Question 2.5.2: Rotating unbalance is also important in rotorcraft such as h......
Rotating unbalance is also important in rotorcraft such as helicopters and prop planes. The tail rotor of a helicopter (the small rotor rotating in a vertical plane at the back of a helicopter used to provide yaw control and torque balance) as sketched in Figure 2.22 can be modeled as a rotating-unbalance problem discussed in this section with stiffness k = 1 × 10^5 N/m (provided by the tail section in the vertical direction) and mass of 20 kg. The tail section providing the vertical stiffness has a mass of 60 kg. Suppose that a 500-g mass is stuck on one of the blades at a distance of 15 cm from the axis of rotation. Calculate the magnitude of the deflection of the tail section of the helicopter as the tail rotor rotates at 1500 rpm. Assume a damping ratio of 0.01. At what rotor speed is the deflection at maximum? Calculate the maximum deflection.
The rotor system is modeled as a machine of mass 20.5 kg attached to a spring, as indicated in Figure 2.23. Here only the vibration of the tail section in the vertical direction is modeled and the helicopter body is modeled as ground. The spring used to represent the tail section has significant mass, so equation (1.76) of Section 1.5 for a heavy beam is used to find the equivalent mass of the system. Using the equivalent mass concept yields that the natural frequency is
\omega_n=\sqrt{\frac{k}{\frac{33}{140} M+m}} (1.76)
\omega_n=\sqrt{\frac{k}{m+\frac{33}{140}\ m_s}}=\sqrt{\frac{10^5\ N / m }{20.5+\frac{33}{140} 60\ kg }}=53.727\ rad / s
The frequency of rotation in rad/s is
\omega_r=1500\ rpm =1500 \frac{ rev }{\min } \frac{ min }{60\ s } \frac{2 \pi\ rad }{ rev }=157\ rad / s
Hence, the frequency ratio, r, becomes
r=\frac{\omega_r}{\omega_n}=\frac{157\ rad / s }{53.727\ rad / s }=2.92
With r=2.92 rad / s and \zeta=0.01, equation (2.84) yields that the magnitude of oscillation of the tail rotor is
X=\frac{m_0 e}{m} \frac{r^2}{\sqrt{\left(1-r^2\right)^2+(2 \zeta r)^2}} (2.84)
\begin{aligned} X & =\frac{m_0 e}{m} \frac{r^2}{\sqrt{\left(1-r^2\right)^2+(2 \zeta r)^2}} \\ & =\frac{(0.5\ kg )(0.15\ m )}{34.64\ kg } \frac{(2.92)^2}{\sqrt{\left[1-(2.92)^2\right]^2-[2(0.01)(2.92)]^2}}=0.002\ m \end{aligned}
Here the equivalent mass is m_{ eq }=m+m_s=34.64\ kg.
The maximum deflection occurs at about r=1 or
\omega_r=\omega_n=53.72\ rad / s =53.72\ \frac{ rad }{ s } \frac{ revs }{2 \pi\ rad } \frac{60\ s }{ min }=513.1\ rpm
In this case, the (maximum) deflection becomes
X=\frac{(0.5\ kg )(0.15)}{34.34\ kg } \frac{1}{2(0.01)}=0.108\ m =10.8\ cm
which represents a large unacceptable deflection of the rotor. Thus the tail rotor should not be allowed to rotate at 513.1 rpm.
