Two blocks, with masses Ma and Mb are connected to each other and to a central post by cords. they rotate about the post at frequency F (rps) on a frictionless horizontal surface at distances Ra and Rb from the post. Derive an algebraic expression for the tension in each segment of the cord (assumed massless)
1 Answer

The tensions must be centripetal so the blocks must be on the same radial line out from the centre.
Let block B be the most distant one.
Centripetal force on B is = Mb . w^2 . Rb = Mb . 4.π^2.F^2 .Rb
That must be the tension on the segment between Ma and Mb
Ma is subject to two forces – the inward tension of the string and the outward tension of the other bit of string.
The total of these two must be the centripetal force needed to make Ma go round the circle.
Tension (in) – Tension(out) = Centripetal force on Ma
So Tension (in) ( which we want to find) = Cent force on Ma + T(out)
Tension in the Ma to centre segment =( Ma . 4.π^2.F^2. Ra ) +
( Mb . 4.π^2.F^2. Rb )
= 4.π^2.F^2 (Ma.Ra + Mb.Rb)
Thats it, I reckon.