PROBLEM 18.137*
The top shown is supported at the fixed Point O. Denoting by
,,
and
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
M
and
0,
z
M
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
(cos)I
where
and
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
of the
angular velocity of the top is constant and that the rate of precession
depends upon the value of the angle of nutation
.
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PROBLEM 18.137* (Continued)
()
O O O Oxyz O
MH H H
z-components:
0( cos)
d
dt
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
and
from the equation obtained and
from the two equations of Problem 18.139, show that the rate of nutation
is defined by the differential equation
2
(),f
where
2
2
1cos
() 2 2 cos sin
fEmgc
II I
(c) Further show, by introducing the auxiliary variable
cos ,
x
that the
maximum and minimum values of
can be obtained by solving for x
the cubic equation
222
1
22(1)()0Emgcxx x
II
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PROBLEM 18.138* (Continued)
Equation (1) of Problem 18.137 gives
2
sin cosI
(C)
cos
2
()f
2
2
1cos
gives
22
1()
x
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
position 030
with a rate of spin 0300 rad/s,
a rate of
precession 020 rad/s,
and a zero rate of nutation. Determine (a) the
maximum value of
in the ensuing motion, (b) the corresponding
values of the rates of spin and precession. [Hint: Use Eq. (2) of Prob.
18.138; you can either solve this equation numerically or reduce it to a
quadratic equation, since one of its roots is known.]
PROBLEM 18.139* (Continued)
After dividing by m, Equation (2) of Problem 18.138 becomes
2
precession: 15.25 rad/s
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
030
with a rate of spin 0300 rad/s,
a rate of precession
04 rad/s,
and a zero rate of nutation. Determine (a) the maximum
value of
in the ensuing motion, (b) the corresponding values of the
rates of spin and precession, (c) the value of
for which the sense of the
precession is reversed. (See hint of Problem 18.139.)
PROBLEM 18.140* (Continued)
After dividing by m, Equation (2) of Problem 18.138 becomes
2
max cos (0.23732) 76.272
(b) By Equation (4) of Problem 18.137,
232
cos 0.257372 (0.320259)(0.23732) 9.6192 rad/s
z
precession: 9.62 rad/s
cos 0
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
with a rate of precession 017 /11
g
a
with no spin or nutation.
Determine the largest value of
in the ensuing motion.
PROBLEM 18.141* (Continued)
We now write constant:
z
H
2
2(sin)constant
5
z
Hma
We have
222
1()
2
25 5 5
xx yy zz
TI I I
and selecting the datum at 0:
From the initial conditions 0, 0, 0,
5.ma
22 2 2 2
0
11 cos 11 ( sin ) 10 sin 11
g
a
(3)
Eq. (3): 22 2 2
0
11 cos 11 10 sin 11
g
a
(3 )
22 2 2 2
00
11( sec ) cos 11 10 sin 11
g
a
(5)
For the maximum value of ,
we have 0
and Eq. (5) yields