Page 34 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

Sec. 2.2 Equations of Motion: Natural Frequency 21

to
10
= 277^q"2 = 2.081 rad/s
The torsional stiffness of the rod is given by the equation K = GIp/l, where
Ip = Trdt^/?>2 = polar moment of inertia of the circular cross-sectional area of the
rod, / = length, and G = 80 X 10^ N/m^ = shear modulus of steel.
/p = ^ (0 .5 X lO -^ / = 0.006136 X 10"* m“

80 X lO“^ X 0.006136 X lO“* ^ , ,

K = ------------------2------------------ = 2.455 N • m /rad
By substituting into the natural frequency equation, the polar moment of inertia of
the wheel and tire is
K 2.455
J = = 0.567 kg • m^
(2.081)^
Example 2.2-4
Figure 2.2-4 shows a uniform bar pivoted about point O with springs of equal stiffness
k at each end. The bar is horizontal in the equilibrium position with spring forces
and jP2- Determine the equation of motion and its natural frequency.
Solution: Under rotation 6, the spring force on the left is decreased and that on the right
is increased. With Jo as the moment of inertia of the bar about G, the moment
equation about O is
= (Pi — kad)a + mgc —(P 2 + kb6)h = JqO

However,
Pjfl + mgc - P2b = 0
in the equilibrium position, and hence we need to consider only the moment of the
forces due to displacement 6, which is
= {-ka^ - kb^)d = Jq^
Thus, the equation of motion can be written as
k(a^ + b^)
d + -8 = 0
and, by inspection, the natural frequency of oscillation is
k(a^ + b^)

Figure 2.2-4.