dynamics18lecture kinetics of rigid bodies.ppt

VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Tenth Edition
CHAPTER
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
18 Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. Self
California Polytechnic State University
Kinetics of Rigid Bodies in
Three Dimensions

Vector Mechanics for Engineers: Dynamics
Tenth
Edition
Contents
18 - 2
Introduction
Rigid Body Angular Momentum in
Three Dimensions
Principle of Impulse and
Momentum
Kinetic Energy
Sample Problem 18.1
Sample Problem 18.2
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
Sample Problem 18.3
Motion of a Gyroscope. Eulerian
Angles
Steady Precession of a Gyroscope
Motion of an Axisymmetrical Body
Under No Force

Tenth
Edition
2 - 3
Three dimensional analyses are needed to determine the forces
and moments on the gimbals of gyroscopes, the rotors of
amusement park rides, and the housings of wind turbines.

Tenth
Edition
Introduction
18 - 4
G
G H
M
a
m
F









• The fundamental relations developed for
the plane motion of rigid bodies may also
be applied to the general motion of three
dimensional bodies.



I
HG 
• The relation which was used
to determine the angular momentum of a
rigid slab is not valid for general three
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.

Tenth
Edition
Rigid Body Angular Momentum in Three Dimensions
18 - 5
• Angular momentum of a body about its mass center,
   
 












n
i
i
i
i
n
i
i
i
i
G m
r
r
m
v
r
H
1
1
Δ








• The x component of the angular momentum,
   
 
   
 
  























n
i
i
i
i
z
n
i
i
i
i
y
n
i
i
i
i
x
n
i
i
i
x
i
z
i
i
y
i
x
i
n
i
i
y
i
i
z
i
i
x
m
x
z
m
y
x
m
z
y
m
z
x
z
x
y
y
m
r
z
r
y
H
1
1
1
2
2
1
1
Δ
Δ
Δ
Δ
Δ













  dm
zx
dm
xy
dm
z
y
H z
y
x
x 

 


 

 2
2
z
xz
y
xy
x
x I
I
I 

 



z
z
y
zy
x
zx
z
z
yz
y
y
x
yx
y
I
I
I
H
I
I
I
H















Tenth
Edition
18 - 6
z
z
y
zy
x
zx
z
z
yz
y
y
x
yx
y
z
xz
y
xy
x
x
x
I
I
I
H
I
I
I
H
I
I
I
H





















• Transformation of into is characterized
by the inertia tensor for the body,


G
H




















z
zy
zx
yz
y
yx
xz
xy
x
I
I
I
I
I
I
I
I
I
• With respect to the principal axes of inertia,













z
y
x
I
I
I
0
0
0
0
0
0
z
z
z
y
y
y
x
x
x I
H
I
H
I
H 







 

 


• 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.
G
H






Tenth
Edition
18 - 7
• The momenta of the particles of a rigid body can
be reduced to:
v
m
L



 momentum
linear
G
HG about
momentum
angular


z
z
y
zy
x
zx
z
z
yz
y
y
x
yx
y
z
xz
y
xy
x
x
x
I
I
I
H
I
I
I
H
I
I
I
H





















• The angular momentum about any other given
point O is
G
O H
v
m
r
H








Tenth
Edition
18 - 8
• The angular momentum of a body constrained to
rotate about a fixed point may be calculated from
G
O H
v
m
r
H







z
z
y
zy
x
zx
z
z
yz
y
y
x
yx
y
z
xz
y
xy
x
x
x
I
I
I
H
I
I
I
H
I
I
I
H





















• Or, the angular momentum may be computed
directly from the moments and products of inertia
with respect to the Oxyz frame.
 
 
 









n
i
i
i
i
n
i
i
i
O
m
r
r
m
v
r
H
1
1
Δ
Δ








Tenth
Edition
Principle of Impulse and Momentum
18 - 9
• 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.

Tenth
Edition
Concept Question
2 - 10
At the instant shown, the
disk shown rotates with an
angular velocity 2 with
respect to arm ABC, which
rotates around the y axis as
shown (1). Determine the
directions of the angular
momentum of the disc
about point A (choose all
that are correct).
+x +y +z
-x -y -z

Tenth
Edition
Concept Question
2 - 11
A homogeneous disk of
mass m and radius r is
mounted on the vertical
shaft AB as shown
Determine the directions of
the angular momentum of
the disc about the mass
center G (choose all that
are correct).
+x +y +z
-x -y -z

Tenth
Edition
Kinetic Energy
18 - 12
• Kinetic energy of particles forming rigid body,
)
2
2
2
(
Δ
Δ
2
2
2
2
1
2
2
1
1
2
2
1
2
2
1
1
2
2
1
2
2
1
x
z
zx
z
y
yz
y
x
xy
z
z
y
y
x
x
n
i
i
i
n
i
i
i
I
I
I
I
I
I
v
m
m
r
v
m
v
m
v
m
T






























• If the axes correspond instantaneously with the
principle axes,
)
( 2
2
2
2
1
2
2
1
z
z
y
y
x
x I
I
I
v
m
T 




 


 


• 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.

Tenth
Edition
Kinetic Energy
18 - 13
• Kinetic energy of a rigid body with a fixed point,
)
2
2
2
( 2
2
2
2
1
x
z
zx
z
y
yz
y
x
xy
z
z
y
y
x
x
I
I
I
I
I
I
T















• If the axes Oxyz correspond instantaneously with
the principle axes Ox’y’z’,
)
( 2
2
2
2
1
z
z
y
y
x
x I
I
I
T 




 

 



Tenth
Edition
Concept Question
2 - 14
2
2
1
2
z
I 
At the instant shown, the disk
shown rotates with an angular
velocity 2 with respect to
arm ABC, which rotates
around the y axis as shown
(1). What terms will
contribute to the kinetic
energy of the disk (choose all
that are correct)?
2
1
1
2
y
I 
2
1
1
2
x
I 
2 2
1
1
2
mL 
2
1
1
2
z
I 

Tenth
Edition
Sample Problem 18.1
18 - 15
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.
SOLUTION:
• 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.

Tenth
Edition
Sample Problem 18.1
18 - 16
SOLUTION:
• 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.
k
v
i
v
v z
x




 j
i y
x





 

Since the x, y, and z axes are principal axes of inertia,
j
ma
i
mb
j
I
i
I
H y
x
y
y
x
x
G








 2
12
1
2
12
1 




Tenth
Edition
Sample Problem 18.1
18 - 17
• 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.
x
mv

0
0

x
v
z
v
m
t
F 
 Δ
m
t
F
vz Δ


 k
m
t
F
v


Δ


x
x
mb
H
t
bF

2
12
1
2
1 Δ


mb
t
F
x Δ
6


y
y
ma
H
t
aF

2
12
1
2
1 Δ



 
ma
t
F
y Δ
6



 
j
b
i
a
mab
t
F 




Δ
6


Tenth
Edition
Sample Problem 18.1
18 - 18
 k
m
t
F
v


Δ

 
j
b
i
a
mab
t
F 




Δ
6

j
ma
i
mb
H y
x
G




 2
12
1
2
12
1 


Tenth
Edition
Sample Problem 18.2
18 - 19
A homogeneous disk of mass m is
mounted on an axle OG of negligible
mass. The disk rotates counter-
clockwise at the rate 1 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.
SOLUTION:
• 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.

Tenth
Edition
Sample Problem 18.2
18 - 20
SOLUTION:
• 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.
j
i



2
1 

 

Noting that the velocity at C is zero,
   
 
L
r
k
r
L
j
r
i
L
j
i
r
v C
C
1
2
1
2
2
1
0
0

























 j
L
r
i



1
1 

 


Tenth
Edition
Sample Problem 18.2
18 - 21
• Compute the angular momentum of the disk using
principle axes of inertia and noting that O is a fixed point.
 j
L
r
i



1
1 

 

k
I
j
I
i
I
H z
z
y
y
x
x
O






 


 
  
  0
0
2
4
1
2
1
2
4
1
2
1
2
2
1










mr
mL
I
H
L
r
mr
mL
I
H
mr
I
H
z
z
z
y
y
y
x
x
x





  j
L
r
r
L
m
i
mr
HO



1
2
4
1
2
1
2
2
1 
 


• The kinetic energy is computed from the angular velocity
and moments of inertia.
 
  
 
2
1
2
4
1
2
2
1
2
2
1
2
2
2
2
1
L
r
r
L
m
mr
I
I
I
T z
z
y
y
x
x












2
1
2
2
2
8
1 6 










L
r
mr
T

Tenth
Edition
Sample Problem 18.2
18 - 22
 j
L
r
i



1
1 

 

• The vector and couple at G are also computed from the
k
mr
v
m


1


 j
L
r
mr
i
mr
k
I
j
I
i
I
H z
z
y
y
x
x
G
















 


2
4
1
1
2
2
1







 j
L
r
i
mr
HG



2
1
2
2
1 

Tenth
Edition
Motion of a Rigid Body in Three Dimensions
18 - 23
G
H
M
a
m
F









• Angular momentum and its rate of change are
taken with respect to centroidal axes GX’Y’Z’ of
fixed orientation.
• Convenient to use body fixed axes Gxyz where
moments and products of inertia are not time
dependent.
• Transformation of into is independent of
the system of coordinate axes.


G
H

• Define rate of change of change of with
respect to the rotating frame,
G
H

  k
H
j
H
i
H
H z
y
x
Gxyz
G











Then,
  













 G
Gxyz
G
G H
H
H

Tenth
Edition
Euler’s Eqs of Motion & D’Alembert’s Principle
18 - 24
  G
Gxyz
G
G H
H
M








 
• With and Gxyz chosen to correspond
to the principal axes of inertia,





 
 
  y
x
y
x
z
z
z
x
z
x
z
y
y
y
z
y
z
y
x
x
x
I
I
I
M
I
I
I
M
I
I
I
M
























Euler’s Equations:
• 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, .
and G
H
a
m 



Tenth
Edition
Group Problem Solving
18 - 25
Two L-shaped arms, each weighing
4 lb, are welded at the third points
of the 2-ft shaft AB. Knowing that
shaft AB rotates at the constant rate
= 240 rpm, determine (a) the
angular momentum of the body
about B, and b) its kinetic energy.
SOLUTION:
• The part rotates about the axis AB.
Determine the mass moment of inertia
matrix for the part.
• Compute the angular momentum of the
disk using the moments inertia.
• Compute the kinetic energy from the

Tenth
Edition
2 - 26
2
4
4 lb, 0.12422 lb s /ft,
32.2
8 in. 0.66667 ft
W m
a
   
 
(2 )(240)
8 rad/s, 0, 0
60
z x y

    
    
Given:
Find: HB, T
There is only rotation about the
z-axis, what relationship(s) can
you use?
( )
B x xz
H I 
  ( )
B y yz
H I 
  ( )
B z z
H I 

Split the part into four different segments, then determine
Ixz, Iyz, and Iz. What is the mass of each segment?
•Mass of each of the four segments 2
0.06211 lb s /ft
2
m
m   

Tenth
Edition
2 - 27
Fill in the table below to help
you determine Ixz and Iyz. 1
2
3 4
Ixz Iyz
1
2
3
4
S
( )( )
m a a
 
( 0.5 )( )
m a a
 
( )(2 )
m a a

a=
(0.5 )(2 )
m a a

(0.5 )( )
m a a

(0)( )
m a

(0)(2 )
m a

(0.5 )(2 )
m a a

   
2
2 2
2 1
1 4 2
1
12
z z
I I m a m a a
 
 
 
   
 
 
2
1.5m a
 2
1.5m a

Determine Iz by using the parallel
axis theorem – do parts 1 and 4 first
Parts 2 and 3 are also
equal to one another
2
2 3
1
3
z z
I I m a
  
 

Tenth
Edition
2 - 28
Calculate Hx
2
2
3
( )
2
3
(0.06211)(0.66667) (8 )
2
1.04067 lb s ft
B x xz
H I m a
 


   
 
   
2
2
3
( )
2
3
(0.06211)(0.66667) (8 )
2
1.04067 lb s ft
B y yz
H I m a
 


   
 
   
2
2
10
( )
3
10
(0.06211)(0.66667) (8 )
3
2.3126 lb s ft
B z z
H I m a
 


 

  
(1.041 lb ft s) (1.041 lb ft s) (2.31 lb ft s)
B          
H i j k
Calculate Hy
Calculate Hz
Total Vector

Tenth
Edition
2 - 29
2 2 2 2
1 1 1 1
2 2 2 2
x x y y z z xy x y xz x z yz y z z
T I I I I I I I
         
      
2 2 2 2
1 10 1 10
(0.06211)(0.66667) (8 )
2 3 2 3
T m a  
   

 
   
   
29.1ft lb
T  
Calculate the kinetic energy T

Tenth
Edition
2 - 30
Retracting the landing gear while the wheels are still spinning
can result in unforeseen moments being applied to the gear.
When turning a motorcycle, you must “steer” in the opposite
direction as you lean into the turn.

Tenth
Edition
Motion About a Fixed Point or a Fixed Axis
18 - 31
• For a rigid body rotation around a fixed point,
  O
Oxyz
O
O
O
H
H
H
M












• For a rigid body rotation around a fixed axis,


 z
z
yz
y
xz
x I
H
I
H
I
H 





 
 
 
    2






i
I
j
I
k
I
j
I
i
I
k
I
j
I
i
I
k
k
I
j
I
i
I
H
H
M
yz
xz
z
yz
xz
z
yz
xz
z
yz
xz
O
Oxyz
O
O











































z
z
xz
yz
y
yz
xz
x
I
M
I
I
M
I
I
M










2
2

Tenth
Edition
Rotation About a Fixed Axis
18 - 32





z
z
xz
yz
y
yz
xz
x
I
M
I
I
M
I
I
M










2
2
• For a rigid body rotation around a fixed axis,
• If symmetrical with respect to the xy plane,

z
z
y
x I
M
M
M 

 

 0
0
• If not symmetrical, the sum of external moments
will not be zero, even if  = 0,
0
2
2


 

 z
xz
y
yz
x M
I
M
I
M 

• A rotating shaft requires both static and
dynamic balancing to avoid excessive
vibration and bearing reactions.
 
0


 
0



Tenth
Edition
Concept Question
2 - 33
The device at the right
rotates about the x axis
with a non-constant
angular velocity. Which
of the following is true
(choose one)?
a) You can use Euler’s equations for the provided x, y, z axes
b) The only non-zero moment will be about the x axis
c) At the instant shown, there will be a non-zero y-axis bearing force
d) The mass moment of inertia Ixx is zero

Tenth
Edition
Sample Problem 18.3
18 - 34
Rod AB with weight W = 40 lb is
pinned at A to a vertical axle which
rotates with constant angular velocity
 = 15 rad/s. The rod position is
maintained by a horizontal wire BC.
Determine the tension in the wire and
the reaction at A.
• 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.
SOLUTION:
• Evaluate the system of effective forces
by reducing them to a vector
attached at G and couple
a
m

.
G
H


Tenth
Edition
Sample Problem 18.3
18 - 35
SOLUTION:
• Evaluate the system of effective forces by reducing them
to a vector attached at G and couple
a
m

.
G
H

 
 I
I
L
I
r
a
a n





2
2
2
1
2
s
ft
450
cos






 


   I
g
a
m


lb
559
450
40




k
I
j
I
i
I
H z
z
y
y
x
x
G






 


0
sin
cos
0 2
2
1
2
2
1







z
y
x
z
y
x mL
I
I
mL
I







i
mL
HG



 cos
2
12
1


 
   
 k
k
mL
i
mL
j
i
H
H
H G
Gxyz
G
G











ft
lb
645
cos
sin
cos
sin
cos
0
2
2
12
1
2
12
1






















Tenth
Edition
Sample Problem 18.3
18 - 36
• 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.
 eff
A
A M
M 
 


     
   K
K
T
K
I
J
J
I
I
T
J









645
1934
80
93
.
6
645
559
46
.
3
40
2
93
.
6












lb
384

T
 eff
F
F 
 


I
J
I
K
A
J
A
I
A Z
Y
X






559
40
384 





   J
I
A



lb
40
lb
175 



Tenth
Edition
18 - 37
A four-bladed airplane propeller has a
mass of 160 kg and a radius of
gyration of 800 mm. Knowing that the
propeller rotates at 1600 rpm as the
airplane is traveling in a circular path
of 600-m radius at 540 km/h,
determine the magnitude of the couple
exerted by the propeller on its shaft
due to the rotation of the airplane
• Determine the mass moment of inertia for
the propeller
• Calculate the angular momentum of the
propeller about its CG
SOLUTION:
• Determine the overall angular velocity of
the aircraft propeller
• Calculate the time rate of change of the
angular momentum of the propeller about
its CG
• Calculate the moment that must be
applied to the propeller and the resulting
moment that is applied on the shaft

Tenth
Edition
2 - 38
Establish axes for the rotations of the
aircraft propeller
Determine the x-component of its
angular velocity
540 km/h 150 m/s
v  
2 rad
1600 rpm
60 s
167.55 rad/s
x


 
  
 

150 m/s
0.25 rad/s
600 m
y
v


  
•
Determine the y-component of its
angular velocity
Determine the mass moment of inertia
2 2 2
(160 kg)(0.8 m) 102.4 kg m
x
I mk
   

Tenth
Edition
2 - 39
Angular momentum about G equation:
Determine the couple that is
exerted on the shaft by the
propeller
Determine the moment that must
be applied to the propeller shaft
G x x y y
I I
 
 
H i j
( ) 0 ( )
G G Gxyz G y x x y y
I I
  
      
H H Ω H j i j
2
(102.4 kg m )(167.55 rad/s)(0.25 rad/s)
G x x y
I  
    
H k k
(4289 N m) = (4.29 kN m)
G     
H k k
Time rate of change of angular
momentum about G :
(4.29 kN m)
G
   
M H k
(4.29 kN m)
on shaft    
M M k

Tenth
Edition
Concept Question
2 - 40
The airplane propeller has a
constant angular velocity +x.
The plane begins to pitch up at
a constant angular velocity x
y
z
The propeller has zero
angular acceleration
TRUE FALSE
Which direction will the pilot have to steer to counteract the
moment applied to the propeller shaft?
a) +x b) +y c) +z
d) -y e) -z


Tenth
Edition
2 - 41
Gyroscopes are used in the navigation system of the Hubble
telescope, and can also be used as sensors (such as in the
Segway PT). Modern gyroscopes can also be MEMS (Micro
Electro-Mechanical System) devices, or based on fiber optic
technology.

Tenth
Edition
Motion of a Gyroscope. Eulerian Angles
18 - 42
• 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 j about AA’,
b) rotation of inner gimbal through q about
c) rotation of the rotor through y about CC’.
• j, q, and y are called the Eulerian Angles and
spin
of
rate
nutation
of
rate
precession
of
rate




q





Tenth
Edition
Motion of a Gyroscope. Eulerian Angles
18 - 43
• The angular velocity of the gyroscope,
k
j
K








q

 


 k
j
i
j
i
K











q


q
q


q
q
cos
sin
cos
sin
with








• Equation of motion,
  O
Oxyz
O
O H
H
M








 
 
j
K
k
I
j
I
i
I
HO













q


q


q
q









 cos
sin
   
   
 
q


q


q

q
q

q
q


q
q

q
q

cos
cos
sin
cos
sin
cos
cos
2
sin
2































dt
d
I
M
I
I
M
I
I
M
z
y
x

Tenth
Edition
Steady Precession of a Gyroscope
18 - 44
Steady precession,
constant
are
,
, y

q 

k
i
k
I
i
I
H
k
i
z
O
z













q

q



q


q


cos
sin
sin
sin










  j
I
I
H
M
z
O
O






q

q



sin
cos






Couple is applied about an axis
perpendicular to the precession
and spin axes
When the precession and spin
axis are at a right angle,
j
I
M O






q




90
Gyroscope will precess about an
axis perpendicular to both the
spin axis and couple axis.

Tenth
Edition
Motion of an Axisymmetrical Body Under No Force
18 - 45
• Consider motion about its mass center of an
axisymmetrical body under no force but its own
weight, e.g., projectiles, satellites, and space craft.
constant
0 
 G
G H
H



• 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.
G
H

x
G
x I
H
H 
q 


 sin
I
HG
x



q

sin
y
y I
H 


 0 0

y

z
G
z I
H
H 
q 
 cos
I
HG
z
q

cos

• q = constant and body is in steady precession.
• Note: q



tan
tan
I
I
z
x





Tenth
Edition
18 - 46
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,
aligned
are
and
0
G
x
x
H
H



 

and body keeps spinning about its axis of
symmetry.
• Body is set to spin about its transverse axis,
aligned
are
and
0
G
z
z
H
H



 

and body keeps spinning about the given
transverse axis.

Tenth
Edition
18 - 47
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.  < q and the
vector  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.  > q and the
vector  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.

Tenth
Edition
Concept Question
2 - 48
The rotor of the gyroscope shown
to the right rotates with a constant
angular velocity. If you hang a
weight on the gimbal as shown,
what will happen?
a) It will precess around to the right
b) Nothing – it will be in equilibrium
c) It will precess around to the left
d) It will tilt down so the rotor is horizontal


dynamics18lecture kinetics of rigid bodies.ppt

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