Question 10.8: A satellite is to be completely despun using a two-mass yo-y...
A satellite is to be completely despun using a two-mass yo-yo device with tangential release. Assume the spin axis of moment of inertia of the satellite is C = 200 kg · m² and the initial spin rate is \omega_{0} = 5 rad/s. The total yo-yo mass is 4 kg, and the radius of the spacecraft is 1 meter. Find
(a) the required cord length l;
(b) the time t to despin;
(c) the maximum tension in the yo-yo cables;
(d) the speed of the masses at release;
(e) the angle rotated by the satellite during the despin;
(f) the cord length required for radial release.
(a) From Equation 10.104,
K=1+\frac{C}{m R^{2}}=1+\frac{200}{4 \cdot 1^{2}}=51 (a)
From Equation 10.119 it follows that the cord length required for complete despin is
\phi=\sqrt{K} \Rightarrow l=R \sqrt{K} Complete despin (10.119)
l=R \sqrt{K}=1 \cdot \sqrt{51}=7.1414 m (b)
(b) The time for complete despin is obtained from Equations 10.116 and 10.119,
\phi=\omega_{0} t (10.116)
\omega_{0} t=\sqrt{K} \Rightarrow t=\frac{\sqrt{K}}{\omega_{0}}=\frac{\sqrt{51}}{5}=1.4283 s(c) A graph of Equation 10.128 is shown in Figure 10.19 , from which we see that
N=\frac{C}{R} \frac{2 K \omega_{0}^{3} t}{\left(K+\omega_{0}^{2} t^{2}\right)^{2}}=\frac{C \omega_{0}^{2}}{R} \frac{2 K \phi}{\left(K+\phi^{2}\right)^{2}} (10.128)
The maximum tension is N 455
which occurs at 0.825 s.
(d) From Equation 10.110, the speed of the yo-yo masses is
According to Equation 10.115, \dot{\phi}=\omega_{0} and at the time of release (ω = 0) Equation 10.117 states that \phi=\sqrt{K} . Thus,
\phi=\sqrt{K \frac{\omega_{0}-\omega}{\omega_{0}+\omega}} Partial despin (10.117)
v=R \sqrt{\omega^{2}+\left(\omega+\omega_{0}\right)^{2} \sqrt{K}^{2}}=1 \cdot \sqrt{0^{2}+(0+5)^{2} \sqrt{51}^{2}}=35.71 m / s(e) The angle through which the satellite rotates before coming to rotational rest is given by Equation 10.125,
\theta=\sqrt{K}\left\lgroup \frac{\pi}{2}-1 \right\rgroup =\sqrt{51}\left\lgroup \frac{\pi}{2}-1 \right\rgroup =4.076 \text { rad }(233.5 \text { degrees })(f) Allowing the cord to detach radially reduces the cord length required for complete despin from 7.141 m to (Equation 10.130)
l=R(\sqrt{K}-1)=1 \cdot(\sqrt{51}-1)=6.141 m