Circular Motion and maximum tension Problem

[desichick07]

Homework Statement

yes this question is a bit ridiculous but stick with it please...

After watching the movie "Corcodile Dundee" you and some friends decide to make a communications device invented by the Austrailian Aborigines. It consists of a noise-maker swung in a vertical circle on the end of a string. Your design calls for a 400 gram noisemaker on a 60 cm string. You are worried about whether the string you have will be strong enough, so you deicde to calculate the tension in the string when the device is swung with an aceleration which has a constant magnitude of 20 m/s2. You and your friends can't agree whether the maximum tension will occur when the noise maker is at the highest point in the circle, at the lowest point in the circle, or is always the same. To settle the argument you decide to calculate the tension at the highest point and at the lowest point and compare them.

Given:
.4 kg
.06 m string
a = 20 m/s^2

Homework Equations

Top: F(n) = mv^2 - mg
r

bottom: F(n) = mv^2 + mg
r

The Attempt at a Solution

T = F(n), where t = tension in the string and F(n) = normal force
F(n)

the centripetal force F(c) = ma(c)

TOP:
mv^2 - mg = ma(c)
r

the masses cancel and you're left with
v^2 - g = a(c) (*+*)
r

and solving for v, the velocity, you get:
v = [r*(a(c) + g)]^.5
v = [.6*(20+9.8)]^.5
= 4.23 m/s^2

T = F(n)
then use the above equation (*+*)with 4.23 for the velocity.
upon solving I got -1.1 N. I triple checked my calculations and they seem right. Can anyone spot my error?
Also is there an easier way to do this?

 

[rock.freak667]

The centripetal acceleration is 20m/s^2 so that the centripetal force $F_c=ma_c$
so just use the mass and multiply by the acceleration to get the force. the weight is always constant;$w=mg$. So then all you need to do is sub it into the two equations. But why did you find the tangential velocity for?EDIT: Even though the question asks for a comparison of values, all you can alternately do is compare the formulas.