and Problems Solutions Section 1.8 (1.90 through 1.93) 1.90 Consider the system of Figure 1.90 and (a) write the equations of motion in terms of the angle, , the bar makes with the vertical. Assume linear deflections of the springs and linearize the equations of motion. Then (b) discuss the stability of the linear systems solutions in terms of the physical constants, m, k, and ! . Assume the mass of the rod acts at the center as indicated in the figure.
Figure P1.90 Solution: Note that from the geometry, the springs deflect a distance kx = k(!sin!) and the cg moves a distance ! 2 cos! . Thus the total potential energy is 1 mg! U = 2 ! k(!sin")2 # cos" 2 2 and the total kinetic energy is 1! 1 m"2 ! 2 T = J O! 2 = ! 2 23 The Lagrange equation (1.64) becomes d # !T & !U d # m"2 ! & 1 % !" ( + !" = dt % 3 " ( + 2k"sin" cos" ) 2 mg"sin" = 0 dt $ ! ' $ ' Using the linear, small angle approximations sin! " ! and cos! " 1 yields m!2 "" # mg! & a) ! + % 2k!2 " !=0 3 2( $ ' Since the leading coefficient is positive the sign of the coefficient of determines the stability. mg mg if 2k! ! > 0 " 4k > " the system is stable 2 ! if 4k = mg " #(t) = at + b " the system is unstable b)
if 2k! ! mg mg < 0 " 4k < " the system is unstable 2 !
Note that physically this results states that the systems response is stable as long as the spring stiffness is large enough to over come the force of gravity. 1.91 Consider the inverted pendulum of Figure 1.37 as discussed in Example 1.8.1. Assume that a dashpot (of damping rate c) also acts on the pendulum parallel to the two springs. How does this affect the stability properties of the pendulum? Solution: The equation of motion is found from the following FBD:
m c l F dash mg 2Fsp k 0 k ! +
Moment about O: !Mo = I"
ml 2! = mgl sin! " 2 kl #l #l #l sin! $ cos! % " c$ ! % $ cos! % & & 2 2 2&2
2 2 + cl ! + # kl " mgl& ! = 0 % ml ! 4 $2 ' 2
When is small, sin and cos ml! 1
+
cl # kl ! + $ " mg& ! = 0 ' 4 2
For stability,
kl > mg and c > 0. 2
The result of adding a dashpot is to make the system asymptotically stable.
1.92
Replace the massless rod of the inverted pendulum of Figure 1.37 with a solid object compound pendulum of Figure 1.20(b). stable. Solution:
m2 m1 2Fsp k 0
! Moment about O: "M o = I!!
Calculate the equations of
vibration and discuss values of the parameter relations for which the system is
m2g + !
k
l kl #l #1 m1 g sin! + m2 gl sin! " 2 sin! $ cos! % = $ m1l 2 + m2 l 2 % ! & & 2 2 2 3
When is small, sin and cos 1.
2 ! m1 + m # l 2% + ! kl & m1 gl & m gl# % = 0 ' 2 2 "3 $ "2 2 $
! m1 + ( kl ! m + m2 # l% + * & " 1 + m2 # g-% = 0 "3 $ $, )2 2
For stability,
kl ! m1 >" + m2 # g. $ 2 2
1.93
A simple model of a control tab for an airplane is sketched in Figure P1.93. The equation of motion for the tab about the hinge point is written in terms of the angle from the centerline to be
!! ! J! + (c " f d )! + k! = 0 .
Here J is the moment of inertia of the tab, k is the rotational stiffness of the hinge,
! c is the rotational damping in the hinge and f d! is the negative damping provided
by the aerodynamic forces (indicated by arrows in the figure). stability of the solution in terms of the parameters c and fd .
Discuss the
Figure P1.93 A simple model of an airplane control tab
! Solution: The stability of the system is determined by the coefficient of ! since the inertia and stiffness terms are both positive. There are three cases Case 1 c - fd > 0 and the systems solution is of the form !(t) = e" at sin(# nt + $) and the solution is asymptotically stable. Case 2 c - fd < 0 and the systems solution is of the form !(t) = eat sin(" nt + #) and the solution is oscillates and grows without bound, and exhibits flutter instability as illustrated in Figure 1.36. Case 3 c = fd and the systems solution is of the form !(t) = Asin(" nt + #) and the solution is stable as illustrated in Figure 1.34.