Torque, mass has to be found with centripetal force

[2&1/2MEN]

Homework Statement

A ball is tied to an elastic string of length 8.0m and swung in a horizontal circle with a velocity if 0.8m/s. When a metallic object is tied to a rope of length 2.75m and swung in a horizontal circle, it makes 1 revolution in 2.9s. The ratio of the centripetal force in the rope is 1/3. Find the mass of the metallic object attached to the rope, if centripetal force in the string is 0.20 N.

given: r1:8m
v1: .8m/s
r2: 2.75m
T: 2.9s
F: .2*(1/3)=1/15

Homework Equations

F=ma(c)=m*(v^2/r)
v=2∏r/T

The Attempt at a Solution

v=2∏(2.75)/2.9m=5.96
a=v^2/r=5.96^2/2.75=12.92m/s^2
F=ma(c)=(1/15)/12.92=.5g
but like i didnt use the values of radius 1 and velocity 1 so something must be wrong, and like .5g for a metallic object isn't that a little bit too lite? thanks for help anyways, i appreciate all topic related answers
P.S. that's due on this friday so please i need it quick please please please.

 

[Delphi51]

F=ma(c)=(1/15)/12.92=.5g

The formula is for force, so you won't get an answer in grams!
Why did you put 1/15 for the mass? It is the unknown, isn't it?
You know acceleration, but you don't know F, so you can't use this formula to find mass.

It might be possible to solve the problem if we knew what
"The ratio of the centripetal force in the rope is 1/3. "
means. Is that the exact wording? A "Ratio" implies a ratio of two things. Probably it means the ratio of two centripetal forces. Could it be the ratio of Fc for the string with the ball to the Fc for the rope with the metal object? Alas, we still can't solve it without knowing the mass of the ball or something about the elasticity of the string. Please check the wording of the question and make sure it is completely correct.