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).
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