PROBLEM 18.132
A homogeneous rectangular plate of mass m and sides c and 2c
is held at A and B by a fork–ended shaft of negligible mass which
is supported by a bearing at C. The plate is free to rotate about
AB, and the shaft is free to rotate about a horizontal axis through
C. Initially the plate lies in the plane of the fork
0
( 0)
θ
=
and the
shaft has an angular velocity
0
10
φ
=
rad/s. If the plate is slightly
disturbed, determine for the ensuring motion (a) the minimum
value of
,
φ
(b) the maximum value of
.
θ
22 2 2
24
1[4 (1 4 sin )]
24
mc
θφ θ
= ++
PROBLEM 18.132 (Continued)
Using the initial conditions, including
0
0,
θ
=
Eq. (3) yields
22222
2
max
2
max
1
4 100 1 5
20
θ
θ
= −
=
max
4.47 rad/s
θ
=
PROBLEM 18.133
A homogeneous square plate of mass m and side c is held at points A
and B by a frame of negligible mass which is supported by bearings at
points C and D. The plate is free to rotate about AB, and the frame is
free to rotate about the vertical CD. Knowing that, initially,
0
45 ,
θ
= °
00,
θ
=
and
0
8
φ
=
rad/s, determine for the ensuing motion (a) the
range of values of
,
θ
(b) the minimum value of
,
φ
(c) the maximum
value of
.
θ
12yy
PROBLEM 18.133 (Continued)
22
2 1 2 0 10
CC C C
θ φ θφ
=−=+
22 2
PROBLEM 18.134
A homogeneous square plate of mass m and side c is held at points A
and B by a frame of negligible mass which is supported by bearings at
points C and D. The plate is free to rotate about AB, and the frame is
free to rotate about the vertical CD. Initially the plate lies in the plane
of the frame
( )
090
θ
= °
and the frame has an angular velocity
08
φ
=
rad/s. If the plate is slightly disturbed, determine for the ensuing
motion (a) the minimum value of
,
φ
(b) the maximum value of
.
θ
12yy
PROBLEM 18.134
(Continued)
22
2 1 2 0 10
CC C C
θ φ θφ
=−=+
22 2
( )
( )
2
12
8
1 cos 90 8 8 rad/s 1 cos
C
φθ
=+ °= =
+
(a)
min 2
84
1 cos
φθ
= =
+
min
4.00 rad/s
φ
=
( )( )
2
2
0 8 8 64 rad/s=+=C
2
21
CC
θφ
= −
(b)
( )
( )( ) ( )
22
max 2 1 min
64 8 4 32 rad/sCC
θφ
=−=−=
max
5.66 rad/s
θ
=
PROBLEM 18.135
A homogeneous disk of radius 9 in. is welded to a rod AG of length 18 in.
and of negligible weight which is connected by a clevis to a vertical shaft
AB. The rod and disk can rotate freely about a horizontal axis AC, and
shaft AB can rotate freely about a vertical axis. Initially rod AG is
horizontal
( )
0
90
θ
= °
and has no angular velocity about AC. Knowing
that the maximum value
m
φ
of the angular velocity of shaft AB in the
ensuing motion is twice its initial value
0
,
φ
determine (a) the minimum
value of
,
θ
(b) the initial angular velocity
0
φ
of shaft AB.
PROBLEM 18.135 (Continued)
Conservation of energy:
,TV E+=
where E is a constant.
2 2 2 2 22
17 1
PROBLEM 18.136
A homogeneous disk of radius 9 in. is welded to a rod AG of length 18 in.
and of negligible weight which is connected by a clevis to a vertical shaft
AB. The rod and disk can rotate freely about a horizontal axis AC, and
shaft AB can rotate freely about a vertical axis. Initially rod AG is
horizontal
( )
0
90
θ
= °
and has no angular velocity about AC. Knowing
that the smallest value of
θ
in the ensuing motion is
30 ,°
determine
(a) the initial angular velocity of shaft AB, (b) its maximum angular
velocity.
PROBLEM 18.136 (Continued)
Conservation of energy:
,TV E+=
where E is a constant.
2 2 2 2 22
17 1
PROBLEM 18.137*
The top shown is supported at the fixed Point O. Denoting by
,,
φθ
and
ψ
the Eulerian angles defining the position of the top with respect to a fixed
frame of reference, consider the general motion of the top in which all
Eulerian angles vary.
(a) Observing that
0
Z
MΣ=
and
0,
z
MΣ=
and denoting by I and
,I′
respectively, the moments of inertia of the top about its axis of
symmetry and about a transverse axis through O, derive the two first–
order differential equations of motion
2
sin ( cos ) cosII
φ θ ψφ θ θα
′++ =
( cos )I
ψφ θ β
+=
where
α
and
β
are constants depending upon the initial conditions. These
equations express that the angular momentum of the top is conserved
about both the Z and z axes, i.e., that the rectangular component of
O
H
along each of these axes is constant.
(b) Use Eqs. (1) and (2) to show that the rectangular component
z
ω
of the
angular velocity of the top is constant and that the rate of precession
φ
depends upon the value of the angle of nutation
.
θ
PROBLEM 18.137* (Continued)
()
sin ( sin ) ( cos )
O O O Oxyz O
dd
mgc I I I
θ φ θ θ ψφ θ
= = +×
′′
=− ++ +
MH H H
j ij k
Ω
PROBLEM 18.138*
(a) Applying the principle of conservation of energy, derive a third differential
equation for the general motion of the top of Problem 18.137.
(b) Eliminating the derivatives
φ
and
ψ
from the equation obtained and
from the two equations of Problem 18.139, show that the rate of nutation
θ
is defined by the differential equation
2
( ),f
θθ
=
where
2
2
1 cos
( ) 2 2 cos sin
f E mgc
II I
β αβ θ
θθ
θ
−
= −− −
′′
(c) Further show, by introducing the auxiliary variable
cos ,x
θ
=
that the
maximum and minimum values of
θ
can be obtained by solving for x
the cubic equation
2
22
1
2 2 (1 ) ( ) 0E mgcx x x
II
βαβ
−− −− − =
′
zz
II
ω
=
(B)
PROBLEM 18.138* (Continued)
Equation (1) of Problem 18.137 gives
2
sin cosI
ϕ θβ θα
′+=
(C)
cos
αβ θ
−
where
( ) 2 2 cos sin
f E mgc
II I
θθ
θ
= −− −
′′
(1)
22
22
1 ()
22 0
( ) (1 )
x
E mgcx
II Ix
β αβ
−
−+ − =
′′−
2
PROBLEM 18.139*
A solid cone of height 180 mm with a circular base of radius 60 mm is
supported by a ball and socket at A. The cone is released from the
18.138; you can either solve this equation numerically or reduce it to a
quadratic equation, since one of its roots is known.]
2
0.396692 m /s
=
PROBLEM 18.139* (Continued)
After dividing by m, Equation (2) of Problem 18.138 becomes
2
2
2
2
22
2
33
22
2
( ) 2 (1 ) 0
(0.342706) (0.396692 0.342706 )
113.0394 (2)(9.81)(0.135) (1 ) 0
1.08 10 19.98 10
(4.29198 2.6487 )(1 ) 5.87825(1.157529 ) 0
m
I
m
E mx
F x gcx x
m Im m
x
xx
xx x
β
αβ
−−
= −− −− − =
′
−
− − −− =
××
− −− −=
(a) Roots are:
cos 0.68170, 0.86603, 2.2919x
θ
= =
1
max
cos (0.68170) 47.023
θ
−
= = °
max
47.0
θ
= °
(b) By Equation (4) of Problem 18.137,
2 32
cos 0.396692 (0.342706)(0.68170) 15.2474
sin (19.98 10 )sin 47.023
317.32 rad/s
cos 317.32 (15.2474)(0.68170)
z
z
I
αβ θ
φθ
ω
ψωφθ
−
−−
= = =
′×°
=
=−=−
spin:
307 rad/s
ψ
=
precession:
15.25 rad/s
φ
=
PROBLEM 18.140*
A solid cone of height 180 mm with a circular base of radius 60 mm is
supported by a ball and socket at A. The cone is released from the position
0
30
θ
= °
with a rate of spin
0
300 rad/s,
ψ
=
a rate of precession
0
4 rad/s,
φ
= −
and a zero rate of nutation. Determine (a) the maximum
value of
θ
in the ensuing motion, (b) the corresponding values of the
rates of spin and precession, (c) the value of
θ
for which the sense of the
precession is reversed. (See hint of Problem 18.139.)
00 0
xy
300 ( 4)cos30 296.536 rad/s
z
ω
= + − °=
22 2
3 2 3 2 22
11
()
22
11
(19.98 10 )((2) 0) (1.08 10 )(296.536) 47.5241 m /s
22
xy z
TI I
mm m
ωω ω
−−
′
= ++
= × ++ × =
22
22
32
cos (9.81)(0.135)cos30 1.1469 m /s
22(47.5241 1.1469) 97.3419 m /s
(1.08 10 )(296.536) 0.320259 m /s
z
Vgc
m
E
m
I
mm
θ
βω
−
= = °=
= +=
==×=
2
00 0
32
2
sin cos
(19.98 10 )( 4)sin 30 0.320259cos30
0.257372 m /s
I
mm
αφ θβθ
−
′
= +
= × − °+ °
=
PROBLEM 18.140* (Continued)
After dividing by m, Equation (2) of Problem 18.138 becomes
2
2
2
2
22
2
33
22
2
( ) 2 (1 ) 0
(0.320259) (0.257372 0.320259 )
97.3419 (2)(9.81)(0.135) (1 ) 0
1.08 10 19.98 10
(2.37324 2.6487 )(1 ) 5.13342(0.80364 ) 0
m
I
m
E mx
F x gcx x
m Im m
x
xx
xx x
β
αβ
−−
= −− −− − =
′
−
− − −− =
××
− −− −=
(a) Roots are:
cos 0.23732, 0.86603, 1.73x
θ
= =
1
max cos (0.23732) 76.272
θ
−
= = °
max
76.3
θ
= °
(b) By Equation (4) of Problem 18.137,
2 32
cos 0.257372 (0.320259)(0.23732) 9.6192 rad/s
sin (19.98 10 )sin 76.272
296.536 rad/s
cos 296.536 (9.6192)(0.23732)
z
z
I
αβ θ
φθ
ω
ψωφθ
−
−−
= = =
′×°
=
=−= −
spin:
294 rad/s
ψ
=
precession:
9.62 rad/s
φ
=
(c)
2
cos 0
sinI
αβ θ
φθ
−
= =
′
0.257372
cos 0.320259
α
θβ
= =
36.5
θ
= °
PROBLEM 18.141*
A homogeneous sphere of mass m and radius a is welded to a rod AB of negligible
mass, which is held by a ball-and-socket support at A. The sphere is released in the
position
0
β
=
with a rate of precession
017 /11ga
φ
=
with no spin or nutation.
Determine the largest value of
β
in the ensuing motion.
0
PROBLEM 18.141* (Continued)
We now write
constant:
z
H=
2
2( sin ) constant
5
z
H ma
ψφ β
= −=
and, from the initial conditions, we find that the constant is zero. Thus,
sin 0
ψφ β
−=
(2)
Conservation of energy.
We have
222
1()
2
xx yy zz
TI I I
ωωω
= ++
11 sin
a