Tension: A light inextensible string AB of length 2L has a particle attached to its mid-point C...?

A light inextensible string AB of length 2L has a particle attached to its mid-point C. The ends A and B of the string are fastened to two fixed points with A at an distance L vertically above B. With both parts of the string taut, the particle describes a horizontal circle about the line AB with constant angular speed omega, w.





If the tension in CA is three times that in CB, show that w = 2√(g/l).





I thought I'd worked it out, but I keep getting back to w = 1.32√(g/l). From that, I think I've resolved the forces wrongly - but if C is a mid-point, that means the length of CA should be = length of CB, right? From there, doesn't it also mean that cos theta = (L/2)/L? I don't really understand; help'd be very much appreciated. Thanks so very much in advance!

Tension: A light inextensible string AB of length 2L has a particle attached to its mid-point C...?
Yes, I think that you are forgetting the vertical component due to the tension (T) in BC.





Resolving vertically: mg + T cos theta = 3T cos theta


mg = 2Tcos theta





As you say cos theta = (L/2)/L = 1/2





Thus mg = T


m = T/g ............Eq (1)





Resolving horizontally: mrw^2 = 3T sin theta + T sin theta


but r = L sin theta


Thus mw^2 = 4T/L





Substitute for m from (1) and rearrange





w^2 = 4g/L





Hope that helps.
Reply:You get the given result only iff the line (not a string) AB has length L.


Summing horizontal forces gives:


m*r*w^2 = (3F + F)*sin(fi)...........(1), and


summing vertical forces gives:


m*g = (3F - F)*cos(fi)..........(2)


where:


F is the lower string force,


fi is the angle between AB axis and the upper string,


r is the radius of the mass (distance from AB axis),


L is the length of AB axis.


Dividing (1) by (2) gives:


(r/g)*w^2 = 2*tan(fi) = 2 * r/(L/2) = 4*r/L giving:


w = 2*sqrt(g/L).