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 ω0 = 5 rad/s. The total yo-yo mass is 4 kg, and the radius of the spacecraft is 1 meter. Find (a)

Question

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 [latex] C = 200 kg·m^{2}[/latex] and the initial spin rate is [latex]\omega_{0}= 5[/latex] 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.

Accepted Answer

(a) From Equation 10.104,
[latex]K=1+\frac{C}{mR^{2}}[/latex] (10.104)
[latex]K=1+\frac{C}{mR^{2}}=1+\frac{200}{4.1^{2}}=51[/latex] (a)
From Equation 10.118 it follows that the cord length required for complete despin is
[latex]l=R\sqrt{K\frac{\omega_{0}-\omega}{\omega_{0}+\omega}}Partial despin.[/latex] (10.118)
[latex]l=R\sqrt{K}=1·\sqrt{5}=\underline{7.1414 m}[/latex] (b)
(b) The time for complete despin is obtained from Equations 10.116 and 10.118,
[latex]\phi=\omega_{0}t[/latex] (10.116)
[latex]\omega_{0}t=\sqrt{K}\Rightarrow t=\frac{\sqrt{K}}{\omega_{0}}=\frac{\sqrt{51}}{5}=\underline{ 1.4283 s}[/latex]
(c) A graph of Equation 10.128 is shown in Figure 10.19. The maximum tension is [latex]\underline{455 N}[/latex] , which occurs at 0.825 s.
[latex]N=\frac{C}{R}\frac{2K\omega _{0}^{3}t}{(K+\omega _{0}^{2}t^{2})}=\frac{C\omega _{0}^{2}}{R}\frac{2K\phi}{(K+\phi ^{2})^{2}}[/latex] (10.128)
(d) From Equation 10.110, the speed of the yo-yo masses is
[latex]v=\sqrt{V·V}=R\sqrt{\omega^{2} +(\omega +\dot{\phi })^{2}\phi ^{2}}[/latex] (10.110)
[latex]v=R\sqrt{\omega ^{2}+(\omega +\dot{\phi} )^{2}\phi ^{2}}[/latex]
According to Equation 10.115, [latex]\dot{\phi} = \omega_{0}[/latex] and at the time of release ([latex]\omega= 0[/latex])
Equation 10.118 states that [latex]\phi=\sqrt{K}[/latex]. Thus
[latex]\dot{\phi}=\omega_{0}[/latex] (10.115)
[latex]v=R\sqrt{\omega ^{2}+(\omega +\omega _{0} )^{2}\sqrt{\sqrt{K} } ^{2}}=1·\sqrt{0^{2}+(0+5)^{2}\sqrt{51}^{2}}=\underline{35.71 m/s}[/latex]
(e) The angle through which the satellite rotates before coming to rotational rest is given by Equation 10.124,
[latex] heta=2\sqrt{K} an^{-1} \frac{\omega_{0}t}{\sqrt{K} }-\omega =2\sqrt{K} an^{-1} \frac{\phi}{\sqrt{K}}-\phi[/latex] (10.124)
[latex] heta=\sqrt{K}(\frac{\pi }{2}-1)=\sqrt{51}(\frac{\pi }{2}-1 )=\underline{4.076 rad (233.5°)}[/latex]
(f) Allowing the cord to detach radially reduces the cord length required for complete despin from 7.141 m to
[latex]l=R(\sqrt{K}-1)=1·(\sqrt{51}-1)=\underline{6.141 m}[/latex]

Question 10.8: A satellite is to be completely despun using a two-mass yo-y...

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