Dsa -- Two masses orbiting their barycenter

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In the discussion, two masses, m1 and m2 (where m2 = a.m1 and m2 > m1), are analyzed for their orbital dynamics around a barycenter. The primary question focuses on determining the relative speed of the masses needed for m1’s orbit to exhibit cusps while the barycenter moves at speed v in an inertial frame. Participants emphasize the necessity for the original poster to provide calculations or equations to facilitate assistance. The thread was locked after the OP attempted to delete their post following the request for additional work. The conversation highlights the importance of showing work in physics discussions to receive meaningful help.
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Homework Statement
frs
Relevant Equations
R1 = (m2/(m1+m2))r
Two bodies with masses m2 and m1, m2=a.m1 and m2>m1 orbit each other in circular orbits. if the barycenter moves at a speed v with respect to an inertial reference frame, What should be the relative speed of the bodies so that m1’s orbit in this inertial frame will have cusps?
 
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You need to show some work before we can offer assistance. You could begin by adding some additional relevant equations.
 
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Mentor Note -- OP tried to delete their post after receiving the reply that they need to show their work first. The OP text has been restored and the thread is now locked.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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