bodies.
• The relation which was used
to determine the angular momentum of a
rigid slab is not valid for general three
18.1 dimensional bodies and motion.
The current chapter is concerned with
evaluation of the angular momentum and
its rate of change for three dimensional
motion and application to effective forces,
the impulse-momentum and the work-
energy principles.
ω
IHG =
GG HM
amF
=
=
∑∑
18.1A Angular Momentum of a Rigid Body in Three Dimensions
Angular momentum of a body about its mass center,
The x component of the angular momentum,
( ) ( )[ ]∑∑==
′××′=×′=n
iiii
n
iiiiG mrrmvrH
11Δ
ω∆
( ) ( )[ ]
( ) ( )[ ]
( ) ∑∑∑
∑
∑
===
=
=
−−+=
−−−=
′×−′×=
n
iiiiz
n
iiiiy
n
iiiix
n
iiixiziiyixi
n
iiyiiziix
mxzmyxmzy
mzxzxyy
mrzryH
111
22
1
1
ΔΔΔ
Δ
Δ
ωωω
ωωωω
ωω
( )2 2x x y zH y z dm xy dm zx dmω ω ω= + − −∫ ∫ ∫
x x xy y xz zI I Iω ω ω= + − −
y yx x y y yz z
z zx x zy y z z
H I I I
H I I I
ω ω ω
ω ω ω
= − + −
= − − +
Transformation of into is characterized by the inertia tensor for the body,
• With respect to the principal axes of inertia,
The angular momentum of a rigid body and its
angular velocity have the same direction if, and only if,
is directed along a principal axis of inertia.
ω
GH
ω
ω
+−−−+−−−+
zzyzx
yzyyx
xzxyx
IIIIIIIII
GH
′
′
′
z
y
x
II
I
000000
x x x y y y z z zH I H I H Iω ω ω′ ′ ′ ′ ′ ′ ′ ′ ′= = =
The momenta of the particles of a rigid body can be reduced
to:
• The angular momentum about any other given point O is
vmL
== momentumlinear
GHG about momentumangular =
zzyzyxzxz
zyzyyxyxy
zxzyxyxxx
IIIH
IIIH
IIIH
ωωω
ωωω
ωωω
+−−=
−+−=
−−+=
GO HvmrH
+×=
The angular momentum of a body constrained to rotate about a
fixed point may be calculated from
• Or, the angular momentum may be
computed directly from the moments
and products of inertia with respect to the Oxyz frame.
( )
( )[ ]∑
∑
=
=
××=
×=
n
iiii
n
iiiO
mrr
mvrH
1
1
Δ
Δ
ω
GO HvmrH
+×=
x x x xy y xz z
y yx x y y yz z
z zx x zy y z z
H I I IH I I IH I I I
ω ω ω
ω ω ω
ω ω ω
= + − −
= − + −
= − − +
18.1B Principle of Impulse and Momentum
• The principle of impulse and momentum can be applied directly to the three-dimensional
motion of a rigid body,
Syst Momenta1 + Syst Ext Imp1-2 = Syst Momenta2
• The free-body diagram equation is used to develop component and moment equations.
• For bodies rotating about a fixed point, eliminate the impulse of the reactions at O by
writing equation for moments of momenta and impulses about O.
18.1C Kinetic Energy
Kinetic energy of particles forming rigid body,
• If the axes correspond instantaneously with the
principle axes,
• With these results, the principles of work and energy and conservation of energy may be
applied to the three-dimensional motion of a rigid body.
)22
2(
Δ
Δ
222212
21
1
2212
21
1
2212
21
xzzxzyyz
yxxyzzyyxx
n
iii
n
iii
II
IIIIvm
mrvm
vmvmT
ωωωω
ωωωωω
ω
−−
−+++=
′×+=
′+=
∑
∑
=
=
)( 222212
21
zzyyxx IIIvmT ′′′′′′ +++= ωωω
• Kinetic energy of a rigid body with a fixed point,
• If the axes Oxyz correspond instantaneously with the principle
axes Ox’y’z’,
)22
2( 22221
xzzxzyyz
yxxyzzyyxx
II
IIIIT
ωωωω
ωωωωω
−−
−++=
)( 22221
zzyyxx IIIT ′′′′′′ ++= ωωω
Sample Problem 18.1
Rectangular plate of mass m that is suspended from two
wires is hit at D in a direction perpendicular to the plate.
Immediately after the impact, determine a) the velocity of
the mass center G, and b) the angular velocity of the plate.
STRATEGY:
• Apply the principle of impulse and momentum. Since the initial momenta is zero, the
system of impulses must be equivalent to the final system of momenta.
• Assume that the supporting cables remain taut such that the vertical velocity and the
rotation about an axis normal to the plate is zero.
• Principle of impulse and momentum yields to two equations for linear momentum and
two equations for angular momentum.
• Solve for the two horizontal components of the linear and angular velocity vectors.
MODELING and ANALYSIS:
• Apply the principle of impulse and momentum. Since the initial momenta is zero, the
system of impulses must be equivalent to the final system of momenta.
• Assume that the supporting cables remain taut such that the vertical velocity and the
rotation about an axis normal to the plate is zero.
Since the x, y, and z axes are principal axes of inertia,
kvivv zx
+= ji yx
ωωω +=
jmaimbjIiIH yxyyxxG
ωωωω 21212
121 +=+=
• Principle of impulse and momentum yields two equations for linear momentum and two
equations for angular momentum.
• Solve for the two horizontal components of the linear and angular velocity vectors.
0 xmv=0xv =
Δ zF t mv− =Δzv F t m= −
12
2112
Δ x
x
bF t H
mb ω
=
=
12
2112
Δ y
y
aF t H
ma ω
− =
=
( )Δv F t m k= −
6 Δx F t mbω = ( )6 Δy F t maω = −
( )6 ΔF t ai b jmab
ω = +
( )kmtFv
Δ−
2 21 112 12G x yH mb i ma jω ω= +
( )jbiamab
tF
+=Δ6ω
REFLECT and THINK:
• Equating the y components
of the impulses and
momenta and their
moments about the z axis,
you can obtain two
additional equations that
yield TA = TB = ½W.
• This verifies that the wires
remain taut and that the
initial assumption was
correct. If the impulse were
at G, this would reduce to a
two-dimensional problem.
Sample Problem 18.2
A homogeneous disk of mass m is mounted on an axle
OG of negligible mass. The disk rotates counter-clockwise
at the rate w1 about OG.
Determine: a) the angular velocity of the disk, b) its
angular momentum about O, c) its kinetic energy, and d)
the vector and couple at G equivalent to the momenta of
the particles of the disk.
STRATEGY:
• The disk rotates about the vertical axis through O as well as about OG. Combine the
rotation components for the angular velocity of the disk.
• Compute the angular momentum of the disk using principle axes of inertia and noting
that O is a fixed point.
• The kinetic energy is computed from the angular velocity and moments of inertia.
• The vector and couple at G are also computed from the angular velocity and moments
of inertia.
MODELING and ANALYSIS:
• The disk rotates about the vertical axis through O as well as
about OG. Combine the rotation components for the angular
velocity of the disk.
Noting that the velocity at C is zero,
ji
21 ωωω +=
( ) ( )( )
LrkrL
jriLjirv CC
12
12
2100
ωωωω
ωωω
=−=
−×+=
=×=
( )1 1i r L jω ω ω= −
• Compute the angular momentum of the disk using principle
axes of inertia and noting that O is a fixed point.
• The kinetic energy is computed from the angular velocity and moments of inertia.
kIjIiIH zzyyxxO
ωωω ++=
( ) jLri
11 ωωω −=
( )( )( )( )
2112
2 2114
2 214 0 0
x x x
y y y
z z z
H I mr
H I mL mr r L
H I mL mr
ω ω
ω ω
ω
= =
= = + −
= = + =
( )( )2 2 21 11 12 4OH mr i m L r r L jω ω= − +
22 21
18 26 rT mrL
ω
= +
( )( )( )[ ]2
12
4122
12
21
22221
LrrLmmr
IIIT zzyyxx
ωω
ωωω
−++=
++=
The vector and couple at G are also computed from
the angular velocity and moments of inertia.
REFLECT and THINK:
• If the mass of the axle were not negligible and it was instead
modeled as a slender rod with a mass Maxle, it would also contribute
to the kinetic energy
Taxle = ½(1/3 MaxleL2) ω22 and the momenta
Haxle = ½(1/3 MaxleL2) ω2 of the system.
kmrvm
1ω=
( ) jLri
11 ωωω −=
( )2 21 112 4
G x x y y z zH I i I j I k
mr i mr r L j
ω ω ω
ω ω′ ′ ′= + +
= + −
2112 2G
rH mr i jL
ω = −
18.2 Motion of a Rigid Body in Three Dimensions
18.2A Rate of Change of Angular Momentum
Angular momentum and its rate of change are taken with
respect to centroidal axes GX’Y’Z’ of fixed orientation.
Transformation of into is independent of the
system of coordinate axes.
Convenient to use body fixed axes Gxyz where moments
and products of inertia are not time dependent.
• Define rate of change of change of with respect
to the rotating frame,
ω GH
GH
GHM
amF
=
=
∑∑
Then,
18.2B Euler’s Eqs of Motion
With and Gxyz chosen to correspond to the
principal axes of inertia,
Euler’s Equations:
ωΩ
=
( ) GGxyzGG HHM
×+=∑ Ω
( ) kHjHiHH zyxGxyzG
++=
( ) ωΩΩ
=×+= GGxyzGG HHH
• System of external forces and effective forces are equivalent for general
three dimensional motion.
• System of external forces are equivalent to the vector and couple,
( )( )( ) yxyxzzz
xzxzyyy
zyzyxxx
IIIM
IIIM
IIIM
ωωω
ωωω
ωωω
−−=
−−=
−−=
∑∑∑
. and GHam
18.2C,D Motion About a Fixed Point or a Fixed Axis
For a rigid body rotation around a fixed point,
For a rigid body rotation around a fixed axis,
x xzH I ω= − y yzH I ω= − z zH I ω=
( ) OOxyzO
OO
HH
HM
×+=
=∑Ω
( )( )
( )( ) ( ) 2
O O OOxyz
xz yz z
xz yz z
xz yz z xz yz
M H H
I i I j I k
k I i I j I k
I i I j I k I j I i
ω
ω
ω ω
α ω
= + ×
= − − +
+ × − − +
= − − + + − +
∑
2
2
x xz yz
y yz xz
z z
M I I
M I I
M I
α ω
α ω
α
= − +
= − +
=
∑∑∑
If symmetrical with respect to the xy plane,
• If not symmetrical, the sum of external moments will not be zero, even if α
= 0,
• A rotating shaft requires both static and dynamic balancing
to avoid excessive vibration and bearing reactions.
αzzyx IMMM === ∑∑∑ 00
022 === ∑∑∑ zxzyyzx MIMIM ωω
( )0=ω ( )0≠ω
Sample Problem 18.3
Rod AB with mass m = 20 kg is pinned at A to a vertical axle which
rotates with constant angular velocity ω = 15 rad/s. The rod
position is maintained by a horizontal wire BC.
Determine the tension in the wire and the reaction at A.
STRATEGY:
• Evaluate the system of effective forces by reducing them to a vector attached at G
and couple
• Expressing that the system of external forces is equivalent to the system of effective
forces, write vector expressions for the sum of moments about A and the summation of
forces.
• Solve for the wire tension and the reactions at A.
am
.GH
MODELING and ANALYSIS:
• Evaluate the system of effective forces by reducing them to
a vector attached at G and couple
am .GH
G x x y y z zH I i I j I kω ω ω= + +
2 21 112 120
cos sin 0x y z
x y z
I mL I I mLω ω β ω ω β ω
= = == − = =21
12 cosGH mL iω β= −
Expressing that the system of external forces is equivalent to
the system of effective forces, write vector expressions for
the sum of moments about A and the summation of forces.
( )effAA MM ∑∑ =
( ) ( ) ( )( ) ( )
1.732 0.5 196.2 0.866 2250 649.5
1.732 98.1 1948.5 649.5
J TI I J J I K
T K K
× − + × − = × − +
− = +
1557 NT =( )
effF F=∑ ∑
1613 196.2 2250X Y ZA I A J A K I J I+ + − − = −
( ) ( )697 N 196.2 NA I J= − +
•
Gyroscopes are used in the navigation system of the Hubble telescope, and can
also be used as sensors . Modern gyroscopes can also be MEMS (Micro Electro-
Mechanical System) devices, or based on fiber optic technology.
* Motion of a Gyroscope. Eulerian Angles *
A gyroscope consists of a rotor with its mass center fixed in space
but which can spin freely about its geometric axis and assume any
orientation.
From a reference position with gimbals and a reference diameter of
the rotor aligned, the gyroscope may be brought to any orientation
through a succession of three steps:
a) rotation of outer gimbal through φ about AA’,
b) rotation of inner gimbal through θ about BB’,
c) rotation of the rotor through ψ about CC’.
• φ θ ψ : Eulerian Angles
spin of rate nutation of rate precession of rate
=
=
=
Ψθφ
The angular velocity of the gyroscope,
Equation of motion,
kjK
Ψθφω ++=
( ) OOxyzOO HHM
×+=∑ Ω
( )with sin cos
sin cos
K i k
i j k
θ θ
ω φ θ θ φ θ
= − +
= − + + Ψ +
( )sin cosOH I i I j I k
K j
φ θ θ φ θ
φ θ
′ ′= − + + Ψ +
Ω = +
( ) ( )( ) ( )( )
2
sin 2 cos cos
sin cos sin cos
cos
x
y
z
M I I
M I I
dM Idt
φ θ θφ θ θ φ θ
θ φ θ θ φ θ φ θ
φ θ
′= − + + Ψ +
′= − + Ψ +
= Ψ +
∑∑
∑
Steady Precession of a Gyroscope
Steady precession,
Couple is applied about an axis perpendicular to the precession and
spin axes
constant are ,, ψφθ
ki
kIiIH
ki
zO
z
θφθφΩ
ωθφ
ωθφω
cossin
sin
sin
+−=
+′−=
+−=
( ) jIIHM
z
OO
θφθφω
Ω
sincos′−=
×=∑
When the precession and spin axis are at a right angle,
Gyroscope will precess about an axis perpendicular to both the
spin axis and couple axis.
jIM O
φΨ
θ
=
°=
∑90
Motion of an Axisymmetrical Body Under No Force
Consider motion about its mass center of an axisymmetrical
body under no force but its own weight, e.g., projectiles,
satellites, and space craft.
Define the Z axis to be aligned with and z in a rotating
axes system along the axis of symmetry. The x axis is chosen
to lie in the Zz plane.
• θ = constant and body is in steady precession.
• Note:
constant 0 == GG HH
GH
xGx IHH ωθ ′=−= sin IHG
x ′−=
θω sin
θγωω tantan
II
z
x′
==−
0y yH I ω′= = 0yω =cosz G zH H Iθ ω= = cosG
zH
Iθω =
• Two cases of motion of an axisymmetrical body which under
no force which involve no precession:
• Body set to spin about its axis of symmetry,
and body keeps spinning
about its axis of symmetry.
• Body is set to spin about its transverse axis,
and body keeps spinning about the given transverse axis.
aligned are and0
G
zz
HH
ω
ω ==
aligned are and
0
G
xx
HH
ω
ω ==
The motion of a body about a fixed point (or its mass center) can be
represented by the motion of a body cone rolling on a space cone.
In the case of steady precession the two cones are circular.
• I < I’. Case of an elongated body. γ θ< and the vector ω
lies inside the angle ZGz. The space cone and body cone are tangent
externally; the spin and precession are both counterclockwise from
the positive z axis. The precession is said to be direct.
• I > I’. Case of a flattened body. γ θ< and the vector ω lies outside the
angle ZGz. The space cone is inside the body cone; the spin and precession
have opposite senses. The precession is said to be retrograde.