How Does Reducing Distance Affect Tangential Speed in Rotating Space Modules?

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In the discussion, two space modules connected by a massless cable rotate around their center of mass, initially moving at a tangential speed of 7.98 m/s. As the modules pull together, reducing their distance by a factor of three, the final tangential speed needs to be determined. The key concept is that angular momentum is conserved during this process, not energy. The initial confusion about using linear kinematics was clarified, leading to the realization that the final speed can be calculated using the conservation of angular momentum. Ultimately, the correct approach was identified, resolving the initial uncertainty.
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In outer space two space modules are joined together by a massless cable. These probes are rotating about their center of mass, which is at the center of the wire, because the modules are identical (see the drawing). In each probe, the cable is connected to a motor, so that the modules can pull each other together. The initial tangential speed of each module is v0 = 7.98 m/s. Then they pull together until the distance between them is reduced by a factor of 3. Determine the final tangential speed, which is vf for each module.


sorry I know it seems like I hav soo many questions but this chapter ha been confusing for me. I amd assuming that I can use the formula from linear inematics which is velocity squared = inivital velocity +2 g 3 I get 8.17 m/s I don;t bleive this is correct though
 

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Hint: What quantity is conserved as the modules are pulled in?
 
Do you mean the energy? or the velocity it self?
 
Hint: Energy is not conserved! (A motor is required to pull the modules in.)
 
Hey thanks I finally found it !
 
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